暴毙场游记

1.前言：

rt，准备回归 WHK 了

2.题解：

T1：

结论题，先猜后证。

结论：走一个直角最优。

证明：

由于整个图是关于对角线对称的，所以说我们走到 ( n , m ) (n, m) (n,m) 和 ( m , n ) (m, n) (m,n) 的最小结果是一样的，所以不妨使 n > m n > m n>m。

有个容易发现的性质

  1. 不走回头路（只向下和向右走）

考虑修改单步。

在这里插入图片描述
如果修改的步数都在对角线内，那么一定变大。

如果修改的步数都在对角线外，这种情况又不如图三。

这样递归到边线，得到某一条边线最短。容易知道靠下的边线最短，如下图。

在这里插入图片描述
向下走的代价是 n + 1 n + 1 n+1
向右走的代价是 ∑ i = 0 m ( n + i n ) = ( n + m + 1 n + 1 ) \sum_{i = 0}^{m} \binom {n + i}{n} = \binom{n + m + 1}{n + 1} ∑i=0m​(nn+i​)=(n+1n+m+1​)
交叉点被算了两次，所以总代价是 ( n + m + 1 n + 1 ) + n \binom{n + m + 1}{n + 1} + n (n+1n+m+1​)+n

n , m n, m n,m 很大，考虑怎么求解组合数。
( n + m + 1 n + 1 ) = ( n + m + 1 m ) \binom{n + m + 1}{n + 1} = \binom{n + m + 1}{m} (n+1n+m+1​)=(mn+m+1​)
所以我们找到 n + 1 , m n + 1, m n+1,m 中的最小值，然后用 ( a b ) = a ! b ! ( a − b ) ! = Π i = a − b + 1 a i b ! \binom{a}{b} = \frac{a!}{b!(a - b)!} = \frac{\Pi_{i = a - b + 1}^{a} i}{ b!} (ba​)=b!(a−b)!a!​=b!Πi=a−b+1a​i​ 即可。

时间复杂度分析， n × m = 1 e 12 n \times m = 1e12 n×m=1e12 则 min ⁡ ( n , m ) m a x = s q r t ( 1 e 12 ) = 1 e 6 \min (n, m)_{max} = sqrt (1e12) = 1e6 min(n,m)max​=sqrt(1e12)=1e6，所以时间复杂度为 O ( 1 e 6 ) O(1e6) O(1e6)

挂分原因： i i i 很大，没有先取模，于是 i ∗ r e s i * res i∗res 就挂了……

T2:

题意转换：相当于用所有非零的数和任意个零组成的小于 n n n 的个数（位数不够用前导零）。（不过我考场时讨论了数的长度，也是一样的道理）

类似于数位dp，枚举到了第 i i i 位，前 i − 1 i - 1 i−1 位都相同，枚举这一位选的数 ( a [ i ] a[i] a[i])，如果这一位小于 n [ i ] n[i] n[i] ，那么后面怎么选都可以了，如果相同，则 a [ ( i + 1 ) a[(i + 1) a[(i+1) ~ n ] n] n] 一定要小于 n [ ( i + 1 ) n[(i + 1) n[(i+1)~ n ] n] n]，这就交给后面的 i i i 了。

挂分原因：

  1. 求质因数的幂次时忘记乘上 o p op op, 就没有区别分子和分母对幂次贡献的正负了……
  2. p i = p i − 1 ∗ p = t m p ∗ p p ^ i = p ^ {i - 1} * p = tmp * p pi=pi−1∗p=tmp∗p，不是 t m p ∗ t m p tmp * tmp tmp∗tmp 啊！！！！！

这个东西却让我把样例和我手玩数据都对了。（样例无语，但是手玩数据时我为了偷懒，只搞了多个零然后其他数字都不一样的数据，所以没测出来。）

T3：

板题，枚举有几个字符的大小超过了 n − 1 n - 1 n−1， 容斥即可。

假设现在有 i i i 个数超过了 n − 1 n - 1 n−1，那么方案数就为

  1. 选出 i i i 个数，让他们一定超过 n − 1 n - 1 n−1，方案数为 ( m i ) \binom{m}{i} (im​)
  2. 那么此时让这几个数都减去 n n n，那么这时所有数加起来为 k − n ∗ i k - n * i k−n∗i，所以方案数为 F k − n ∗ i m = ( k − n ∗ i + m − 1 m − 1 ) F_{k - n * i}^{m} = \binom{k - n * i + m - 1}{m - 1} Fk−n∗im​=(m−1k−n∗i+m−1​)。
  3. 系数为 ( − 1 ) i (-1)^i (−1)i

乘法原理，这个 i i i 的贡献为： ( − 1 ) i ( k − n ∗ i + m − 1 m − 1 ) ( m i ) (-1)^i\binom{k - n * i + m - 1}{m - 1} \binom{m}{i} (−1)i(m−1k−n∗i+m−1​)(im​)

a n s = ∑ i = 0 m ( − 1 ) i ( k − n ∗ i + m − 1 m − 1 ) ( m i ) ans = \sum_{i = 0}^{m} (-1)^i\binom{k - n * i + m - 1}{m - 1} \binom{m}{i} ans=∑i=0m​(−1)i(m−1k−n∗i+m−1​)(im​)

挂分原因：如果 k − n ∗ i + m − 1 < m − 1 k - n * i + m - 1 < m - 1 k−n∗i+m−1<m−1 就会出现负数的阶乘，没有特判啊。

本机数组下标小于零的话，他聪明的给我搞了个零而不是卡住，笑死。

T4:

板题。

容斥，考虑这个人有多少人和他不是敌对关系。

加上和 i i i 第{1},{2},{3},{4},{5}爱好相同的人
减去和 i i i 第{1,2},{1,3},{1,4},{1,5},{2,3}…爱好相同的人。
…

把这些爱好的值用桶记录即可。

这里也可以用 T r i e + H a s h Trie + Hash Trie+Hash 的做法。(卡常数）

H a s h Hash Hash 作用有两个：

  1. 将一个人的爱好子集搞成一个 ULL
  2. H a s h Hash Hash 找到 p p p 节点的 c h ch ch 这个儿子的下标

T r i e Trie Trie 就是模板了

挂分原因：卡常数……

3.考场源码及挂分点标注。

T1:

#include <cmath>
#include <queue>
#include <vector>
#include <cstdio>
#include <cstdlib> 
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define fi first
#define se second
#define db double
#define LL long long
#define ULL unsigned long long
#define PII pair <int, int>
#define MP(x,y) make_pair (x, y)
#define rep(i,j,k) for (int i = (j); i <= (k); ++i)
#define per(i,j,k) for (int i = (j); i >= (k); --i)

template <typename T> T Max (T x, T y) { return x > y ? x : y; }
template <typename T> T Min (T x, T y) { return x < y ? x : y; }
template <typename T> T Abs (T x) { return x > 0 ? x : -x; }
template <typename T>
void read (T &x) {
	x = 0; T f = 1;
	char ch = getchar ();
	while (ch < '0' || ch > '9') {
		if (ch == '-') f = -1;
		ch = getchar ();
	}
	while (ch >= '0' && ch <= '9') {
		x = (x << 3) + (x << 1) + ch - '0';
		ch = getchar ();
	}
	x *= f;
}
char For_Print[25];
template <typename T>
void write (T x) {
	if (x == 0) { putchar ('0'); return; }
	if (x < 0) { putchar ('-'); x = -x; }
	int poi = 0;
	while (x) {
		For_Print[++poi] = x % 10 + '0';
		x /= 10;
	}
	while (poi) putchar (For_Print[poi--]);
}
template <typename T>
void print (T x, char ch) {
	write (x); putchar (ch);
}

const LL Mod = 1e9 + 7;
const LL Limit = 2 * 1e6;

LL n, m;
LL inv[Limit + 5];

void del (LL &x, LL y) { ((x -= y) < 0) && (x += Mod); }
void add (LL &x, LL y) { ((x += y) >= Mod) && (x -= Mod); }

LL C (LL x, LL y) {
    if (y > Limit) y = x - y;
    LL res = 1;
    for (LL i = x; i > x - y; i--) res = (res * i /* 就是这里！！！！！！！*/) % Mod;
    for (LL i = y; i > 0; i--) res = (res * inv[i]) % Mod;
    return res;
}

int main () {
	freopen ("jump.in", "r", stdin);
	freopen ("jump.out", "w", stdout);

    inv[1] = 1; rep (i, 2, Limit) inv[i] = (Mod - Mod / i) * inv[Mod % i] % Mod;

    read (n); read (m);
    if (n < m) swap (n, m);

    write ((n + C (m + n + 1, n + 1)) % Mod);
	return 0;
}

T2

#include <cmath>
#include <queue>
#include <vector>
#include <cstdio>
#include <cstdlib> 
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define fi first
#define se second
#define db double
#define LL long long
#define ULL unsigned long long
#define PII pair <int, int>
#define MP(x,y) make_pair (x, y)
#define rep(i,j,k) for (int i = (j); i <= (k); ++i)
#define per(i,j,k) for (int i = (j); i >= (k); --i)

template <typename T> T Max (T x, T y) { return x > y ? x : y; }
template <typename T> T Min (T x, T y) { return x < y ? x : y; }
template <typename T> T Abs (T x) { return x > 0 ? x : -x; }
template <typename T>
void read (T &x) {
	x = 0; T f = 1;
	char ch = getchar ();
	while (ch < '0' || ch > '9') {
		if (ch == '-') f = -1;
		ch = getchar ();
	}
	while (ch >= '0' && ch <= '9') {
		x = (x << 3) + (x << 1) + ch - '0';
		ch = getchar ();
	}
	x *= f;
}
char For_Print[25];
template <typename T>
void write (T x) {
	if (x == 0) { putchar ('0'); return; }
	if (x < 0) { putchar ('-'); x = -x; }
	int poi = 0;
	while (x) {
		For_Print[++poi] = x % 10 + '0';
		x /= 10;
	}
	while (poi) putchar (For_Print[poi--]);
}
template <typename T>
void print (T x, char ch) {
	write (x); putchar (ch);
}

const int Maxn = 200;

int lenn, lenm;
char n[Maxn + 5], m[Maxn + 5];
int bak[Maxn + 5];

int cnt, primes[Maxn + 5];
bool vis[Maxn + 5];
void Euler () {
	rep (i, 2, Maxn) {
		if (vis[i] == 0) {
			vis[i] = 1;
			primes[++cnt] = i;
		}
		rep (j, 1, cnt) {
			if (primes[j] > Maxn / i) break;
			vis[i * primes[j]] = 1;
			if (i % primes[j] == 0) break;
		}
	}
}
int tot[Maxn + 5];
ULL quick_pow (ULL x, int y) {
    ULL res = 1;
    while (y) {
        if (y & 1) res = res * x;
        x = x * x; y >>= 1;
    }
    return res;
}
void Calc (int x, int op) {
	rep (i, 1, cnt) {
		int tmp = primes[i];
		while (tmp <= x) {
			tot[i] += x / tmp /* 我的 "*op" 呢？？？？ */;
			tmp *= tmp; /* tmp应该 *= primes[i]啊！！！ */
		}
	}
}

ULL Get_Res () {
	ULL res = 1;
	memset (tot, 0, sizeof tot);
	int sum = 0;
    rep (j, 0, 9) { Calc (bak[j], -1); sum += bak[j]; }
    Calc (sum, 1);
    rep (j, 1, cnt) res *= quick_pow (primes[j], tot[j]);
    return res;
}

ULL fac[Maxn + 5];

int main () {
	// freopen ("C:\\Users\\Administrator\\Desktop\\lihan\\1.in", "r", stdin);
	// freopen ("C:\\Users\\Administrator\\Desktop\\lihan\\1.out", "w", stdout);
	freopen ("count.in", "r", stdin);
	freopen ("count.out", "w", stdout);
	fac[1] = 1; rep (i, 2, Maxn) fac[i] = fac[i - 1] * i;
	
	Euler ();
	
    scanf ("%s", n + 1);
    lenn = strlen (n + 1);
    rep (i, 1, lenn)
        if (n[i] != '0')
            m[++lenm] = n[i];
    
    ULL ans = 0;

    //数位小于 n
    rep (i, 1, lenm) bak[m[i] - '0']++;
    /*
    rep (i, 0, lenn - lenm - 1) {
        ULL res1 = 0, res2 = 0;
        
        bak[0] = i;
        res1 = Get_Res ();
        
        if (i != 0) {
        	bak[0] = i - 1;
    		res2 = Get_Res ();
		}
        ans += res1 - res2;
    }
    */

//	/*
	if (lenn - lenm - 1 >= 0) {
	    bak[0] = lenn - lenm - 1;
		ans += Get_Res ();
	}
//	*/
    
	// 数位等于 n
    bak[0] = lenn - lenm;
    rep (i, 1, lenn) {
    	rep (j, 0, n[i] - '0' - 1) {
    		if (!bak[j]) continue;
    		if (i == 1 && j == 0) continue;
    		bak[j]--;
    		ans += Get_Res ();
    		bak[j]++;
		}
    	
    	bak[n[i] - '0']--;
	}
    
    write (ans);
	return 0;
}

T3:

#include <cmath>
#include <queue>
#include <vector>
#include <cstdio>
#include <cstdlib> 
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define fi first
#define se second
#define db double
#define LL long long
#define ULL unsigned long long
#define PII pair <int, int>
#define MP(x,y) make_pair (x, y)
#define rep(i,j,k) for (int i = (j); i <= (k); ++i)
#define per(i,j,k) for (int i = (j); i >= (k); --i)

template <typename T> T Max (T x, T y) { return x > y ? x : y; }
template <typename T> T Min (T x, T y) { return x < y ? x : y; }
template <typename T> T Abs (T x) { return x > 0 ? x : -x; }
template <typename T>
void read (T &x) {
	x = 0; T f = 1;
	char ch = getchar ();
	while (ch < '0' || ch > '9') {
		if (ch == '-') f = -1;
		ch = getchar ();
	}
	while (ch >= '0' && ch <= '9') {
		x = (x << 3) + (x << 1) + ch - '0';
		ch = getchar ();
	}
	x *= f;
}
char For_Print[25];
template <typename T>
void write (T x) {
	if (x == 0) { putchar ('0'); return; }
	if (x < 0) { putchar ('-'); x = -x; }
	int poi = 0;
	while (x) {
		For_Print[++poi] = x % 10 + '0';
		x /= 10;
	}
	while (poi) putchar (For_Print[poi--]);
}
template <typename T>
void print (T x, char ch) {
	write (x); putchar (ch);
}

const int Maxn = 2 * 1e5;
const LL Mod = 998244353;

void del (LL &x, LL y) { ((x -= y) < 0) && (x += Mod); }
void add (LL &x, LL y) { ((x += y) >= Mod) && (x -= Mod); }

int t, n, m, k;

LL fac[Maxn + 5], inv_fac[Maxn + 5];
LL quick_pow (LL x, LL y) {
    LL res = 1;
    while (y) {
        if (y & 1) res = (res * x) % Mod;
        x = (x * x) % Mod; y >>= 1;
    }
    return res;
}
LL inv (LL x) {
    return quick_pow (x, Mod - 2);
}
LL C (LL x, LL y) {
    if (x < 0 || y < 0) return 0; //x - y < 0 怎么办？？？？
    if (y == 0 || y == x) return 1;
    return fac[x] * inv_fac[y] % Mod * inv_fac[x - y] % Mod;
}

int main () {
	freopen ("encoding.in", "r", stdin);
	freopen ("encoding.out", "w", stdout);

    fac[1] = 1; rep (i, 2, Maxn) fac[i] = fac[i - 1] * i % Mod;
    inv_fac[Maxn] = inv (fac[Maxn]); per (i, Maxn - 1, 1) inv_fac[i] = inv_fac[i + 1] * (i + 1) % Mod;

    read (t);
    while (t--) {
        read (n); read (m); read (k);

        LL res = 0;
        rep (i, 0, m) {
            if (i & 1) del (res, C (k - i * n + m - 1, m - 1) * C (m, i) % Mod);
            else add (res, C (k - i * n + m - 1, m - 1) * C (m, i) % Mod);
        }
        print (res, '\n');
    }
	return 0;
}

#include <map>
#include <cmath>
#include <queue>
#include <vector>
#include <cstdio>
#include <cstdlib> 
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define fi first
#define se second
#define db double
#define LL long long
#define ULL unsigned long long
#define PII pair <int, int>
#define MP(x,y) make_pair (x, y)
#define rep(i,j,k) for (int i = (j); i <= (k); ++i)
#define per(i,j,k) for (int i = (j); i >= (k); --i)

template <typename T> T Max (T x, T y) { return x > y ? x : y; }
template <typename T> T Min (T x, T y) { return x < y ? x : y; }
template <typename T> T Abs (T x) { return x > 0 ? x : -x; }
template <typename T>
void read (T &x) {
	x = 0; T f = 1;
	char ch = getchar ();
	while (ch < '0' || ch > '9') {
		if (ch == '-') f = -1;
		ch = getchar ();
	}
	while (ch >= '0' && ch <= '9') {
		x = (x << 3) + (x << 1) + ch - '0';
		ch = getchar ();
	}
	x *= f;
}
char For_Print[25];
template <typename T>
void write (T x) {
	if (x == 0) { putchar ('0'); return; }
	if (x < 0) { putchar ('-'); x = -x; }
	int poi = 0;
	while (x) {
		For_Print[++poi] = x % 10 + '0';
		x /= 10;
	}
	while (poi) putchar (For_Print[poi--]);
}
template <typename T>
void print (T x, char ch) {
	write (x); putchar (ch);
}

const int Maxn = 5 * 1e4;
const int Maxm = 5;
const ULL P = 1333331;

int n, m = 5;
int a[Maxn + 5][Maxm + 5];
map <ULL, int> bak;

int main () {
	// freopen ("C:\\Users\\Administrator\\Desktop\\lihan\\1.in", "r", stdin);
	// freopen ("C:\\Users\\Administrator\\Desktop\\lihan\\1.out", "w", stdout);
	freopen ("against.in", "r", stdin);
	freopen ("against.out", "w", stdout);

    read (n);
    LL res = 0;
    rep (i, 1, n) {
        rep (j, 1, m) {
            read (a[i][j]);
            // if (a[i][j] == 0) printf ("FUCK");
        }
        sort (a[i] + 1, a[i] + 1 + m);
        rep (j, 1, (1 << m) - 1) {
            int sum = 0; ULL tmp = 0;
            rep (k, 1, m) {
                if ((j >> (k - 1)) & 1) {
                    tmp = tmp * P + a[i][k];
                    sum++;
                }
            }
            if (bak.find (tmp) == bak.end ()) bak[tmp] = 0;
            bak[tmp]++;

            if (sum & 1) {
                res = res + (bak[tmp] - 1);
            }
            else {
                res = res - (bak[tmp] - 1);
            }
            // printf ("i = %d, j = %d, res = %lld\n", i, j, res);
        }
    }
    write ((LL)n * (n - 1) / 2 - res);
	return 0;
}

4.考场改后代码

T1:

#include <cmath>
#include <queue>
#include <vector>
#include <cstdio>
#include <cstdlib> 
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define fi first
#define se second
#define db double
#define LL long long
#define ULL unsigned long long
#define PII pair <int, int>
#define MP(x,y) make_pair (x, y)
#define rep(i,j,k) for (int i = (j); i <= (k); ++i)
#define per(i,j,k) for (int i = (j); i >= (k); --i)

template <typename T> T Max (T x, T y) { return x > y ? x : y; }
template <typename T> T Min (T x, T y) { return x < y ? x : y; }
template <typename T> T Abs (T x) { return x > 0 ? x : -x; }
template <typename T>
void read (T &x) {
	x = 0; T f = 1;
	char ch = getchar ();
	while (ch < '0' || ch > '9') {
		if (ch == '-') f = -1;
		ch = getchar ();
	}
	while (ch >= '0' && ch <= '9') {
		x = (x << 3) + (x << 1) + ch - '0';
		ch = getchar ();
	}
	x *= f;
}
char For_Print[25];
template <typename T>
void write (T x) {
	if (x == 0) { putchar ('0'); return; }
	if (x < 0) { putchar ('-'); x = -x; }
	int poi = 0;
	while (x) {
		For_Print[++poi] = x % 10 + '0';
		x /= 10;
	}
	while (poi) putchar (For_Print[poi--]);
}
template <typename T>
void print (T x, char ch) {
	write (x); putchar (ch);
}

const LL Mod = 1e9 + 7;
const LL Limit = 2 * 1e6;

LL n, m;
LL inv[Limit + 5];

void del (LL &x, LL y) { ((x -= y) < 0) && (x += Mod); }
void add (LL &x, LL y) { ((x += y) >= Mod) && (x -= Mod); }

LL C (LL x, LL y) {
    if (y > Limit) y = x - y;
    LL res = 1;
    for (LL i = x; i > x - y; i--) res = (res * (i % Mod)) % Mod;
    for (LL i = y; i > 0; i--) res = (res * inv[i]) % Mod;
    return res;
}

int main () {
	freopen ("jump.in", "r", stdin);
	freopen ("jump.out", "w", stdout);

    inv[1] = 1; rep (i, 2, Limit) inv[i] = (Mod - Mod / i) * inv[Mod % i] % Mod;

    read (n); read (m);
    if (n < m) swap (n, m);

    write ((n + C (m + n + 1, n + 1)) % Mod);
	return 0;
}

T2:

#include <cmath>
#include <queue>
#include <vector>
#include <cstdio>
#include <cstdlib> 
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define fi first
#define se second
#define db double
#define LL long long
#define ULL unsigned long long
#define PII pair <int, int>
#define MP(x,y) make_pair (x, y)
#define rep(i,j,k) for (int i = (j); i <= (k); ++i)
#define per(i,j,k) for (int i = (j); i >= (k); --i)

template <typename T> T Max (T x, T y) { return x > y ? x : y; }
template <typename T> T Min (T x, T y) { return x < y ? x : y; }
template <typename T> T Abs (T x) { return x > 0 ? x : -x; }
template <typename T>
void read (T &x) {
	x = 0; T f = 1;
	char ch = getchar ();
	while (ch < '0' || ch > '9') {
		if (ch == '-') f = -1;
		ch = getchar ();
	}
	while (ch >= '0' && ch <= '9') {
		x = (x << 3) + (x << 1) + ch - '0';
		ch = getchar ();
	}
	x *= f;
}
char For_Print[25];
template <typename T>
void write (T x) {
	if (x == 0) { putchar ('0'); return; }
	if (x < 0) { putchar ('-'); x = -x; }
	int poi = 0;
	while (x) {
		For_Print[++poi] = x % 10 + '0';
		x /= 10;
	}
	while (poi) putchar (For_Print[poi--]);
}
template <typename T>
void print (T x, char ch) {
	write (x); putchar (ch);
}

const int Maxn = 200;

int lenn, lenm;
char n[Maxn + 5], m[Maxn + 5];
int bak[Maxn + 5];

int cnt, primes[Maxn + 5];
bool vis[Maxn + 5];
void Euler () {
	rep (i, 2, Maxn) {
		if (vis[i] == 0) {
			vis[i] = 1;
			primes[++cnt] = i;
		}
		rep (j, 1, cnt) {
			if (primes[j] > Maxn / i) break;
			vis[i * primes[j]] = 1;
			if (i % primes[j] == 0) break;
		}
	}
}
int tot[Maxn + 5];
ULL quick_pow (ULL x, int y) {
    ULL res = 1;
    while (y) {
        if (y & 1) res = res * x;
        x = x * x; y >>= 1;
    }
    return res;
}
void Calc (int x, int op) {
	rep (i, 1, cnt) {
		int tmp = primes[i];
		while (tmp <= x) {
			tot[i] += (x / tmp) * op;
			tmp *= primes[i];
		}
	}
}

ULL Get_Res () {
	ULL res = 1;
	memset (tot, 0, sizeof tot);
	int sum = 0;
    rep (j, 0, 9) { Calc (bak[j], -1); sum += bak[j]; }
    Calc (sum, 1);
    rep (j, 1, cnt) res *= quick_pow (primes[j], tot[j]);
    return res;
}

ULL fac[Maxn + 5];

int main () {
	// freopen ("C:\\Users\\Administrator\\Desktop\\lihan\\1.in", "r", stdin);
	// freopen ("C:\\Users\\Administrator\\Desktop\\lihan\\1.out", "w", stdout);
	freopen ("count.in", "r", stdin);
	freopen ("count.out", "w", stdout);
	fac[1] = 1; rep (i, 2, Maxn) fac[i] = fac[i - 1] * i;
	
	Euler ();
	
    scanf ("%s", n + 1);
    lenn = strlen (n + 1);
    rep (i, 1, lenn)
        if (n[i] != '0')
            m[++lenm] = n[i];
    
    ULL ans = 0;

    //数位小于 n
    rep (i, 1, lenm) bak[m[i] - '0']++;
    /*
    rep (i, 0, lenn - lenm - 1) {
        ULL res1 = 0, res2 = 0;
        
        bak[0] = i;
        res1 = Get_Res ();
        
        if (i != 0) {
        	bak[0] = i - 1;
    		res2 = Get_Res ();
		}
        ans += res1 - res2;
    }
    */

//	/*
	if (lenn - lenm - 1 >= 0) {
	    bak[0] = lenn - lenm - 1;
		ans += Get_Res ();
	}
	else {
		bak[0] = 0;
	}
//	*/
    
	// 数位等于 n
    bak[0] = lenn - lenm;
    rep (i, 1, lenn) {
    	rep (j, 0, n[i] - '0' - 1) {
    		if (!bak[j]) continue;
    		if (i == 1 && j == 0) continue;
    		bak[j]--;
    		ans += Get_Res ();
    		bak[j]++;
		}
    	
    	bak[n[i] - '0']--;
	}
    
    write (ans);
	return 0;
}


#include <cmath>
#include <queue>
#include <vector>
#include <cstdio>
#include <cstdlib> 
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define fi first
#define se second
#define db double
#define LL long long
#define ULL unsigned long long
#define PII pair <int, int>
#define MP(x,y) make_pair (x, y)
#define rep(i,j,k) for (int i = (j); i <= (k); ++i)
#define per(i,j,k) for (int i = (j); i >= (k); --i)

template <typename T> T Max (T x, T y) { return x > y ? x : y; }
template <typename T> T Min (T x, T y) { return x < y ? x : y; }
template <typename T> T Abs (T x) { return x > 0 ? x : -x; }
template <typename T>
void read (T &x) {
	x = 0; T f = 1;
	char ch = getchar ();
	while (ch < '0' || ch > '9') {
		if (ch == '-') f = -1;
		ch = getchar ();
	}
	while (ch >= '0' && ch <= '9') {
		x = (x << 3) + (x << 1) + ch - '0';
		ch = getchar ();
	}
	x *= f;
}
char For_Print[25];
template <typename T>
void write (T x) {
	if (x == 0) { putchar ('0'); return; }
	if (x < 0) { putchar ('-'); x = -x; }
	int poi = 0;
	while (x) {
		For_Print[++poi] = x % 10 + '0';
		x /= 10;
	}
	while (poi) putchar (For_Print[poi--]);
}
template <typename T>
void print (T x, char ch) {
	write (x); putchar (ch);
}

const int Maxn = 3 * 1e5;
const LL Mod = 998244353;

void del (LL &x, LL y) { ((x -= y) < 0) && (x += Mod); }
void add (LL &x, LL y) { ((x += y) >= Mod) && (x -= Mod); }

int t; LL n, m, k;

LL fac[Maxn + 5], inv_fac[Maxn + 5];
LL quick_pow (LL x, LL y) {
    LL res = 1;
    while (y) {
        if (y & 1) res = (res * x) % Mod;
        x = (x * x) % Mod; y >>= 1;
    }
    return res;
}
LL inv (LL x) {
    return quick_pow (x, Mod - 2);
}
LL C (LL x, LL y) {
    if (x < 0 || y < 0) return 0;
    if (y == 0 || y == x) return 1;
    if (x - y < 0) return 0;
    return fac[x] * inv_fac[y] % Mod * inv_fac[x - y] % Mod;
}

int main () {
	freopen ("encoding.in", "r", stdin);
	freopen ("encoding.out", "w", stdout);

    fac[1] = 1; rep (i, 2, Maxn) fac[i] = fac[i - 1] * i % Mod;
    inv_fac[Maxn] = inv (fac[Maxn]); per (i, Maxn - 1, 1) inv_fac[i] = inv_fac[i + 1] * (i + 1) % Mod;

    read (t);
    while (t--) {
        read (n); read (m); read (k);

        LL res = 0;
        rep (i, 0, m) {
            if (i & 1) del (res, C (k - i * n + m - 1, m - 1) * C (m, i) % Mod);
            else add (res, C (k - i * n + m - 1, m - 1) * C (m, i) % Mod);
        }
        print (res, '\n');
    }
	return 0;
}

T4:

#include <map>
#include <cmath>
#include <queue>
#include <vector>
#include <cstdio>
#include <cstdlib> 
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define fi first
#define se second
#define db double
#define LL long long
#define ULL unsigned int
#define PII pair <int, int>
#define MP(x,y) make_pair (x, y)
#define rep(i,j,k) for (int i = (j); i <= (k); ++i)
#define per(i,j,k) for (int i = (j); i >= (k); --i)

template <typename T> T Max (T x, T y) { return x > y ? x : y; }
template <typename T> T Min (T x, T y) { return x < y ? x : y; }
template <typename T> T Abs (T x) { return x > 0 ? x : -x; }
template <typename T>
void read (T &x) {
	x = 0; T f = 1;
	char ch = getchar ();
	while (ch < '0' || ch > '9') {
		if (ch == '-') f = -1;
		ch = getchar ();
	}
	while (ch >= '0' && ch <= '9') {
		x = (x << 3) + (x << 1) + ch - '0';
		ch = getchar ();
	}
	x *= f;
}
char For_Print[25];
template <typename T>
void write (T x) {
	if (x == 0) { putchar ('0'); return; }
	if (x < 0) { putchar ('-'); x = -x; }
	int poi = 0;
	while (x) {
		For_Print[++poi] = x % 10 + '0';
		x /= 10;
	}
	while (poi) putchar (For_Print[poi--]);
}
template <typename T>
void print (T x, char ch) {
	write (x); putchar (ch);
}

const int Maxn = 5 * 1e4;
const int Maxm = 5;
const int Maxt = 1e7;
const int MaxTire = 114514;
const ULL P = 133331;

int n, m = 5;
int a[Maxn + 5][Maxm + 5];

int cnt, len, x[Maxm + 5];
int End[Maxt + 5];
struct Node {
    int p, ch, to;
    Node () {}
    Node (int _p, int _ch, int _to) {
        p = _p; ch = _ch; to = _to;
    }
};
bool operator == (Node x, Node y) {
    return (x.p == y.p) && (x.ch == y.ch);
}
vector <Node> Tire[MaxTire + 5];
int Find (int p, int ch) {
    int Hash = (p * P + ch) % MaxTire;

//	if (Tire[Hash].size () > 10000)
//    	printf ("A Tire[Hash].size () = %d\n", (int)Tire[Hash].size ());

    for (int i = 0; i < (int)Tire[Hash].size (); i++) {
        if (Tire[Hash][i] == Node (p, ch, -1))
            return Tire[Hash][i].to;
    }
    return -1;
}
int Insert (int p, int ch) {
    int Hash = (p * P + ch) % MaxTire;
    
//	if (Tire[Hash].size () > 10000)
//    	printf ("B Tire[Hash].size () = %d\n", (int)Tire[Hash].size ());
    
    Tire[Hash].push_back (Node (p, ch, ++cnt));
    
//	if (Tire[Hash].size () > 10000)
//    	printf ("C Tire[Hash].size () = %d\n", (int)Tire[Hash].size ());
    	
    return cnt;
}
void Update () {
	int p = 0;
	rep (i, 1, len) {
		int ch = x[i];
        int to = Find (p, ch);
		if (to == -1) to = Insert (p, ch);
        p = to;
	}
	
//	printf ("cnt = %d\n", cnt);
	
	End[p]++;
}
int Query () {
	int p = 0;
	rep (i, 1, len) {
		int ch = x[i];
        int to = Find (p, ch);
        if (to == -1) return 0;
		p = to;
	}
	return End[p];
}

int main () {
	// freopen ("C:\\Users\\Administrator\\Desktop\\lihan\\1.in", "r", stdin);
	// freopen ("C:\\Users\\Administrator\\Desktop\\lihan\\1.out", "w", stdout);
	freopen ("against.in", "r", stdin);
	freopen ("against.out", "w", stdout);

    read (n);
    LL res = 0;
    rep (i, 1, n) {
        rep (j, 1, m) {
            read (a[i][j]);
            // if (a[i][j] == 0) printf ("FUCK");
        }
        sort (a[i] + 1, a[i] + 1 + m);
        rep (j, 1, (1 << m) - 1) {
            len = 0; memset (x, 0, sizeof x);
            rep (k, 1, m) {
                if ((j >> (k - 1)) & 1) {
					x[++len] = a[i][k];
                }
            }
            Update ();

            if (len & 1) {
                res = res + (Query () - 1);
            }
            else {
                res = res - (Query () - 1);
            }
        }
    }
    write ((LL)n * (n - 1) / 2 - res);
	return 0;
}
posted @ 2021-11-27 14:05  C2022lihan  阅读(11)  评论(0)  编辑  收藏  举报