codeforces 1096 题解

A：

发现最优的方案一定是选 $ l $ 和 $ 2 * l $，题目保证有解，直接输出即可

#include <bits/stdc++.h>
#define Fast_cin ios::sync_with_stdio(false), cin.tie();
#define rep(i, a, b) for(register int i = a; i <= b; i++)
#define per(i, a, b) for(register int i = a; i >= b; i--)
#define DEBUG(x) cerr << "DEBUG" << x << " >>> " << endl;
using namespace std;

typedef unsigned long long ull;
typedef long long ll;

template <typename _T>
inline void read(_T &f) {
    f = 0; _T fu = 1; char c = getchar();
    while(c < '0' || c > '9') { if(c == '-') fu = -1; c = getchar(); }
    while(c >= '0' && c <= '9') { f = (f << 3) + (f << 1) + (c & 15); c = getchar(); }
    f *= fu;
}

template <typename T>
void print(T x) {
    if(x < 0) putchar('-'), x = -x;
    if(x < 10) putchar(x + 48);
    else print(x / 10), putchar(x % 10 + 48);
}

template <typename T>
void print(T x, char t) {
    print(x); putchar(t);
}

int T, l, r; 

int main() {
	read(T);
	while(T--) {
		read(l); read(r);
		print(l, ' '); print(2 * l, '\n');
	}
	return 0;
}

B：

情况 1：所有字母都相同，输出 $ n * (n - 1) / 2 $ 即可
情况 2：左边有连续 $ x $ 个字母相同，右边有 $ y $ 个，第一个字母和最后一个字符相同，输出 $ (x + 1) * (y + 1) $
情况 3：左边有连续 $ x $ 个字母相同，右边有 $ y $ 个，第一个字母和最后一个字符不同，输出 $ x + y + 2 - 1 $，最后的 $ -1 $ 是因为整个串被算了 $ 2 $ 次

#include <bits/stdc++.h>
#define Fast_cin ios::sync_with_stdio(false), cin.tie();
#define rep(i, a, b) for(register int i = a; i <= b; i++)
#define per(i, a, b) for(register int i = a; i >= b; i--)
#define DEBUG(x) cerr << "DEBUG" << x << " >>> " << endl;
using namespace std;

typedef unsigned long long ull;
typedef long long ll;

template <typename _T>
inline void read(_T &f) {
    f = 0; _T fu = 1; char c = getchar();
    while(c < '0' || c > '9') { if(c == '-') fu = -1; c = getchar(); }
    while(c >= '0' && c <= '9') { f = (f << 3) + (f << 1) + (c & 15); c = getchar(); }
    f *= fu;
}

template <typename T>
void print(T x) {
    if(x < 0) putchar('-'), x = -x;
    if(x < 10) putchar(x + 48);
    else print(x / 10), putchar(x % 10 + 48);
}

template <typename T>
void print(T x, char t) {
    print(x); putchar(t);
}

const int N = 2e5 + 5, md = 998244353;

char c[N];
int cnt[23333];
int n, cut1 = -1, cut2 = -1;

inline int mul(int x, int y) { return (ll)x * y % md; }

int main() {
	read(n); scanf("%s", c + 1);
	for(register int i = 2; i <= n; i++) if(c[i] != c[i - 1]) { cut1 = i - 1; break; }
	for(register int i = n - 1; i >= 1; i--) if(c[i] != c[i + 1]) { cut2 = i + 1; break; }
	if(cut1 == -1) cout << (ll)n * (n - 1) / 2 % md << endl;
	else if(c[1] == c[n]) cout << mul(cut1 + 1, n - cut2 + 2) << endl;
	else cout << cut1 + 1 + n - cut2 + 2 - 1 << endl;
	return 0;
}

C：

答案本质上是把一个圆切成答案份

那么圆心角确定了，就可以算出一个圆周角的大小，如果切成 360 份可以拼出任意角度，所以枚举这个角度即可

#include <bits/stdc++.h>
#define Fast_cin ios::sync_with_stdio(false), cin.tie();
#define rep(i, a, b) for(register int i = a; i <= b; i++)
#define per(i, a, b) for(register int i = a; i >= b; i--)
#define DEBUG(x) cerr << "DEBUG" << x << " >>> " << endl;
using namespace std;

typedef unsigned long long ull;
typedef long long ll;

template <typename _T>
inline void read(_T &f) {
    f = 0; _T fu = 1; char c = getchar();
    while(c < '0' || c > '9') { if(c == '-') fu = -1; c = getchar(); }
    while(c >= '0' && c <= '9') { f = (f << 3) + (f << 1) + (c & 15); c = getchar(); }
    f *= fu;
}

template <typename T>
void print(T x) {
    if(x < 0) putchar('-'), x = -x;
    if(x < 10) putchar(x + 48);
    else print(x / 10), putchar(x % 10 + 48);
}

template <typename T>
void print(T x, char t) {
    print(x); putchar(t);
}

int T;

int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }

int main() {
	read(T); while(T--) {
		int d; read(d); int ans = -1;
		for(register int i = 3; i <= 23333; i++) {
			if(d * i % 180 == 0 && d <= 180 - 360 / (double)i) {
				ans = i; break;
			}
		}
		print(ans, '\n');
	}
	return 0;
}

D：

$ f[i][j] $ 表示到了第 $ i $ 个字母，$ hard $ 已经匹配到了第 $ j $ 个字母的最小代价

直接 $ dp $ 即可

#include <bits/stdc++.h>
#define int long long
#define Fast_cin ios::sync_with_stdio(false), cin.tie();
#define rep(i, a, b) for(register int i = a; i <= b; i++)
#define per(i, a, b) for(register int i = a; i >= b; i--)
#define DEBUG(x) cerr << "DEBUG" << x << " >>> " << endl;
using namespace std;

typedef unsigned long long ull;
typedef long long ll;

template <typename _T>
inline void read(_T &f) {
    f = 0; _T fu = 1; char c = getchar();
    while(c < '0' || c > '9') { if(c == '-') fu = -1; c = getchar(); }
    while(c >= '0' && c <= '9') { f = (f << 3) + (f << 1) + (c & 15); c = getchar(); }
    f *= fu;
}

template <typename T>
void print(T x) {
    if(x < 0) putchar('-'), x = -x;
    if(x < 10) putchar(x + 48);
    else print(x / 10), putchar(x % 10 + 48);
}

template <typename T>
void print(T x, char t) {
    print(x); putchar(t);
}

const int N = 1e5 + 5;

int f[N][5], w[N];
char c[N];
int n;

int calc(char t) {
	if(t == 'h') return 1;
	if(t == 'a') return 2;
	if(t == 'r') return 3;
	if(t == 'd') return 4;
	return 0;
}

#undef int
int main() {
#define int long long
	memset(f, -1, sizeof(f));
	read(n); scanf("%s", c + 1);
	for(register int i = 1; i <= n; i++) read(w[i]);
	f[0][0] = 0;
	for(register int i = 0; i < n; i++) {
		int val = calc(c[i + 1]);
		for(register int j = 0; j <= 3; j++) {
			if(f[i][j] == -1) continue;
			if(val == j + 1) {
				if(f[i + 1][j + 1] == -1) f[i + 1][j + 1] = f[i][j];
				else f[i + 1][j + 1] = min(f[i + 1][j + 1], f[i][j]);
				if(f[i + 1][j] == -1) f[i + 1][j] = f[i][j] + w[i + 1];
				else f[i + 1][j] = min(f[i + 1][j], f[i][j] + w[i + 1]);
			} else {
				if(f[i + 1][j] == -1) f[i + 1][j] = f[i][j];
				else f[i + 1][j] = min(f[i + 1][j], f[i][j]);
			}
		}
	}
	int ans = f[n][0];
	for(register int i = 0; i <= 3; i++) if(f[n][i] != -1) ans = min(ans, f[n][i]);
	cout << ans << endl;
	return 0;
}

E：

没写出来，先咕了

update：2019.1.2

去年不会这题，今年来补

$ n $ 为人数，总分为 $ s $，自己的下限 $ r $ , $ c $ 为组合数，枚举有多少个人跟自己得分相同，自己的得分 $ j $，对答案的贡献是 $ C[n - 1][i - 1] * \frac{1}{i} * calc(n - i, s - i * j, j) $

$ calc( n, s, big ) $ 表示有 $ n $ 个人，总分为 $ s $，每个人的得分都 $ < big $ 的方案数，这个可以通过容斥求出

总方案数可以用隔板法求出是 $ C[s - r + n - 1][n - 1] $，乘上这个的逆元即可

#include <bits/stdc++.h>
#define CIOS ios::sync_with_stdio(false);
#define rep(i, a, b) for(register int i = a; i <= b; i++)
#define per(i, a, b) for(register int i = a; i >= b; i--)
#define DEBUG(x) cerr << "DEBUG" << x << " >>> ";
using namespace std;

typedef unsigned long long ull;
typedef long long ll;

template <typename T>
inline void read(T &f) {
	f = 0; T fu = 1; char c = getchar();
	while (c < '0' || c > '9') { if (c == '-') fu = -1; c = getchar(); }
	while (c >= '0' && c <= '9') { f = (f << 3) + (f << 1) + (c & 15); c = getchar(); }
	f *= fu;
}

template <typename T>
void print(T x) {
	if (x < 0) putchar('-'), x = -x;
	if (x < 10) putchar(x + 48);
	else print(x / 10), putchar(x % 10 + 48);
}

template <typename T>
void print(T x, char t) {
	print(x); putchar(t);
}

const int N = 5005, md = 998244353;

inline int mul(int x, int y) { return (ll)x * y % md; }

inline int add(int x, int y) {
	x += y;
	if(x >= md) x -= md;
	return x;
}

inline int sub(int x, int y) {
	x -= y;
	if(x < 0) x += md;
	return x;
}

inline int fpow(int x, int y) {
	int ans = 1;
	while(y) {
		if(y & 1) ans = mul(ans, x);
		y >>= 1; x = mul(x, x);
	}
	return ans;
}

int C[N + 105][105], inv[N];
int n, s, r, ans;

int calc(int n, int s, int big) {
	if(n == 0) return s == 0;
	int ans = 0;
	for(register int i = 0, opt = 1; i <= n && i * big <= s; i++, opt = md - opt)
		ans = add(ans, mul(opt, mul(C[s - i * big + n - 1][n - 1], C[n][i])));
	return ans;
}

int main() {
	read(n); read(s); read(r);
	for(register int i = 0; i <= s + n; i++) {
		C[i][0] = 1;
		for(register int j = 1; j <= i && j <= n; j++)
			C[i][j] = add(C[i - 1][j - 1], C[i - 1][j]);
	}
	for(register int i = 1; i <= n; i++) inv[i] = fpow(i, md - 2);
	for(register int i = 1; i <= n; i++) {
		for(register int j = r; j <= s; j++) {
			if(s - i * j < 0) break;
			ans = add(ans, mul(mul(C[n - 1][i - 1], calc(n - i, s - i * j, j)), inv[i])); 
		}
	}
	print(mul(ans, fpow(C[s - r + n - 1][n - 1], md - 2)), '\n');
	return 0;
}

// Rotate Flower Round.

F：

分别计算 $ -1 $ 和 $ -1 $ 的贡献，$ -1 $ 和数字的贡献，数字和数字的贡献

第一个用 $ dp $ 求出，第二个用前缀和求出，第三个用树状数组求出

$ sb $ 的我能用前缀和的地方写了树状数组

#include <bits/stdc++.h>
#define Fast_cin ios::sync_with_stdio(false), cin.tie();
#define rep(i, a, b) for(register int i = a; i <= b; i++)
#define per(i, a, b) for(register int i = a; i >= b; i--)
#define DEBUG(x) cerr << "DEBUG" << x << " >>> " << endl;
using namespace std;

typedef unsigned long long ull;
typedef long long ll;

template <typename _T>
inline void read(_T &f) {
    f = 0; _T fu = 1; char c = getchar();
    while(c < '0' || c > '9') { if(c == '-') fu = -1; c = getchar(); }
    while(c >= '0' && c <= '9') { f = (f << 3) + (f << 1) + (c & 15); c = getchar(); }
    f *= fu;
}

template <typename T>
void print(T x) {
    if(x < 0) putchar('-'), x = -x;
    if(x < 10) putchar(x + 48);
    else print(x / 10), putchar(x % 10 + 48);
}

template <typename T>
void print(T x, char t) {
    print(x); putchar(t);
}

const int N = 2e5 + 5, md = 998244353;

inline int mul(int x, int y) { return (ll)x * y % md; }

inline int add(int x, int y) {
	x += y;
	if(x >= md) x -= md;
	return x;
}

inline int sqr(int x) { return mul(x, x); }

inline int fpow(int x, int y) {
	int ans = 1;
	while(y) {
		if(y & 1) ans = mul(ans, x);
		y >>= 1; x = sqr(x);
	}
	return ans;
}

int a[N], s[N], f[N], fac[N];
int n, inv2 = (md + 1) / 2, ans1;

inline int lowbit(int x) { return x & -x; }

void change(int x, int y) {
	for(register int i = x; i <= n; i += lowbit(i))
		f[i] += y;
}

int query(int x) {
	int ans = 0;
	for(register int i = x; i; i -= lowbit(i))
		ans += f[i];
	return ans;
}

int main() {
	read(n); fac[0] = 1;
	for(register int i = 1; i <= n; i++) read(a[i]), fac[i] = mul(fac[i - 1], i), s[i] = 1;
	for(register int i = n; i >= 1; i--) {
		if(a[i] == -1) continue;
		ans1 = add(ans1, query(a[i]));
		change(a[i], 1);
	}
	memset(f, 0, sizeof(f));
	for(register int i = 1; i <= n; i++) {
		if(a[i] != -1) s[a[i]] = 0;
	}
	for(register int i = 1; i <= n; i++) s[i] += s[i - 1];
	ans1 = mul(ans1, fac[s[n]]);
	for(register int i = 1; i <= s[n]; i++) f[i] = add(f[i - 1], mul(mul(fac[s[n]], fpow(i, md - 2)), mul(i - 1, mul(i, inv2))));
	ans1 = add(ans1, f[s[n]]); memset(f, 0, sizeof(f));
	for(register int i = 1; i <= n; i++) if(a[i] == -1) change(i, 1);
	for(register int i = 1; i <= n; i++) {
		if(a[i] == -1) continue;
		int val = mul(s[a[i]], query(n) - query(i));
		val = add(val, mul(s[n] - s[a[i]], query(i)));
		val = mul(val, fac[s[n] - 1]);
		ans1 = add(ans1, val);
	}
	print(mul(ans1, fpow(fac[s[n]], md - 2)), '\n');
	return 0;
}

G：

看到背包问题就会想到卷积，因为背包的转移和卷积的形式相同

对于最开始的 $ k $ 个数构造生成函数，计算 $ n / 2 $ 次幂，每一位的平方加起来就是答案啦

#pragma GCC target("avx")
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
#define Fast_cin ios::sync_with_stdio(false), cin.tie();
#define rep(i, a, b) for(register int i = a; i <= b; i++)
#define per(i, a, b) for(register int i = a; i >= b; i--)
#define DEBUG(x) cerr << "DEBUG" << x << " >>> " << endl;
using namespace std;

typedef unsigned long long ull;
typedef long long ll;

template <typename _T>
inline void read(_T &f) {
    f = 0; _T fu = 1; char c = getchar();
    while(c < '0' || c > '9') { if(c == '-') fu = -1; c = getchar(); }
    while(c >= '0' && c <= '9') { f = (f << 3) + (f << 1) + (c & 15); c = getchar(); }
    f *= fu;
}

template <typename T>
void print(T x) {
    if(x < 0) putchar('-'), x = -x;
    if(x < 10) putchar(x + 48);
    else print(x / 10), putchar(x % 10 + 48);
}

template <typename T>
void print(T x, char t) {
    print(x); putchar(t);
}

const int P = 998244353;

inline int add(int x, int y) {
    x += y;
    if(x >= P) x -= P;
    return x;
}

inline int sub(int x, int y) {
    x -= y;
    if(x < 0) x += P;
    return x;
}

inline int mul(int x, int y) {
    return (ll)x * y % P;
}

inline int fpow(int x, int y) {
    int ans = 1;
    while(y) {
        if(y & 1) ans = mul(ans, x);
        y >>= 1; x = mul(x, x);
    }
    return ans;
}

namespace ntt {
    int base = 1, root = -1, maxbase = -1;
    vector <int> roots = {0, 1}, rev = {0, 1};
    
    void init() {
        int tmp = P - 1; maxbase = 0;
        while(!(tmp & 1)) {
            tmp >>= 1;
            maxbase++;
        }
        root = 2;
        while(1) {
            if(fpow(root, 1 << maxbase) == 1 && fpow(root, 1 << (maxbase - 1)) != 1) break;
            root++;
        }
    }
    
    void ensure_base(int nbase) {
        if(maxbase == -1) init();
        if(nbase <= base) return;
        assert(nbase <= maxbase);
        rev.resize(1 << nbase);
        for(register int i = 1; i < (1 << nbase); i++) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (nbase - 1));
        roots.resize(1 << nbase);
        while(base < nbase) {
            int z = fpow(root, 1 << (maxbase - base - 1));
            for(register int i = (1 << (base - 1)); i < (1 << base); i++) {
                roots[i << 1] = roots[i];
                roots[i << 1 | 1] = mul(roots[i], z);
            }
            base++;
        } 
    }
    
    void dft(vector <int> &a) {
        int n = a.size(), zeros = __builtin_ctz(n);
        ensure_base(zeros);
        int shift = base - zeros;
        for(register int i = 0; i < n; i++) if(i < (rev[i] >> shift)) swap(a[i], a[rev[i] >> shift]);
        for(register int mid = 1; mid < n; mid <<= 1) {
            for(register int i = 0; i < n; i += (mid << 1)) {
                for(register int j = 0; j < mid; j++) {
                    int x = a[i + j], y = mul(a[i + j + mid], roots[mid + j]);
                    a[i + j] = add(x, y); a[i + j + mid] = sub(x, y);
                }
            }
        }
    }
    
    vector <int> pmul(vector <int> a, vector <int> b, bool is_sqr = false) {
        int need = a.size() + b.size() - 1, nbase = 0;
        while((1 << nbase) < need) nbase++;
        ensure_base(nbase); int size = 1 << nbase;
        a.resize(size); dft(a); if(!is_sqr) b.resize(size), dft(b); else b = a;
        int inv = fpow(size, P - 2);
        for(register int i = 0; i < size; i++) a[i] = mul(a[i], mul(b[i], inv));
        reverse(a.begin() + 1, a.end()); dft(a); a.resize(need); return a;
    }
    
    vector <int> psqr(vector <int> a) { return pmul(a, a, 1); }
    
    vector <int> inv(vector <int> a, int size) {
    	if(size == 1) return vector <int> { fpow(a[0], P - 2) };
    	vector <int> b = inv(a, size >> 1); a = pmul(a, psqr(b)); b.resize(size);
    	for(register int i = 0; i < size; i++) b[i] = sub(add(b[i], b[i]), a[i]);
    	return b;
    }
    
    vector <int> pinv(vector <int> a) {
    	int nbase = 0; while((1 << nbase) < a.size()) nbase++;
    	return inv(a, 1 << nbase);
    }
}
using ntt::pmul;
using ntt::psqr;

vector <int> wxw, ans;
int n, k, sum;

int main() {
	read(n); read(k); wxw.resize(10); 
	for(register int i = 1; i <= k; i++) {
		int t; read(t); wxw[t] = 1;
	}
	n >>= 1; ans = wxw; n--;
	while(n) {
		if(n & 1) ans = pmul(ans, wxw);
		n >>= 1; wxw = psqr(wxw);
	}
	for(register int i = 0; i < ans.size(); i++) sum = add(sum, mul(ans[i], ans[i]));
	cout << sum << endl;
	return 0;
}