10.2模拟赛总结
10.2 模拟赛总结
T1.
数位dp：
一个非常非常非常非常显然的数位 DP
\([L,R] = [1,R]-[1,L-1]\)
所以是分别求两次小于等于某个数字的方案数
\(f(i,j,k)\) 表示从低位数起的第 \(i\) 位，按照规则计算后答案为 \(j\quad (j=0,1)\)
\(k\) 表示只考虑后面结尾和 \(lmt\)后面几位 的大小关系 \((k=0,1)\)
考虑第 \(i+1\) 位，算一下新构成的数字并判断下大小就可以了
注意到 \(L,R\) 数据范围特别大，需要用高精度，最后结果要以二进制输出，所以可以对高精度压位
（以上扒的题解）
这题是个正常人就会想到找规律：
然后就有打表：（1~100）
然后就没了（啥？还有高精呢）
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#define LL long long
using namespace std;
int n, q, v, t, L, R, len;
char s[208];
struct bigint
{
int len, zz;
int v[1005];
bigint(){len = 0; memset(v, 0, sizeof v); zz = 1;}
bigint(int x)
{
if(x >= 0) zz = 1;
else x = -x, zz = 0;
len = 0;
memset(v, 0, sizeof v);
while(x)
{
v[ ++len] = x % 10;
x /= 10;
}
}
friend bool operator < (const bigint &a, const bigint &b)
{
if(a.len < b.len) return 1;
if(a.len > b.len) return 0;
for(int i = a.len ; i >= 1; i -- )
{
if(a.v[i] < b.v[i]) return 1;
if(a.v[i] > b.v[i]) return 0;
}
return 0;
}
friend bool operator == (const bigint &a, const bigint &b)
{
if(a.len != b.len ) return 0;
for(int i = a.len; i >= 1; i --)
{
if(a.v[i] != b.v[i]) return 0;
}
return 1;
}
friend bool operator <= (const bigint &a, const bigint &b)
{
if(a < b) return 1;
else if(a == b) return 1;
else return 0;
}
friend bool operator != (const bigint &a, const bigint &b)
{
if(a.len != b.len) return 1;
for(int i = a.len; i >= 1; i --)
{
if(a.v[i] != b.v[i]) return 1;
}
return 0;
}
}x, y, res;
bigint operator + (bigint a, bigint b)
{
int len = a.len + b.len;
bigint c;
c.len = len;
for(int i = 1; i <= len; i ++)
c.v[i] = a.v[i] + b.v[i];
for(int i = 1; i <= len; i ++)
{
if(c.v[i] >= 10)
{
++c.v[i+1];
c.v[i] -= 10;
}
}
while(c.len&&!c.v[c.len]) c.len --;
return c;
}
bigint operator - (bigint a, bigint b)
{
int len = max(a.len, b.len);
bigint c;
for(int i = 1; i <= len; i ++)
c.v[i] = a.v[i] - b.v[i];
c.len = len;
for(int i = 1; i <= c.len; i ++)
{
if(c.v[i] < 0)
{
c.v[i+1]--;
c.v[i] += 10;
}
}
while(c.len&&!c.v[c.len]) c.len --;
return c;
}
bigint operator *(bigint a,bigint b)
{
bigint c;
for(int i = 1; i <= a.len; ++ i)
for(int j = 1; j <= b.len; ++ j)
c.v[i+j-1] += a.v[i] * b.v[j];
c.len = a.len + b.len;
for(int i = 1; i <= c.len - 1; ++ i)
{
if(c.v[i] >= 10)
{
c.v[i+1] += c.v[i] / 10;
c.v[i] %= 10;
}
}
while(c.v[c.len] == 0&&c.len > 1) -- c.len;
return c;
}
bigint operator /(bigint a,long long b)
{
bigint c;int d = 0;
for(int i = a.len; i >= 1; -- i)
c.v[++ c.len] = ((d * 10 + a.v[i]) / b),d=(d*10+a.v[i])%b;
for(int i=1;i<=c.len/2;++i)swap(c.v[i],c.v[c.len-i+1]);
while(c.v[c.len]==0&&c.len>1)--c.len;
return c;
}
bigint Min(bigint a, bigint b)
{
if(a < b) return a;
else return b;
}
bigint work(bigint x)
{
if(x < bigint(4)) return bigint(1);
bigint l = bigint(4), r = Min(x, bigint(7)), res = bigint(1);
int opt = 1;
for(; ; l = r + bigint(1), r = Min(r * bigint(2) + bigint(1), x), opt ^= 1 )
{
if(opt)
res = res + (r - l + bigint(1));
if(r == x) break;
}
return res;
}
void out(bigint x)
{
if(!x.len) return (void)printf("0");
bigint qwq = bigint(1);
while(qwq <= x) qwq = qwq * bigint(2);
qwq = qwq / 2;
for(; ; qwq = qwq /2)
{
if(qwq <= x)
{
printf("1");
x = x - qwq;
}
else printf("0");
if(qwq == bigint(1))break;
}
}
void solve()
{
x = y = res = bigint(0);
scanf("%s", s + 1);
for(int i = 1; i <= n; i ++)
{
x = x * 2 + bigint(s[i] - '0');
}
scanf("%s", s + 1);
for(int i = 1; i <= n; i ++)
{
y = y * 2 + bigint(s[i] - '0');
}
res = work(y) - work(x - bigint(1));
if((n&1) == (q&1)) res = y - x + 1 - res;
out(res);
puts("");
}
signed main()
{
// freopen("a.in", "r", stdin);
// freopen("a.out", "w", stdout);
scanf("%d", &t);
while(t --)
{
scanf("%d%d",&n, &q);
solve();
}
return 0;
}
T2
一sb题， 没啥总结的。。
。
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#define LL long long
#define N 100005
using namespace std;
struct node
{
int w1, w2, l1, l2;
}q[N];
int n, m, t, minw = 2e9, maxw = -233, minl = 2e9, maxl = -233, flag;
signed main()
{
freopen("b.in", "r",stdin);
freopen("b.out", "w", stdout);
scanf("%d", &t);
while(t -- )
{
flag = 0; minw = minl = 2e9; maxw = maxl = -233;
scanf("%d", &n);
for(int i = 1; i <= n; i ++)
scanf("%d%d%d%d", &q[i].w1, &q[i].w2, &q[i].l1, &q[i].l2);
for(int i = 1; i <= n; i ++)
{
minw = min(minw, q[i].w1);
maxw = max(maxw, q[i].w2);
minl = min(minl, q[i].l1);
maxl = max(maxl, q[i].l2);
}
for(int i = 1; i <= n; i ++)
{
if(q[i].w1 <= minw&&q[i].w2 >= maxw&&q[i].l1 <= minl&&q[i].l2 >= maxl)
{flag = 1; break; }
}
if(flag)printf("TAK\n");
else printf("NIE\n");
}
return 0;
}
T3
毒瘤数据结构+数论题
--给定1, n, d, v 给序列所有满足\(gcd(x, n)=d\)的\(x\), 给\(a[x]+=v\);
就相当于\(a[x]+=v[gcd(x, n)==d]\)
然后就可以愉快的推式子了
\(\ \ \ \ v[\gcd(x,n) = d]\)
$ = v [\gcd(\frac{x}{d},\frac{n}{d})=1]$
$ = v\sum\limits_{k|\gcd(\frac{x}{d},\frac{n}{d})} \mu(k)$ (日常反演)
$ = v\sum\limits_{k|\frac{x}{d},k|\frac{n}{d}} \mu(k)$
\(=\sum\limits_{k|\frac{n}{d},kd|x} v\mu(k)\)
暴力做法显然是要枚举\(x\), 对于每一个\({k|\frac{n}{d}且kd|x}\), 都加上\(v\mu(k)\),
可以等价于
对于一个合法的\(k|\dfrac{n}{d}\), 则\(x =kd,2kd,3kd...\), 枚举\(k\), 把所有\(kd\), 的倍数都加上\(v\mu(k)\);
这样虽然\(O(1)\)查询， 但修改的复杂度太大
考虑均摊复杂度
我们开一个数组\(f\) 表示所有是\(i\), 的倍数的位置都加上\(f[i]\)
修改时只需找出合法的\(k\), 然后\(f[kd]+=v\mu(k)\), 省去了枚举\(kd\) 的倍数；
然后查询时 查询一个数\(i\) 时， 就成了\(\sum_{d|i}f(d)\)
则\(x\), 的前缀和就是
\(\sum\limits_{i=1}^x\sum\limits_{d|i} f(d)=\sum\limits_{d=1}^x f(d)\lfloor \frac{x}{d}\rfloor\)
然后就可以整除分块， 对与每一块需要求出那一块的\(f\)的和；单点修改区间求和树状数组可以维护；
时间复杂度$O（q\sqrt{l}\log l+ l \log l） $
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#define int long long
#define N 200005
using namespace std;
int val[N], a[N], l, m, tot;
void add(int pos, int v)
{
for(int i = pos; i <= l; i += i&(-i))
val[i] += v;
}
int ask(int pos)
{
int res = 0;
for(int i = pos; i ; i -= i&(-i))
res += val[i];
return res;
}
int mu[N], prime[N], vis[N];
void yych()
{
mu[1] = 1;
for(int i = 2;i <= N; i ++)
{
if(!vis[i])
{
mu[i] = -1;
prime[++tot] = i;
}
for(int j = 1; j <= tot&& i * prime[j]<= N; j ++)
{
vis[i * prime[j]] = 1;
if(i%prime[j] == 0)
{
mu[i * prime[j]] = 0;
break;
}
mu[i*prime[j]] = -mu[i];
}
}
}
signed main()
{
freopen("c.in", "r",stdin);
freopen("c.out", "w", stdout);
int cas = 0, opt, x, y, z, n, d, v;
yych();
while(scanf("%lld%lld", &l, &m) && l && m)
{
for(int i = 1; i <= l; i ++) val[i] = 0;
printf("Case #%lld:\n", ++cas);
for(int i = 1; i <= m ;i ++)
{
scanf("%lld",&opt);
if(opt&1)
{
scanf("%lld%lld%lld", &n, &d, &v);
if(n%d!=0)continue;
int q = n/d;
for(int p = 1; p * p <= q; p ++)
{
if(q % p == 0)
{
add(d*p, v * mu[p]);
if(p * p != q)
add(d*(q/p), v * mu[q/p]);
}
}
}
else
{
scanf("%lld", &x);
int ans = 0;
for(int l = 1, r; l <= x; l = r + 1)
{
r = min(x / (x / l), x);
ans += (x/l) * (ask(r) - ask(l - 1));
}
printf("%lld\n", ans);
}
}
}
return 0;
}
可以撒花了
然后这个柿子的理解
\(\sum\limits_{i=1}^x\sum\limits_{d|i}1=\sum\limits_{d=1}^{x}\lfloor\frac{x}{d}\rfloor\)
1到x每一个数的所有约数的个数
就相当于枚举一个约数， 这个约数的倍数的个数， x以内d的倍数的个数就是\(\lfloor\frac{x}{d}\rfloor\);
