HDU 5768Lucky7-中国剩余定理+容斥原理

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http://acm.hdu.edu.cn/showproblem.php?pid=5768

注意爆ll

容斥原理求一下【1，x】里有多少个满足的即可

#include <cstdio>#include <cmath>#include <cstring>#include <string>#include <algorithm>#include <queue>#include <map>#include <set>#include <vector>#include <iostream>using namespace std;long long m[20],a[20],mm[20],aa[20];void extend_Euclid(long long a, long long b, long long &x, long long &y){    if(b == 0)    {        x = 1;        y = 0;        return;    }    extend_Euclid(b, a % b, x, y);    long long tmp = x;    x = y;    y = tmp - (a / b) * y;}long long CRT(long long a[],long long  m[],int n,long long & M){    M = 1;    long long ans = 0;    for(int i=1; i<=n; i++)        M *= m[i];    for(int i=1; i<=n; i++)    {        long long x, y;        long long Mi = M / m[i];        extend_Euclid(Mi, m[i], x, y);        x = (x%m[i]+m[i])%m[i];        ans = (ans + ((Mi % M) * (x % M) % M* a[i]) % M) % M;    }    if(ans < 0) ans += M;    return ans;} int main(){    //freopen("input.txt","r",stdin);    //freopen("output.txt","w",stdout);    int cnt=1;    int tt; scanf("%d",&tt);    while (tt--)    {        int n;        long long x,y,ans=0;        scanf("%d%lld%lld",&n,&x,&y);        for (int i=1; i<=n; i++)            scanf("%lld%lld",&m[i],&a[i]);        for (int state=0; state<(1<<n); state++)        {            int nn=0;            for (int j=1; j<=n; j++)            {                if (state & (1<<(j-1)))                {                    mm[++nn]=m[j];                    aa[nn]=a[j];                }            }            mm[++nn]=7;            aa[nn]=0;            long long M;            long long tmp=CRT(aa,mm,nn,M);            long long numx,numy;            if (x-1<tmp) numx=0;            else numx=(x-tmp-1)/M+1;            if (y<tmp) numy=0;            else numy=(y-tmp)/M+1;            long long num=numy-numx;            if (nn % 2)                ans+=num;            else ans-=num;        }        printf("Case #%d: %lld\n",cnt++,ans);    }    return 0;}

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