HDU 5768 Lucky7 (中国剩余定理+容斥)

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

给你n个同余方程组，然后给你l,r,问你l，r中有多少数%7=0且%ai != bi.

比较明显的中国剩余定理+容斥，容斥的时候每次要加上个(%7=0)这一组。

中间会爆longlong，所以在其中加上个快速乘法(类似快速幂)。因为普通的a*b是直接a个b相加，很可能会爆。但是你可以将b拆分为二进制来加a，这样又快又可以防爆。

//#pragma comment(linker, "/STACK:102400000, 102400000")
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cmath>
#include <ctime>
#include <list>
#include <set>
#include <map>
using namespace std;
typedef long long LL;
typedef pair <int, int> P;
const int N = 1e5 + 5;
LL a[17] , b[17], fuck = 1e18;

LL Fmul(LL a , LL n , LL mod) { //快速乘法
    LL res = 0;
    while(n) {
        if(n & 1) {
            res = (res + a) % mod;
        }
        a = a * 2 % mod;
        n >>= 1;
    }
    return res;
}

LL exgcd(LL a , LL b , LL &x , LL &y) {
    LL res = a;
    if(!b) {
        x = 1;
        y = 0;
    }
    else {
        res = exgcd(b , a % b , x , y);
        LL temp = x;
        x = y;
        y = temp - a / b * y;
    }
    return res;
}

LL CRT(LL a[] , LL m[] , LL n) {
    LL M = 1 , res = 0;
    for(LL i = 1 ; i <= n ; i++) {
        M = M * m[i];
    }
    for(LL i = 1 ; i <= n ; i++) {
        LL x , y , Mi = M / m[i];
        exgcd(Mi , m[i] , x , y);
        x = (x % m[i] + m[i]) % m[i];
        res = (res + Fmul(x * a[i] % M , Mi , M) + M) % M;
    }
    if(res < 0) 
        res += M;
    return res;
}

LL Get(LL n, LL x, LL y) {
    if(x > n)
        return 0;
    else {
        n -= x;
        return (LL)(1 + n / y);
    }
}

int main()
{
    int t, n;
    LL l , r , md[17], di[17];
    scanf("%d", &t);
    for(int ca = 1; ca <= t; ++ca) {
        scanf("%d %lld %lld", &n, &l, &r);
        di[1] = 7, md[1] = 0;
        for(int i = 1; i <= n; ++i) {
            scanf("%lld %lld", a + i, b + i);
        }
        LL res1 = 0, res2 = 0;
        int limit = (1 << n) - 1;
        for(int i = 1; i <= limit; ++i) {
            int temp = i, index = 1;
            LL mul = 7;
            for(int j = 1 ; temp ; ++j) {
                if(temp & 1) {
                    di[++index] = a[j];
                    md[index] = b[j];
                    mul *= a[j];
                }
                temp >>= 1;
            }
            if(index % 2) {
                res1 -= Get(l - 1, CRT(md, di, index), mul);
                res2 -= Get(r, CRT(md, di, index), mul);
            }
            else {
                res1 += Get(l - 1, CRT(md, di, index), mul);
                res2 += Get(r, CRT(md, di, index), mul);
            }
        }
        printf("Case #%d: %lld
",ca, r/7 - (l-1)/7 - (res2 - res1));
    }
    return 0;
}