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 + ;
 LL a[] , b[], fuck = 1e18;

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

 LL exgcd(LL a , LL b , LL &x , LL &y) {
     LL res = a;
     if(!b) {
         x = ;
         y = ;
     }
     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 =  , res = ;
     for(LL i =  ; i <= n ; i++) {
         M = M * m[i];
     }
     for(LL i =  ; 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 < )
         res += M;
     return res;
 }

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

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