HDU5768Lucky7（中国剩余定理+容斥定理）(区间个数统计)

When ?? was born, seven crows flew in and stopped beside him. In its childhood, ?? had been unfortunately fall into the sea. While it was dying, seven dolphins arched its body and sent it back to the shore. It is said that ?? used to surrounded by 7 candles when he faced a extremely difficult problem, and always solve it in seven minutes. 
?? once wrote an autobiography, which mentioned something about himself. In his book, it said seven is his favorite number and he thinks that a number can be divisible by seven can bring him good luck. On the other hand, ?? abhors some other prime numbers and thinks a number x divided by pi which is one of these prime numbers with a given remainder ai will bring him bad luck. In this case, many of his lucky numbers are sullied because they can be divisible by 7 and also has a remainder of ai when it is divided by the prime number pi. 
Now give you a pair of x and y, and N pairs of ai and pi, please find out how many numbers between x and y can bring ?? good luck.

InputOn the first line there is an integer T(T≤20) representing the number of test cases. 
Each test case starts with three integers three intergers n, x, y(0<=n<=15,0<x<y<10181018) on a line where n is the number of pirmes. 
Following on n lines each contains two integers pi, ai where pi is the pirme and ?? abhors the numbers have a remainder of ai when they are divided by pi. 
It is guranteed that all the pi are distinct and pi!=7. 
It is also guaranteed that p1*p2*…*pn<=10181018 and 0<ai<pi<=105105for every i∈(1…n). 
OutputFor each test case, first output "Case #x: ",x=1,2,3...., then output the correct answer on a line.Sample Input

2
2 1 100
3 2
5 3
0 1 100

Sample Output

Case #1: 7
Case #2: 14



        
 

Hint

For Case 1: 7,21,42,49,70,84,91 are the seven numbers.
For Case2: 7,14,21,28,35,42,49,56,63,70,77,84,91,98 are the fourteen numbers.

 题意：

求区间[X,Y]中模7为0，为满足n对关系：膜m[i]不为r[i]，问这样的数字有多少。满足m[]为素数，且不为7。

思路：

• 容斥定理，保证了结果中不多算，不少算，不重复算。
• 中国剩余定理，求出最小的x=c1满足线性同余方程组，则变成求以c1为起始量，M=∏m[]为等差的数列，在[L,R]中的个数，结合抽屉原理，这里是奇加偶减。 
• 保险起见，全部是long long 
• 这里其实是用的线性同余方程组求解的，如果用中国剩余定理，得用快速除法。

 

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#define ll long long
using namespace std;
ll n,ans,tmp,m[20],r[20];
ll BIT(ll x) { ll res=0; while(x){if(x&1LL) res++;x>>=1;} return res&1LL?-1LL:1LL;}
void Ex_gcd(ll a,ll b,ll &d,ll &x,ll &y)
{
    if(b==0){ d=a; x=1; y=0; return ;};
    Ex_gcd(b,a%b,d,y,x);y-=a/b*x;
}
ll Ex_CRT(ll L,ll R,ll N)
{
    ll a,b,c,c1,c2,x,y,d,M=7;
    a=7;  c1=0;
    for(int i=0;i<n;i++){
        if(!(1LL<<i&N)) continue;// 状态 
        M*=m[i];
        b=m[i];c2=r[i]; c=c2-c1;
        Ex_gcd(a,b,d,x,y);
        x=((c/d*x)%(b/d)+b/d)%(b/d);//最小正单元 
        c1=a*x+c1;a=a*b/d;
    }
    return (R-c1+M)/M - (L-1-c1+M)/M;//以c1为起始量，M为等差的数列，在[L,R]中的个数。 
}
int main()
{
    ll T,x,y,i,Case=0; scanf("%lld",&T);
    while(T--){
        ans=0;tmp=0;
        scanf("%lld%lld%lld",&n,&x,&y);
        for(i=0;i<n;i++)
           scanf("%lld%lld",&m[i],&r[i]);
        for(i=0;i<1LL<<n;i++) ans+=BIT(i)*Ex_CRT(x,y,i);
        printf("Case #%lld: %lld\n",++Case,ans);
    }   return 0;
}