hdu1573(中国剩余定理定解的个数)

X问题

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4338    Accepted Submission(s): 1389

Problem Description

求在小于等于N的正整数中有多少个X满足：X mod a[0] = b[0], X mod a[1] = b[1], X mod a[2] = b[2], …, X mod a[i] = b[i], … (0 < a[i] <= 10)。

 

Input

输入数据的第一行为一个正整数T，表示有T组测试数据。每组测试数据的第一行为两个正整数N，M (0 < N <= 1000,000,000 , 0 < M <= 10)，表示X小于等于N，数组a和b中各有M个元素。接下来两行，每行各有M个正整数，分别为a和b中的元素。

 

Output

对应每一组输入，在独立一行中输出一个正整数，表示满足条件的X的个数。

 

Sample Input

   
   
   
   

    
    
    
    3
10 3
1 2 3
0 1 2
100 7
3 4 5 6 7 8 9
1 2 3 4 5 6 7
10000 10
1 2 3 4 5 6 7 8 9 10
0 1 2 3 4 5 6 7 8 9
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    1
0
3
   
   
   
   

 

Author

lwg

 

Source

HDU 2007-1 Programming Contest

/**
中国剩余定理（不互质）
*/
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
typedef __int64 int64;
int64 Mod;

int64 gcd(int64 a, int64 b){
    if(b==0)
        return a;
    return gcd(b,a%b);
}

int64 Extend_Euclid(int64 a, int64 b, int64&x, int64& y){
    if(b==0)
    {
        x=1,y=0;
        return a;
    }
    int64 d = Extend_Euclid(b,a%b,x,y);
    int64 t = x;
    x = y;
    y = t - a/b*y;
    return d;
}

//a在模n乘法下的逆元，没有则返回-1
int64 inv(int64 a, int64 n){
    int64 x,y;
    int64 t = Extend_Euclid(a,n,x,y);
    if(t != 1)
        return -1;
    return (x%n+n)%n;
}

//将两个方程合并为一个
bool merge(int64 a1, int64 n1, int64 a2, int64 n2, int64& a3, int64& n3){
    int64 d = gcd(n1,n2);
    int64 c = a2-a1;
    if(c%d)
        return false;
    c = (c%n2+n2)%n2;
    c /= d;
    n1 /= d;
    n2 /= d;
    c *= inv(n1,n2);
    c %= n2;
    c *= n1*d;
    c += a1;
    n3 = n1*n2*d;
    a3 = (c%n3+n3)%n3;
    return true;
}

//求模线性方程组x=ai(mod ni),ni可以不互质
int64 China_Reminder2(int len, int64* a, int64* n){
    int64 a1=a[0],n1=n[0];
    int64 a2,n2;
    for(int i = 1; i < len; i++)
    {
        int64 aa,nn;
        a2 = a[i],n2=n[i];
        if(!merge(a1,n1,a2,n2,aa,nn))
            return -1;
        a1 = aa;
        n1 = nn;
    }
    Mod = n1;
    return (a1%n1+n1)%n1;
}
int64 a[1000],b[1000];

int main(){
    int i;
    int k;
    int n;
    int t;
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&n,&k);
        for(i = 0; i < k; i++)
            scanf("%I64d",&a[i]);
        for(int i=0;i<k;i++)
            scanf("%I64d",&b[i]);
        __int64 ans=China_Reminder2(k,b,a);
       // cout<<"PPP"<<ans<<endl;
        if(ans>n||ans==-1)
            printf("0\n");
        else
            printf("%d\n",(n-ans)/Mod+1-(ans==0?1:0));
    }
    return 0;
}