中国剩余定理 （求某一区间内有多少个解的问题） HDU X问题

 X问题       
         Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1551    Accepted Submission(s): 443



 
  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
         
        
 


#include <iostream>
#include <cstdio>
#include <cstring>
#include <string.h>
using namespace std;
typedef __int64 LL;

LL GCD(LL a, LL b) {
    return (b == 0) ? a : GCD(b, a % b);
}

LL extend_Euclid(LL a, LL b, LL &x, LL &y) {
    if (b == 0) {
        x = 1;
        y = 0;
        return a;
    }
    LL gcd = extend_Euclid(b, a % b, x, y);
    LL temp = x;
    x = y;
    y = temp - (a / b) * x;
    return gcd;
}


LL CRT_2(LL a, LL x, LL b, LL y) {
    LL xx, yy, gcd;
    gcd = extend_Euclid(a, b, xx, yy);
    LL c = y - x;
    while (c < 0) {
        c += a;
    }
    if (c % gcd != 0) {
        return -1;
    }
    xx *= c / gcd;
    yy *= c / gcd;
    LL t = yy / (a / gcd);
    while (yy - t * (a / gcd) > 0) t++;
    while (yy - (t - 1)*(a / gcd) <= 0) t--;
    return (t * (a / gcd) - yy)*b + y;
}




LL CRT(LL crt[][2], int n, LL &m) {
    m = crt[0][0] / GCD(crt[0][0], crt[1][0]) * crt[1][0];
    LL ans = CRT_2(crt[0][0], crt[0][1], crt[1][0], crt[1][1]) % m;
    for (int i = 2; i < n && ans != -1; i++) {
        ans = CRT_2(m, ans, crt[i][0], crt[i][1]);
        m *= crt[i][0] / GCD(m, crt[i][0]);
        ans %= m;
    }
    return ans;
}

//
//int main() {
//    LL crt[11][2];
//    int n;
//    while (scanf("%d", &n) != EOF) {
//        for (int i = 0; i < n; i++) {
//            scanf("%lld%lld", &crt[i][0], &crt[i][1]);
//        }
//
//        
//        if (n == 1) {
//            printf("%lld\n", crt[0][1]);
//        } else {
//            printf("%lld\n", CRT(crt, n));
//        }
//    }
//    return 0;
//}

int main() {
    int T;
    int n;
    LL m;
    LL crt[23][2];
    scanf("%d", &T);
    while (T--) {
        scanf("%I64d%d", &m, &n);
        for (int i = 0; i < n; i++) {
            scanf("%I64d", &crt[i][0]);
        }
        for (int i = 0; i < n; i++) {
            scanf("%I64d", &crt[i][1]);
        }
        LL ans;
        
        if (n == 1) {
            if (crt[0][0] == 0) {
                ans = 0;
            } else {
                if (crt[0][1] > m) {
                    ans = 0;
                } else if (crt[0][1] == 0) {
                    ans = m / crt[0][0];
                } else {
                    ans = (m - crt[0][1]) / crt[0][0] + 1;
                }
            }
        } else {
            LL LCM;
            int res = CRT(crt, n, LCM);
            if (res == -1 || res > m) {
                ans = 0;
            } else if (res == 0) {
                ans = m / LCM;
            } else {
                ans = (m - res) / LCM + 1;
            }
        }
        printf("%I64d\n", ans);
    }
    return 0;
}