hdu X问题 (中国剩余定理不互质)
http://acm.hdu.edu.cn/showproblem.php?pid=1573
X问题
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4439 Accepted Submission(s): 1435
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
这道题除数不是两两互质的,也就不能直接套用中国剩余定理模板,既然不能直接套用就两两合成
#include<stdio.h> #include<algorithm> #include<string.h> #include<stdlib.h> using namespace std; const int M = 20; typedef long long ll; int r; ll n[M], b[M], N; void gcd(ll a, ll b, ll &x, ll &y) { if(b == 0) { x = 1; y = 0; r = a; return ; } gcd(b, a % b, x, y); ll t = x; x = y; y = t - a / b * y; } ll CRT2(ll b[], ll n[], int m) { int f = 0; ll n1 = n[0], n2, b1 = b[0], b2, c, t, k, x, y; for(int i = 1 ; i < m ; i++) { n2 = n[i]; b2 = b[i]; c = b2 - b1; gcd(n1, n2, x, y); if(c % r != 0) { f = 1; break; } k = c / r * x; t = n2 / r; k = (k % t + t) % t; b1 = b1 + n1 * k; n1 = n1 * t; } if(f)//无解 return 0; if(b1 == 0) b1 = n1; if(b1 > N) return 0; return (N - b1) / n1 + 1; } int main() { int t, m; scanf("%d", &t); while(t--) { scanf("%lld%d", &N, &m); for(int i = 0 ; i < m ; i++) scanf("%lld", &n[i]); for(int i = 0 ; i < m ; i++) scanf("%lld", &b[i]); printf("%lld\n", CRT2(b, n, m)); } return 0; }