标签:hid 技术 nbsp long play problem isp ber display
嗯...
题目链接:http://poj.org/problem?id=1995
快速幂模板...
AC代码:
1 #include<cstdio> 2 #include<iostream> 3 4 using namespace std; 5 6 int main(){ 7 long long N, M, n, a, b, c, sum = 0; 8 scanf("%lld", &N); 9 while(N--){ 10 scanf("%lld%lld", &M, &n); 11 sum = 0; 12 for(int i = 1; i <= n; i++){ 13 c = 1; 14 scanf("%lld%lld", &a, &b); 15 while(b){ 16 if(b & 1) c = c * a % M; 17 a = a * a % M; 18 b /= 2; 19 } 20 sum += c % M; 21 } 22 printf("%lld\n", sum % M); 23 } 24 return 0; 25 }
POJ 1995 Raising Modulo Numbers(快速幂)
标签:hid 技术 nbsp long play problem isp ber display
原文地址:https://www.cnblogs.com/New-ljx/p/11515343.html