题意: 思路: 对于每个幂次方,将幂指数的二进制形式表示,从右到左移位,每次底数自乘,循环内每步取模. #include <cstdio> typedef long long LL; LL Ksm(LL a, LL b, LL p) { LL ans = 1; while(b) { if(b & 1) { ans = (ans * a) % p; } a = (a * a) % p; b >>= 1; } return ans; } int main() { LL p, a…
Raising Modulo Numbers Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 5532 Accepted: 3210 Description People are different. Some secretly read magazines full of interesting girls' pictures, others create an A-bomb in their cellar, oth…
Raising Modulo Numbers Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 5477 Accepted: 3173 Description People are different. Some secretly read magazines full of interesting girls' pictures, others create an A-bomb in their cellar, oth…
Raising Modulo Numbers Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 6347 Accepted: 3740 Description People are different. Some secretly read magazines full of interesting girls' pictures, others create an A-bomb in their cellar, oth…
题目链接:http://poj.org/problem?id=1995 解题思路:用整数快速幂算法算出每一个 Ai^Bi,然后依次相加取模即可. #include<stdio.h> long long quick_mod(long long a,long long b,long long c) { long long ans=1; while(b) { if(b&1) { ans=ans*a%c; } b>>=1; a=a*a%c; } return ans; } int…
快速幂取模 #include<cstdio> int mod_exp(int a, int b, int c) { int res, t; res = % c; t = a % c; while (b) { ) res = res * t % c; t = t * t % c; b >>= ; } return res; } int main() { int T; scanf("%d",&T); while(T--) { int m,h; scanf(&…
ZOJ2150 快速幂,但是用递归式的好像会栈溢出. #include<cstdio> #include<cstdlib> #include<iostream> #include<cmath> using namespace std; long long M,i; #define LL long long int _work(LL a,LL n) { LL ans=1; while(n){ if(n&1){ ans=(ans*a)%M; n--; }…
