POJ1995 Raising Modulo Numbers 计算(A1B1+A2B2+ ... +AHBH)mod M. 快速幂,套模板 /* * Created: 2016年03月30日 23时01分45秒 星期三 * Author: Akrusher * */ #include <cstdio> #include <cstdlib> #include <cstring> #include <cmath> #include <ctime> #…

题目:http://poj.org/problem?id=1995 题目解析:求(A1B1+A2B2+ ... +AHBH)mod M. 大水题. #include <iostream> #include <stdio.h> #include <string.h> #include <algorithm> #include <math.h> using namespace std; int n,mod,sum; int main() { ],b[…

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…

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--; }…

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: 5500   Accepted: 3185 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: 6373   Accepted: 3760 Description People are different. Some secretly read magazines full of interesting girls' pictures, others create an A-bomb in their cellar, oth…

Description People are different. Some secretly read magazines full of interesting girls' pictures, others create an A-bomb in their cellar, others like using Windows, and some like difficult mathematical games. Latest marketing research shows, that…

题目链接: https://vjudge.net/problem/POJ-1995 题目大意: 求一堆ab的和模上m 思路: 直接上模板 #include<iostream> #include<vector> #include<queue> #include<algorithm> #include<cstring> #include<cstdio> #include<set> #include<cmath> u…

#include <cstdio> typedef long long ll; int quick_pow(ll a,ll b,ll mod){ ll ans=; ))ans=(ans*a)%mod; return ans; } int main(){ int z,m,h,a,b,ans; for(scanf("%d",&z);z--;){ scanf(; while(h--)scanf("%d%d",&a,&b),ans=(an…