POJ 1995 Raising Modulo Numbers【快速幂】

Raising Modulo Numbers

Time Limit: 1000MS Memory Limit: 30000KTotal 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, others like using Windows, and some like difficult mathematical games. Latest marketing research shows, that this market segment was so far underestimated and that there is lack of such games. This kind of game was thus included into the KOKODáKH. The rules follow: 

Each player chooses two numbers Ai and Bi and writes them on a slip of paper. Others cannot see the numbers. In a given moment all players show their numbers to the others. The goal is to determine the sum of all expressions Ai^Bi from all players including oneself and determine the remainder after division by a given number M. The winner is the one who first determines the correct result. According to the players' experience it is possible to increase the difficulty by choosing higher numbers. 

You should write a program that calculates the result and is able to find out who won the game. 

Input

The input consists of Z assignments. The number of them is given by the single positive integer Z appearing on the first line of input. Then the assignements follow. Each assignement begins with line containing an integer M (1 <= M <= 45000). The sum will be divided by this number. Next line contains number of players H (1 <= H <= 45000). Next exactly H lines follow. On each line, there are exactly two numbers Ai and Bi separated by space. Both numbers cannot be equal zero at the same time.

Output

For each assingnement there is the only one line of output. On this line, there is a number, the result of expression 

(A₁^B₁+A₂^B₂+ ... +A_H^B_H)mod M.

Sample Input

31642 33 44 55 63612312374859 30293821713 18132

Sample Output

21319513

题目比较难懂，其实就是快速幂取模运算，然后把给出的数求出后，别忘再次取模........

关键是要读懂题意.....

#include<cstdio>long long mod,sum;long long mi(int n,int m)//调用函数...{long long s=1;//保险起见，64位while(m){if(m&1){s=(s*n)%mod;}n=(n*n)%mod;m>>=1;}return s;}int main(){int t,n,i;//freopen("shuju.txt","r",stdin);scanf("%d",&t);while(t--){sum=0;scanf("%lld%d",&mod,&n);for(i=0;i<n;++i){int a,b;scanf("%d%d",&a,&b);sum+=mi(a%mod,b);//运算之前先取模}printf("%lld\n",sum%mod);//结果}return 0;}

//2016年4月15日16:04

记得做过这道题，差点博客写重复了....

时隔这么久，代码写的基本还没有变，只是看起来更专业了，哈哈...

/*http://blog.csdn.net/liuke19950717*/#include<cstdio>#include<cstring>#include<algorithm>using namespace std;typedef long long ll;ll quick_mod(ll n,ll m,ll mod){ll ans=1;while(m){if(m&1){ans=ans*n%mod;}n=n*n%mod;m>>=1;}return ans;}int main(){int t;scanf("%d",&t);while(t--){ll n,ans=0,mod;scanf("%lld%lld",&mod,&n);for(ll i=0;i<n;++i){ll a,b;scanf("%lld%lld",&a,&b);ans+=quick_mod(a%mod,b,mod);}printf("%lld\n",ans%mod);}return 0;}