zoukankan      html  css  js  c++  java
  • poj 1995 Raising Modulo Numbers【快速幂】

    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, 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 AiBi 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 

    (A1B1+A2B2+ ... +AHBH)mod M.

    Sample Input

    3
    16
    4
    2 3
    3 4
    4 5
    5 6
    36123
    1
    2374859 3029382
    17
    1
    3 18132
    

    Sample Output

    2
    13195
    13

    #include<stdio.h>
    #include<string.h>
    #include<algorithm>
    #include<math.h>
    #define DD double
    #define LL long long
    #define INF 0x3f3f3f
    #define MAX 1010
    using namespace std;
    int fun(int a,int b,int c)
    {
        int ans=1;
        a=a%c;
        while(b)
        {
            if(b&1)
                ans=(a*ans)%c;
            b=b/2;
            a=(a*a)%c;
        }
        return ans;
    }
    int main()
    {
        int t,i,j;
        int n,m,sum;
        int x,y;
        scanf("%d",&t);
        while(t--)
        {
            scanf("%d%d",&m,&n);
            sum=0;
            for(i=1;i<=n;i++)
            {
                scanf("%d%d",&x,&y);        
                sum=fun(x,y,m)+sum;
            }
            printf("%d
    ",sum%m);
        }
        return 0;
    }
    

      

  • 相关阅读:
    spring 整合 shiro框架
    Kafka常见问题及解决方法
    设计模式之解释器模式规则你来定(二十五)
    设计模式之原型模式简单即复杂(二十四)
    设计模式之访问者模式层次操作(二十三)
    设计模式之状态模式IFORNOIF(二十二)
    设计模式之职责链模式永不罢休(二十一)
    设计模式之组合模式透明实用(二十)
    设计模式之享元模式高效复用(十九)
    设计模式之迭代器模式解析学习源码(十八)
  • 原文地址:https://www.cnblogs.com/tonghao/p/4750224.html
Copyright © 2011-2022 走看看