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hdu 4861 Couple doubi

时间:2017-05-07 11:39:40      阅读:168      评论:0      收藏:0      [点我收藏+]

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Couple doubi

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 607    Accepted Submission(s): 435


Problem Description
DouBiXp has a girlfriend named DouBiNan.One day they felt very boring and decided to play some games. The rule of this game is as following. There are k balls on the desk. Every ball has a value and the value of ith (i=1,2,...,k) ball is 1^i+2^i+...+(p-1)^i (mod p). Number p is a prime number that is chosen by DouBiXp and his girlfriend. And then they take balls in turn and DouBiNan first. After all the balls are token, they compare the sum of values with the other ,and the person who get larger sum will win the game. You should print “YES” if DouBiNan will win the game. Otherwise you should print “NO”.
 

Input
Multiply Test Cases. 
In the first line there are two Integers k and p(1<k,p<2^31).
 

Output
For each line, output an integer, as described above.
 

Sample Input
2 3 20 3
 

Sample Output
YES NO
 

Author
FZU
 

Source
 

题解:水平有点渣,比赛时打表看规律写的。

技术分享

#include <iostream>
#include <cstdio>
using namespace std;

int main()
{
    int k,p;
    while(cin>>k>>p)
    {
        if(p)

        if(k/(p-1)%2==0) cout<<"NO"<<endl;
        else cout<<"YES"<<endl;
    }
    return 0;
}




hdu 4861 Couple doubi

标签:ble   bottom   tle   hose   mode   art   mon   sub   clu   

原文地址:http://www.cnblogs.com/liguangsunls/p/6819914.html

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