sicily 猴子选大王

题目描述

猴子选大王，有N只猴子，从1～N进行编号。它们按照编号的顺时针方向，排成一个圆圈，然后从第一只猴子开始报数。第一只猴子报1，以后每只猴子报的数字都是它前面猴子所报数字加1。如果一只猴子报的数字是M，则该猴子出列，下一只猴子重新从1开始报数。剩下的猴子继续排成一个圆圈报数，直到全部的猴子都出列为止。最后一个出列的猴子胜出。

输入格式

The first line is an integer t, indicating the number of test cases. Then there are t lines and each line contains two positive integer N(0<N<=100) and M(0<M<=100).

输出格式

For each test case, print out the number of the Monkey King.

样例输入

aaarticlea/jpeg;base64,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" alt="" /> 将样例输入复制到剪贴板

2
5 2
4 3

样例输出

3
1
--------------------------我不是分割线--------------------------------------------------------------

分析：开始时并没有考虑太多，就选择了用struct的数组来简单实现，不过时间上耗费略大

#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;

struct data
{
    int count;
    bool is;
};

int cmp(data a[],int n){
    int count1=,out;
    for (int i = ; i < n; ++i)
    {
        if (a[i].is==)
            {
                ++count1;
                out=a[i].count;
            }
    }
    if (count1==)
    {
        return out;
    }
    else{
        return ;
    }
}
int main(int argc, char const *argv[])
{
    int t;
    cin>>t;
    for (int i = ; i < t; ++i){
        int m,n;
        cin>>m>>n;
        data monkey[m];
        for (int i = ; i <m; ++i)
        {
            monkey[i].count=i+;
            monkey[i].is=;
        }
        int ji=;int count2=;int i=;
        int isf=cmp(monkey,m);
        while(!isf){
                if (monkey[i].is==)
                {
                    ++count2;
                    if (count2==n)
                    {
                        monkey[i].is=;
                        count2=;
                    }
                }
                ++i;
                if(i==m){
                    i=;/*实现循环，保证由count2来决定循环撤出的有效性*/
                }
                isf=cmp(monkey,m);
        }
        cout<<isf<<endl;
    }
    return ;
}