HDU 5158 Have meal

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5158

Have meal

Problem Description

I have been in school for several years, so I have visited all messes here. Now I have lost intersts in all of the foods. So when during the meal time, I don’t know which mess I should go to. So I came up with a solution. 
There are 4 messes in our school, I number them from 0 to 3. Then I says “Big Bing Small Jiang, Point Who Is Who!”, when I say the first word I point to the mess which is numbered 0, when I say the i-th (i>1) word I point to the mess whose number is one larger than the previous one. In case of the number of previous mess is 3, I will point to 0 again. I will go to the mess which I point to last time. Thus in this case I will go to the mess which is numbered 3. The following table explains the course of my solution to this case.
 Word I say Mess id I point to Big 0 Bing 1 Small 2 Jiang 3 Point 0 Who 1 Is 2 Who 3
I will go to university after several days, I have heard that there are so many messes in it. So I will apply my solution again. Surpose there are n messes which are numberd through 0 to n-1, and I will say m words. When I say the first word I point to the mess which is numbered 0, when I say the i-th (i>1) word I point to the mess whose number is one larger than the previous one. In case of the number of previous mess is n-1, I will point to 0 again. I will go to the mess which I point to last time. So which mess will I point to?.
It is so time-consuming to count it through manual work. So I want you to write a program to help me. Would you help me?

 

Input

Multi test cases (about 10000), every case contain two integers n and m in a single line.

[Technical Specification]
1<=n, m<=100

 

Output

For each case, output the number of the mess which I should go to.

 

Sample Input

  

   4 3
1 100
  

 

Sample Output

  

   2
0
  
  


  

题意：

     题意很简单，n个数字，标记为0，1，2，……n-1，组成一个环，为0，1，2，3，……，n-1，0，1，2,……,  

m次报数，问第m次报到哪一个

思路：

           如果n>=m，那么报到的就是m-1

           如果 n<m，算出循环的轮数，减去循环节 x=m-(m/n)*n ,如果刚刚循环结束，那么报到的就是n-1,否则就是 x-1

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
int main()
{
    int n,m,i;
    while(~scanf("%d%d",&n,&m))
    {
        if(n>=m)
        {
            printf("%d\n",m-1);continue;
        }
        if(n==1)
        {
            printf("0\n");continue;
        }
        int x=m/n;
        x=m-x*n;
        if(x==0)
        {
            printf("%d\n",n-1);continue;
        }
        /*
        int k=1;
        for(i=0;i<n;i++)
        {
            if(k==x)
            {
                 printf("%d\n",i);break;
            }
            k++;
        }
        */
        printf("%d\n",x-1);
    }
    return 0;
}