【CUGBACM15级BC第26场 A】hdu 5158 Have meal

Have meal

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 951    Accepted Submission(s): 635

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 31 100

 

Sample Output

20

 

#include<iostream>using namespace std;int main(){    int n, m;    while (cin >> n >> m)    {        int x = m % n;        if (x == 0)        {            cout << n - 1 << endl;        }        else        {            cout << x - 1 << endl;        }    }}