hdu 5158(水题)
Have meal
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 756 Accepted Submission(s): 507
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?
[Technical Specification]
1<=n, m<=100
1 100
0
#include<stdio.h>
#include<iostream>
#include<string.h>
#include <stdlib.h>
#include<math.h>
#include<algorithm>
#include <queue>
using namespace std;
typedef long long LL; int main()
{
int n,m;
while(scanf("%d%d",&n,&m)!=EOF){
int t = m%n;
if(t==) printf("%d\n",n-);
else printf("%d\n",t-);
}
return ;
}
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