Codeforces Round #346 (Div. 2) A. Round House

A. Round House
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

Vasya lives in a round building, whose entrances are numbered sequentially by integers from 1 to n. Entrance n and entrance 1 are adjacent.

Today Vasya got bored and decided to take a walk in the yard. Vasya lives in entrance a and he decided that during his walk he will move around the house b entrances in the direction of increasing numbers (in this order entrance n should be followed by entrance 1). The negative value of b corresponds to moving |b| entrances in the order of decreasing numbers (in this order entrance 1 is followed by entrance n). If b = 0, then Vasya prefers to walk beside his entrance.

Codeforces Round #346 (Div. 2) A. Round House_第1张图片 Illustration for n = 6a = 2b =  - 5.

Help Vasya to determine the number of the entrance, near which he will be at the end of his walk.

Input

The single line of the input contains three space-separated integers na and b (1 ≤ n ≤ 100, 1 ≤ a ≤ n,  - 100 ≤ b ≤ 100) — the number of entrances at Vasya's place, the number of his entrance and the length of his walk, respectively.

Output

Print a single integer k (1 ≤ k ≤ n) — the number of the entrance where Vasya will be at the end of his walk.

Examples
input
6 2 -5
output
3
input
5 1 3
output
4
input
3 2 7
output
3
Note

The first example is illustrated by the picture in the statements.

比较简单,就是选定一个点,问顺时针或逆时针走几步后,到哪个点?

对于大于总点数的进行取模,对于小于等于0的,把答案转到正数上;

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int n,a,b;
    scanf("%d %d %d",&n,&a,&b);
    int sum=a+b;
    if(sum<=0)
    {
        while(sum<=0)
        {
            sum+=n;
        }
    }
    else
    {
        sum=sum%n;
        if(sum==0)
            sum=n;
    }
    printf("%d\n",sum);
    return 0;
}


你可能感兴趣的:(codeforces)