codeforces 659A Round House

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本文介绍了一个简单的算法问题，主人公瓦夏居住在一个圆形建筑中，从任意入口出发，按指定方向行走特定数量的入口后，确定最终停留的位置。文章通过一个C语言程序解决了这一问题。

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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.

$\text{[math]}$ Illustration for n = 6, a = 2, b = - 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 n, a 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.

题意：走路，可正可反，问最后的位置。

#include<stdio.h>
int main()
{
    int n,a,b;
    scanf("%d%d%d",&n,&a,&b);
    if(n==1)
    {
        printf("%d\n",a);
        return 0;
    }
    if(b>=0)
    {
        a+=b%n;
        if(a>n)
            a=a%n;
        printf("%d\n",a);
    }
    else
    {
        a=a+b;
        while(a<=0)
            a+=n;
        printf("%d\n",a);
    }
    return 0;
}