傳送門
A. Round House
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard 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.

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.

題目大意：
有n個數從小到大排列，需要注意的首尾相接 也就是一個環，然後有一個人站在a處，他需要走b步，看能夠到達什麼地方，如果b是<0的那麼就是倒著走，否則就是正著走（語言表達能力太差。。。）

解題思路：
其實這個題目我沒就是分析一下就行，注意分情況討論一下，假設b>0，那麼我們只需要(a+b)%n取餘就行了，注意取餘之後如果是0的話就輸出n，當b<0的時候，我們首先判斷一下他能不能到達1，然後如果到達了1在走的話就是

然後就可以寫程式碼了：
上程式碼：

#include <iostream>

using namespace std;

int main()
{
    int n, a, b;
    while(cin>>n>>a>>b)
    {
        if(b == 0)
            cout<<a<<endl;
        else if(b < 0)
        {
            if(a+b >= 1)
            {
                cout<<(a+b)<<endl;
            }
            else
            {
                cout<<n+((a+b)%n)<<endl;
            }

        }
        else
        {
            if((a+b)%n)
                cout<<(a+b)%n<<endl;
            else
                cout<<n<<endl;
        }
    }
    return 0;
}