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

A. Round House

题目链接http://codeforces.com/contest/659/problem/A

Description

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.

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.

Sample Input

6 2 -5

Sample Output

3

题意：

给你一个环，你一开始在a位置，然后你走b步，求走完后地址。

题解：

先将最环节全部去掉，剩下绝对值在a以内的数。这样就可以通过环的性质求出数据。

代码：

#include<bits/stdc++.h>
using namespace std;
const int maxn = 1123456789;
int s[1000];
int main()
{
    int key = 0;
    int n,a,b;
    cin>>n>>a>>b;
    if (b >= 0){
        if ((a+b)%n == 0)
            cout<<n<<endl;
        else cout<<((a+b)%n)<<endl;;
    }else {
        int sb = b;
        while (sb < 0){
            sb+=n;
        }
        if ((a+sb)%n == 0)
            cout<<n<<endl;
        else cout<<((a+sb)%n)<<endl;;

    }

    return 0;
}