Codeforces Round #346 (Div. 2)--A. Round House-优快云博客

本文链接：https://blog.youkuaiyun.com/Inite/article/details/51027139

本文介绍了一个简单的算法问题，即在一座圆形建筑中，从任意入口出发，按指定方向行走特定的入口数量后，最终停留在哪个入口。通过取余运算进行模拟，解决了这一问题，并提供了清晰的代码实现。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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：有 n 个入口围成一个圈，加入你在第 a 个入口，要向编号增大的方向

走 b 个单位(b)或向编号减少的方向走 b 个单位(-b),求最后停在哪个

入口处，给出编号。

思路：直接模拟，求余运算。

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;
#define size 120
#define inf 0x3f3f3f3f
int main()
{
#ifdef OFFLINE
	freopen("t.txt", "r", stdin);
#endif
	int n, m, a, b, k;
	while (~scanf("%d%d%d", &n, &a, &b))
	{
		if (b >= 0)
		{
			k = a + (b%n);
		}
		else
		{
			k = (a - (abs(b) % n) < 0) ? (a - (abs(b) % n) + n) : (a - (abs(b) % n));
		}
		k = (k%n == 0) ? n : k%n;//注意取余！
		printf("%d\n", k);
	}
	return 0;
}