A. Round House 模拟

圆楼漫步算法题解析

最新推荐文章于 2025-07-13 15:14:36 发布

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本文介绍了一道关于计算在圆形建筑中漫步最终位置的趣味算法题。通过给出的输入参数，包括楼层数、起始楼层及步行的相对楼层数，文章详细解释了如何确定走完指定步数后所在的楼层。

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.

直接模拟正向反向走路，可以一步一步模拟，数据很水，暴力直接过；

#include <stdio.h>
#include <algorithm>
int a[1000005] , b[1000005];
int main() {
	int n , m;
	int t = 0, k = 0;
	scanf("%d%d",&n,&m);
	for(int i = 0; i < n; i++) {
		scanf("%d",&a[i]);
	}
	std::sort(a,a+n);
	for(int i = 1; ; i++) {
		if(i == a[k]) {
			k++;
			continue;
		}
		if(m >= i) {
			m -= i;
			b[t++] = i;
		}
		else break;
	}
	printf("%d\n",t);
	for(int i = 0; i < t; i++) {
		printf("%d ",b[i]);
	}
	printf("\n");
	return 0;
}