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 bcorresponds 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.
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.
Print a single integer k (1 ≤ k ≤ n) — the number of the entrance where Vasya will be at the end of his walk.
6 2 -5
3
5 1 3
4
3 2 7
3
题意：1~n个数围成一个圆形，a为起点，b为将要走的步数，b>0则顺时针走b步，b<0逆时针走b步问最后终点是哪个点
#include<stdio.h>
#include<string.h>
#include<vector>
#include<map>
#include<queue>
#include<stack>
#include<cstdio>
#include<string>
#include<math.h>
#include<algorithm>
#define LL long long
#define PI atan(1.0)*4
#define II __int64
#define DD double
#define MAX 10010
#define mod 10003
#define INF 0x3f3f3f3f
using namespace std;
int main()
{
int n,m,a,b,i,j;
while(scanf("%d%d%d",&n,&a,&b)!=EOF)
{
// printf("%d
",-5%3);
int sum=0;
if(b==0) sum=a;
else if(b<0)
{
b=b%n;
if(a+b<=0)
sum=n+a+b;
else
sum=a+b;
}
else if(b>0)
{
b=b%n;
if(a+b<=n)
sum=a+b;
else
sum=a+b-n;
}
printf("%d
",sum);
}
return 0;
}