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CodeForces - 659A Round House (形成环)水

2016-04-02 22:07 483 查看
CodeForces
- 659A

Round House

Time Limit: 1000MS Memory Limit: 262144KB 64bit IO Format: %I64d & %I64u
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Description

Vasya lives in a round building, whose entrances are numbered sequentially by integers from 1 to n.
Entrance n and entrance 1are 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.

Sample Input

Input
6 2 -5


Output
3


Input
5 1 3


Output
4


Input
3 2 7


Output
3

//题意:输入n,a,b;

有n个房子,现在小明在第a个房子出,如果b<0,那么小明逆时针走过|b|个房子,若b>0则顺时针走b个房子,问小明最终在第几个房子处。

//在赋值的时候直接顺序赋值,在输出时就好输出了。一个小技巧

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#include<iostream>
#define INF 0x3f3f3f3f
#define ll long long
#define IN __int64#define N 110
#define M 1000000007
using namespace std;
int s
;
int main()
{
int n,a,b;
int i,j;
while(scanf("%d%d%d",&n,&a,&b)!=EOF)
{
if(b>0)
{
for(i=0;i<n;i++)
{
if(a>n)
a=1;
s[i]=a++;
}
}
else
{
for(i=0;i<n;i++)
{
if(a)
s[i]=a;
else
{
a=n;
s[i]=a;
}
a--;
}
}
int p=abs(b)%n;
printf("%d\n",s[p]);
}
return 0;
}


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