Codeforces 659A Round House 【小数学】_659a0296de68368b-优快云博客

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本文深入解析了Codeforces平台上的RoundHouse竞赛题，详细解释了如何通过给定的入口数量、居住入口及行走方向计算最终停留的入口。通过实例演示，帮助读者理解并解决类似路径规划问题。

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A. Round House
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
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.

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：给定一个有n个点的环，知道了起点s和走的步数b，若b为正逆时针走，反之顺时针走。

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <map>
#include <stack>
#define PI acos(-1.0)
#define CLR(a, b) memset(a, (b), sizeof(a))
#define fi first
#define se second
#define ll o<<1
#define rr o<<1|1
using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
const int MAXN = 1e3 + 10;
const int pN = 1e6;// <= 10^7
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
void add(LL &x, LL y) { x += y; x %= MOD; }
int main()
{
    int n, a, b; cin >> n >> a >> b;
    int ans;
    if(b > 0) {
        ans = (b % n + a) % n;
    }
    else {
        ans = ((a - abs(b) % n) % n + n) % n;
    }
    if(ans == 0) ans = n;
    cout << ans << endl;
    return 0;
}