Kagome Kagome（简单）

3492 - Kagome Kagome

时间限制:	Java: 2000 ms / Others: 2000 ms
内存限制:	Java: 65536 KB / Others: 65536 KB

问题描述

Kagome kagome, kago no naka no tori wa Itsu itsu deyaru? Yoake no ban ni Tsuru to kame to subetta. Ushiro no shoumen daare? Translation: Kagome kagome, the bird in the cage, when will you come out? In the evening of the dawn, the crane and turtle slipped. Who stands right behind you now?

Kagome Kagome is a Japanese children's game. One child is chosen as the oni (literally demon or ogre, but similar to the concept of "it" in tag) and sits blindfolded (or with their eyes covered). The other children join hands and walk in circles around the oni while singing the song for the game. When the song stops, the oni speaks aloud the name of the person behind him, and if he is correct, the person behind will exchange places with the oni.

Higurashi Tewi is playing Kagome Kagome with her n (n is even) friends as the oni now. She peeps to know who is right in front of her. Knowing the order of the children in circle and assuming that they keep distance evenly, it's easy to derive who is right behind her.

输入说明

There are multiple test cases. The first line of input is an integer T ≈ 100 indicating the number of test cases.

The first line of each test case starts with an even number 1 ≤ n ≤ 100, followed by the name of the child who is right in front of Higurashi Tewi. The second line contains exactly n different names, listed in counterclockwise order. Name is an alphanumeric string whose length never exceeds 20. It's guaranteed that the child in front of Higurashi Tewi is always contained in the list exactly once.

输出说明

For each test case, output the name of the child who is right behind Higurashi Tewi.

输入样例

3
2 Alice
Alice Bob
4 inu
inu neko usagi kizune
4 cat
dog cat rabbit fox

输出样例

Bob
usagi
fox


References

	http://en.wikipedia.org/wiki/Kagome_Kagome

来源

The 8th Zhejiang Provincial Collegiate Programming Contest

 

#include <iostream>
#include <string>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <map>
#include <vector>
#include <set>
#include <queue>
#include <stack>
#define LL long long
#define MAXI 2147483647
#define MAXL 9223372036854775807
#define eps (1e-8)
#define dg(i) cout << "*" << i << endl;
  
using namespace std;
  
int main()
{
    int t, n, pos;
    string fro;
    string name[105];
    cin >> t;
    while(t--)
    {
        cin >> n >> fro;
        for(int i = 0; i < n; i++)
        {
            cin >> name[i];
            if(name[i] == fro) pos = i;
        }
        cout << name[(pos + n / 2) % n] << endl;  //注意%n
    }
    return 0;
}