1200. Stick 匹配相同长度筷子_输出无法配对的一根筷子的长度-优快云博客

Anthony收集了奇数根棍子用于制作筷子。通过每次选取两根等长棍子进行加工，最终总是剩下一根棍子。此篇博客将介绍如何通过输入棍子长度集合，计算并输出剩余棍子的长度。

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1200. Stick

Constraints

Time Limit: 1 secs, Memory Limit: 32 MB

Description

Anthony has collected a large amount of sticks for manufacturing chopsticks. In order to simplify his job, he wants to fetch two equal-length sticks for machining at a time. After checking it over, Anthony finds that it is always possible that only one stick is left at last, because of the odd number of sticks and some other unknown reasons. For example, Anthony may have three sticks with length 1, 2, and 1 respectively. He fetches the first and the third for machining, and leaves the second one at last. You task is to report the length of the last stick.

Input

The input file will consist of several cases.

Each case will be presented by an integer n (1<=n<=100, and n is odd) at first. Following that, n positive integers will be given, one in a line. These numbers indicate the length of the sticks collected by Anthony.

The input is ended by n=0.

Output

For each case, output an integer in a line, which is the length of the last stick.

Sample Input

Sample Output

Problem Source

ZSUACM Team Member

#include <iostream>

using namespace std;

int main () {
	int n;
	int a[105] = {0};
	while (cin>>n && n) {
		int b;
		for (int i = 0; i < n; i++) {
			cin>>b;
			bool flag = false;
			for (int j = 0; j <= i; j++) {  //从头查找有没有跟新输入匹配的 如果有就把原来的设为0
				if (a[j] == b) {
					a[j] = 0;
					flag = true;
					break;
				}
			}
			if (flag == false)
				a[i] = b;
		}
		for (int i = 0; i < n; i++) { //把没匹配的输出
			if ( a[i]!= 0 )
				cout<<a[i]<<endl;
		}
	}
	//system("pause");
	return 0;
}