ZOJ 3869-- Ace of Aces
来源:互联网 发布:精华乳和精华液 知乎 编辑:程序博客网 时间:2024/05/19 22:28
Description
There is a mysterious organization called Time-Space Administrative Bureau (TSAB) in the deep universe that we humans have not discovered yet. This year, the TSAB decided to elect an outstanding member from its elite troops. The elected guy will be honored with the title of "Ace of Aces".
After voting, the TSAB received N valid tickets. On each ticket, there is a number Ai denoting the ID of a candidate. The candidate with the most tickets nominated will be elected as the "Ace of Aces". If there are two or more candidates have the same number of nominations, no one will win.
Please write program to help TSAB determine who will be the "Ace of Aces".
Input
There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case:
The first line contains an integer N (1 <= N <= 1000). The next line contains N integers Ai (1 <= Ai <= 1000).
Output
For each test case, output the ID of the candidate who will be honored with "Ace of Aces". If no one win the election, output "Nobody" (without quotes) instead.
Sample Input
352 2 2 1 151 1 2 2 31998
Sample Output
2Nobody998
题意:求出出现次数最多的且是唯一的数,若有多个则输出 Nobody。
代码如下:
#include <iostream>#include <cstring>#include <cstdio>#include <map>#include <functional>#include <algorithm>using namespace std;#define N 1020#define inf 0x3f3f3f3fstruct pp{int num, v;}p[N];int Num[N];int main(){#ifdef OFFLINEfreopen("t.txt", "r", stdin);#endifint t, i, j, k, n, m;scanf("%d", &t);while (t--){memset(p, 0, sizeof(p));memset(Num, 0, sizeof(Num));scanf("%d", &n);m = 0;for (i = 0; i < n; i++){scanf("%d", &k);Num[k]++;}if (n == 1){printf("%d\n", k); continue;}j = 0;for (i = 1; i <= 1000; i++){if (Num[i]){p[j].v = i;p[j++].num = Num[i];}}int max_1 = 0, max_v1 = 0, max_2 = 0, max_v2 = 0;for (i = 0; i<j; i++){if (p[i].num > max_1){//最大max_2 = max_1;max_v2 = max_v1;max_1 = p[i].num;max_v1 = p[i].v;}else if (p[i].num > max_2){//次大max_2 = p[i].num;max_v2 = p[i].v;}}if (max_1 == max_2)puts("Nobody");elseprintf("%d\n", max_v1);}return 0;}
- zoj 3869 Ace of Aces
- ZOJ 3869 Ace of Aces
- ZOJ-3869 Ace of Aces
- ZOJ 3869Ace of Aces
- ZOJ 3869-- Ace of Aces
- zoj-3869-Ace of Aces
- Ace of Aces ZOJ
- ZOJ Problem Set - 3869||Ace of Aces
- ZOJ - 3869 Ace of Aces (水)
- ZOJ 3869-Ace of Aces【模拟众数】
- ZOJ 3869:Ace of Aces【水】
- zoj 3869 Ace of Aces (水题)
- A - Ace of Aces——ZOJ
- ZOJ 3869 Ace of Aces (The 12th Zhejiang Provincial Collegiate Programming Contest)
- Ace of Aces
- ZOJ3869 Ace of Aces
- Ace of Aces
- Ace of Aces
- iOS 摇一摇功能的实现
- java中的进制转换
- HDOJ1712
- 数据结构(树)
- 奇异值分解(SVD) --- 几何意义
- ZOJ 3869-- Ace of Aces
- jdk+adt+eclipse安装
- hduoj1232(并查集)
- mac svn升级 亲测可用
- hadoop2.6.0版本集群环境搭建
- 正则表达式的用法详解(字符判读的利器)
- 文件查找源码(linux环境下)
- discuz论坛不登录状态(游客浏览)模式下,门户页面显示keyword关键词,description描述
- Netty in Action (十九) 第九章节 单元测试