HDU5873
Football Games
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 514 Accepted Submission(s): 188
Problem Description
At the first phase of the championships, teams are divided into M groups using the single round robin rule where one and only one game will be played between each pair of teams within each group. The winner of a game scores 2 points, the loser scores 0, when the game is tied both score 1 point. The schedule of these games are unknown, only the scores of each team in each group are available.
When those games finished, some insider revealed that there were some false scores in some groups. This has aroused great concern among the pubic, so the the Association of Credit Management (ACM) asks you to judge which groups' scores must be false.
Input
For each case, the first line contains a positive integers M, which is the number of groups.
The i-th of the next M lines begins with a positive integer Bi representing the number of teams in the i-th group, followed by Bi nonnegative integers representing the score of each team in this group.
number of test cases <= 10
M<= 100
B[i]<= 20000
score of each team <= 20000
Output
Sample Input
Sample Output
//2016.9.11
#include <iostream>
#include <cstdio>
#define N 20005 using namespace std; int b[N]; int C(int m, int n)
{
if(n-m < m)m = n-m;
int ans = ;
for(int i = ; i < m; i++)
{
ans *= n;
n--;
}
for(int i = ; i <= m; i++)
ans /= i;
return ans;
} int main()
{
int m, n;
while(scanf("%d", &m)!=EOF)
{
while(m--)
{
scanf("%d", &n);
int sum = , max = ;
for(int i = ; i < n; i++)
{
scanf("%d", &b[i]);
sum += b[i];
if(max<b[i])max = b[i];
}
int c = C(, n);
if(sum==*c && max<=*(n-))
printf("T\n");
else printf("F\n");
}
} return ;
}
HDU5873的更多相关文章
- HDU5873:Football Games
题目链接: Football Games 分析: 先将分数排序,然后 设当前队编号为p,设个指针为p+1,然后p>1,每次p-=2,指针右移一位p==1,指针指向的队-=1p==0,从指针开始到 ...
随机推荐
- The Willpower Instinct
https://book.douban.com/subject/7043452/ 1.冥想2.健康饮食(低GI.素食为主,未加工食物为主).低GI食物使血糖稳定(蛋白.麦片.粗纤谷类.豆类.水果蔬菜) ...
- css3常用样式集锦
控制线显示0.5px .line:after{ content:""; display:block; position:absolute; width:200%; left:0; ...
- N皇后问题——递归求解
比较简单,废话不说,上代码: public class NQueen { //比如:position[1]=3,表示第一行的第三列有一个皇后 private int [] position; //总的 ...
- 推荐几个web中常用js图表插件
作者:zccst 我自己用过fusioncharts和highchart. jQuery插件有: TufteGraph flot js charts jqchart elycharts jquery ...
- 128階數的Shunt音量控制器
源:128階數的Shunt音量控制器 紅外線遙控 - 256階Shunt音量及控制及音源 選擇器
- FZU 1064 教授的测试
递归构造答案. 根据当前整颗树的编号,可以计算左右子树有几个节点以及编号.因此,不断dfs下去就可以了. #include<cstdio> #include<cstring> ...
- IOS开发之按钮控件Button详解
reference:http://mxcvns.lofter.com/post/1d23b1a3_685d59d 首先是继承问题,UIButton继承于UIControl,而UIControl继承于U ...
- Jmeter 新手
转载:http://www.cnblogs.com/TankXiao/p/4059378.html 什么是压力测试 顾名思义:压力测试,就是 被测试的系统,在一定的访问压力下,看程序运行是否稳定/服 ...
- mySql 分段查询
准备: 创建一个成绩表 Create table grade (id integer, score integer); 插入数据(只有id每次加一,score是1到100的随机数,java生成): p ...
- 开始Java学习(Java之负基础实战)
开发平台: JavaSE:java标准平台,一般用于桌面程序开发 JavaEE:开发web(如网站+Sping) JavaME:开发移动应用 开发环境: JVM:跨平台核心. JRE:java运行时, ...