HDU 5873-Football Games（分情况/Landau定理）

Football Games

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 729    Accepted Submission(s): 275

Problem Description

A mysterious country will hold a football world championships---Abnormal Cup, attracting football teams and fans from all around the world. This country is so mysterious that none of the information of the games will be open to the public till the end of all the matches. And finally only the score of each team will be announced. 
  
  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

Multiple test cases, process till end of the 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

For each test case, output M lines. Output ``F" (without quotes) if the scores in the i-th group must be false, output ``T" (without quotes) otherwise. See samples for detail.

 

Sample Input

23 0 5 12 1 1

 

Sample Output

FT

 

Source

2016 ACM/ICPC Asia Regional Dalian Online

 

Recommend

wange2014   |   We have carefully selected several similar problems for you:  5877 5876 5875 5874 5873 

 

题目意思：

有N支足球队，每两支队伍都要比一场，赢了得2分，输了得0分，平局双方各得1分。

解题思路：

①分情况考虑，把不合法的全部都筛掉（不过测评数据太水，自己写的测试数据都木有过居然还能AC…o(╯□╰)o）//奇数分数必须是偶数个
//最大分数是2*(n-1) 
//所有分数之和是偶数
/所有分数之和是(n-1)*n
//零分只能有一个
//最大值分数只能有一个

②如果没有平手选项, 赢得加一分的话, 可以用Landau's Theorem（兰道定理）判定。 

这题稍微修改下这个定理就好了。令s1,s2,...,sns_1,s_2,...,s_ns​1​​,s​2​​,...,s​n​​是他们的得分序列, 从小到大排个序，

使得s1≤s2≤...≤sns_1 \le s_2 \le ... \le s_ns​1​​≤s​2​​≤...≤s​n​​, 那么这个序列合法, 当且仅当:

1. s1+s2+...+si≥i(i−1)s_1+s_2+...+s_i \ge i(i-1)s​1​​+s​2​​+...+s​i​​≥i(i−1), 对于所有$1\le i\le n-1$
2. s1+s2+...+sn=n(n−1)s_1+s_2+...+s_n=n(n-1)s​1​​+s​2​​+...+s​n​​=n(n−1).

#include <iostream>#include <cstring>#include <string>#include <vector>#include <queue>#include <cstdio>#include <set>#include <cmath>#include <map>#include <algorithm>#define INF 0x3f3f3f3f#define MAXN 20010using namespace std;int a[MAXN];int main(){    int t;    while(scanf("%d",&t)!=EOF)    {        while(t--)        {            memset(a,0,sizeof(0));            int n,i,j=0,sum=0,l=0,summax=0,Max=-1;            scanf("%d",&n);            for(i=0; i<n; ++i)            {                scanf("%d",&a[i]);                sum+=a[i];                if(a[i]%2) ++j;                if(a[i]==0) ++l;                if(a[i]==2*(n-1)) ++summax;                Max=max(Max,a[i]);            }            if(j%2!=0) puts("F");//奇数分数必须是偶数个            else if(Max>2*(n-1))puts("F");//最大分数是2*(n-1)            else if(sum%2!=0)puts("F");//所有分数之和是偶数            else if(sum!=(n-1)*n)puts("F");//所有分数之和是(n-1)*n            else if(l>1)puts("F");//零分只能有一个            else if(summax>1)puts("F");//最大值分数只能有一个            else puts("T");        }    }    return 0;}/**993 0 5 12 1 16 10 1 1 1 8 96 10 3 4 5 8 0**/

#include <iostream>#include <cstring>#include <string>#include <vector>#include <queue>#include <cstdio>#include <set>#include <cmath>#include <map>#include <algorithm>#define INF 0x3f3f3f3f#define MAXN 20010using namespace std;int a[MAXN];int main(){    int t;    while(scanf("%d",&t)!=EOF)    {        while(t--)        {            memset(a,0,sizeof(0));            int n,i,sum=0;            bool flag=true;            scanf("%d",&n);            for(i=0; i<n; ++i)            {                scanf("%d",&a[i]);                sum+=a[i];            }            if(sum!=n*(n-1))//判断总分是否合法            {                flag=false;                puts("F");                continue;            }            sort(a,a+n);            sum=0;            for(i=0; i<n-1; ++i)//排序后分段求和判断            {                sum+=a[i];                if(sum>=(i+1)*i) continue;//注意这里的下标                else                {                    flag=false;                    puts("F");                    break;                }            }            if(flag) puts("T");        }    }    return 0;}/**993 0 5 12 1 16 10 1 1 1 8 96 10 3 4 5 8 0**/