昨天做了一道簡單但很麻煩的題,我只能想到結構體,并用了STL的sort方法解決了它.不過從中有許多細節問題.

題目:

?

Problem Description

Lcy wanted to choose 50 ACMers from m players to join HDU-ACM team.
He made n competitions , and now is your task to make the ranklist.
Here’re some instructions :
Effective score : the sum of the best n-2 competitions’ score.
One’s score in one competition : number of problems he(or she) solved divided the sum of problems all players solved.

The ranklist is made , of course , by the effective score , what if some players have the same score ?Follow these rules : girls always come first , same again and younger first , again ? In lexicographic order.

Input

For each case , two integers in the first line : n (3<= n <= 10)and m(m<=1000)
The next m lines are someone’s information :
name , sex (F，M), grade and n integer numbers which means one’s number of solved problem in the ith competition.

Output

Output the first fifty players in order . Output all if m<50.

Sample Input

3 2 WXL F 08 3 4 5 HH M 08 3 4 5

Sample Output

WXL HH

?

代碼如下:

?

Code
#include<iostream>
#include<algorithm>
#include<string>
using?namespace?std;
bool?dic(char?*a,?char?*b)
{
????int?p;
????p=strcmp(a,b);
????if(p>0)
????????return?false;
????else?return?true;
}
typedef?struct
{
????char?name[21];
????char?sex;
????int?grade;
????float?score;
????int?s[11];
}node;
node?a[1001];

bool?comp(node?a,?node?b)
{
????if(a.score?!=?b.score)
????????return?a.score>b.score;
????else
????{
????????if(a.sex?!=?b.sex)
????????????return?a.sex<b.sex;
????????else?if(a.grade?!=?b.grade)
????????????return?a.grade>b.grade;
????????else
????????????return?dic(a.name,b.name);
????}
}

int?main()
{
????int?n,m,i,j,k;
????float?b[11],p,f[11];
????while(scanf("%d%d",&n,&m)!=EOF)
????{
????????for(i=0;?i<m;?i++)
????????{
????????????getchar();
????????????scanf("%s?%c?%d",&a[i].name,&a[i].sex,&a[i].grade);
????????????for(j=0;?j<n;?j++)
????????????{
????????????????scanf("%d",&a[i].s[j]);
????????????}
????????}

????????for(i=0;?i<n;?i++)
????????{
????????????b[i]=0;
????????????for(j=0;?j<m;?j++)
????????????{
????????????????b[i]+=a[j].s[i];
????????????}
????????}
????????for(i=0;?i<m;?i++)
????????{
????????????a[i].score=0;
????????????for(j=0;?j<n;?j++)
????????????{
????????????????p=(float)a[i].s[j]/b[j];
????????????????f[j]=p;
????????????}
????????????sort(f,f+n);
????????????for(k=n-1;?k>1;?k--)
????????????{
????????????????a[i].score+=f[k];
????????????}
????????}
????????sort(a,a+m,comp);
????????for(i=0;?i<m;?i++)
????????{
????????????cout<<a[i].name<<endl;
????????????if(i==49)break;
????????}
????}
????return?0;
}

轉載于:https://www.cnblogs.com/chenjunyu/archive/2009/05/04/1448920.html

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