排序,结构体排序，大小排序

Problem F

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 3 Accepted Submission(s) : 2

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Problem Description

“Point, point, life of student!” This is a ballad（歌谣）well known in colleges, and you must care about your score in this exam too. How many points can you get? Now, I told you the rules which are used in this course. There are 5 problems in this final exam. And I will give you 100 points if you can solve all 5 problems; of course, it is fairly difficulty for many of you. If you can solve 4 problems, you can also get a high score 95 or 90 (you can get the former(前者) only when your rank is in the first half of all students who solve 4 problems). Analogically（以此类推）, you can get 85、80、75、70、65、60. But you will not pass this exam if you solve nothing problem, and I will mark your score with 50. Note, only 1 student will get the score 95 when 3 students have solved 4 problems. I wish you all can pass the exam! Come on!

Input

Input contains multiple test cases. Each test case contains an integer N (1<=N<=100, the number of students) in a line first, and then N lines follow. Each line contains P (0<=P<=5 number of problems that have been solved) and T（consumed time）. You can assume that all data are different when 0<p. A test case starting with a negative integer terminates the input and this test case should not to be processed.

Output

Output the scores of N students in N lines for each case, and there is a blank line after each case.

Sample Input

Sample Output

100
90
90
95

100

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
int main()
{
    char a[100][9];
    int b[100],c[100],d[6],N,i,j,f;
    while(cin>>N)
    if(N>-1)
    {
        memset(d,0,sizeof(d));
        for(i=0;i<N;i++)
        {cin>>b[i]>>a[i];
        switch(b[i])
        {
        case 5:d[5]++;break;
        case 4:d[4]++;break;
        case 3:d[3]++;break;
        case 2:d[2]++;break;
        case 1:d[1]++;break;
        }
        memset(c,0,sizeof(c));
        }
        for(i=0;i<N;i++)
            for(int j=0;j<N;j++)
                if(strcmp(a[i],a[j])>0&&b[i]==b[j])c[i]++;
        for(i=0;i<N;i++)
        {
        switch(b[i])
        {
            case 5:cout<<100<<endl;break;
            case 4:if(c[i]<d[4]/2)cout<<95<<endl;
                else cout<<90<<endl;break;
            case 3:if(c[i]<d[3]/2)cout<<85<<endl;
                else cout<<80<<endl;break;
            case 2:if(c[i]<d[2]/2)cout<<75<<endl;
                else cout<<70<<endl;break;
            case 1:if(c[i]<d[1]/2)cout<<65<<endl;
                else cout<<60<<endl;break;
            default:cout<<50<<endl;break;
        }
  }
        cout<<endl;
    }
    return 0;
}
我写的

#include <iostream>

#include <stdio.h>

#include <string.h>

#include <algorithm>

using namespace std

; int main()

{

int t,save[6][2]={50,50,60,65,70,75,80,85,90,95,100,100},j,i,h,g,temp[100],a1[100],a[100],ans[1000],k;

char c[100][1000];

while(scanf("%d",&t))

{

if(t<=0)

break;

memset(a1,0,sizeof(a1));

memset(temp,0,sizeof(temp));

for(h=0;h<t;h++)

{

scanf("%d",&a[h]);

j=a[h];

a1[j]++;

cin>>c[h];

}

for(i=0;i<t;i++)

for(g=0;g<t;g++)

if(strcmp(c[i],c[g])>0&&a[i]==a[g])temp[i]++;

for(i=0;i<t;i++)

{

k=a[i];

if(temp[i]<a1[k]/2)

ans[i]=save[k][1];

else

ans[i]=save[k][0];

}

for(i=0;i<t;i++)

cout<<ans[i]<<endl;

cout<<endl;

}

return 0;

}

#include <iostream>

#include <stdio.h>

#include<string.h>

#include <algorithm>

using namespace std;

struct node {

int index; int score; int p,t;

bool operator<(const node&a)const;

};node a[103];

bool node:: operator<(const node&b)const

{ if(b.p==p) return t<b.t; return p>b.p; }

bool cmp1(const node&a,const node&b)

{ return a.index<b.index;

} int hash[6];

int main() {

int i,j,n;

int h,m,s;

while(scanf("%d",&n),n!=-1)

{

memset(hash,0,sizeof(hash));

for(i=0;i<n;i++)

{

scanf("%d %d:%d:%d",&a[i].p,&h,&m,&s);

hash[a[i].p]++;

a[i].t=h*3600+m*60+s;

a[i].index=i;

}

sort(a,a+n);

i=0;

while(a[i].p==5)

a[i++].score=100;

for(j=0;j<hash[4]/2;j++)

a[i++].score=95;

for(;j<hash[4];j++)

a[i++].score=90;

for(j=0;j<hash[3]/2;j++)

a[i++].score=85;

for(;j<hash[3];j++)

a[i++].score=80;

for(j=0;j<hash[2]/2;j++)

a[i++].score=75;

for(;j<hash[2];j++)

a[i++].score=70;

for(j=0;j<hash[1]/2;j++)

a[i++].score=65;

for(;j<hash[1];j++)

a[i++].score=60;

for(;i<n;i++)

a[i].score=50;

sort(a,a+n,cmp1);

for(i=0;i<n;i++)

printf("%d\n",a[i].score);

printf("\n");

}

return 0;

}