题目：
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Problem Description “Point, point, life of student!” Come on! |
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Input Input contains multiple test cases. Each test case |
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Output Output the scores of N students in N lines for each |
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Sample Input 4 5 06:30:17 4 07:31:27 4 08:12:12 4 05:23:13 1 5 06:30:17 -1 |
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Sample Output 100 90 90 95 100 |
比较简单的一道题目……首先根据耗时和解决题目的数量进行排序，然后根据题目要求进行分数换算，这里面if语句的顺序弄反了让我提交了好几次才ac...
最后再根据输入位置进行排序输出...
写的有点麻烦
#include "stdafx.h"
#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
using namespace std;
struct Student
{
int p;
string s;
int grade;
int order;
};
bool cmp(const Student&a,const Student&b)
{
if(a.p!=b.p)
return a.p>b.p;
else
return a.s<b.s;
}
bool cmp2(const Student&a,const Student&b)
{
return a.order<b.order;
}
int main()
{
int n;
while(scanf_s("%d",&n)&&n>0)
{
vector<Student> t;
Student temp;
int k=n;
temp.p=100;
t.push_back(temp);
int j=0;
int count[6]={0};
while(n--)
{
cin>>temp.p>>temp.s;
temp.order=j++;
t.push_back(temp);
}
sort(t.begin()+1,t.end(),cmp);
//temp.p=100;
//t.push_back(temp);
for (int i = 1; i < k+1; i++)
{
count[t[i].p]++;
}
for (int i = 0; i < 6; i++)
{
count[i]==1?count[i]=-5:count[i]/=2;
}
for (int i = 1; i < k+1; i++)
{
if(count[t[i].p]==-5)
t[i].grade=50+t[i].p*10+5;
if (count[t[i].p]==0)
{
t[i].grade=50+t[i].p*10;
}
if(count[t[i].p]>0)
{
t[i].grade=50+t[i].p*10+5;
count[t[i].p]--;
}
if(t[i].p==5||t[i].p==0)
t[i].grade=50+t[i].p*10;
}
sort(t.begin()+1,t.end(),cmp2);
for (int i = 1; i < k+1; i++)
{
cout<<t[i].grade<<endl;
}
cout<<endl;
}
return 0;
}