What Is Your Grade?(水，排序)

最新推荐文章于 2025-09-10 22:21:01 发布

转载最新推荐文章于 2025-09-10 22:21:01 发布 · 76 阅读

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原文链接：http://www.cnblogs.com/handsomecui/p/5102432.html

文章标签：

#java #数据结构与算法

本文详细解析了一道关于比赛编程中如何根据解决的问题数量及时间来分配分数的题目。介绍了输入输出格式、评分规则，并提供了完整的C++代码实现。

What Is Your Grade?

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 9935 Accepted Submission(s): 3041

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

题解：虽然水题，但是还是贴一下吧，毕竟刚开始没读清题错了几次，贴下长个记性；刚开始读成相同题目的第一名+5了，最后又读了下，发现是前一半的+5。。。

对了，还有像time，rank了都不能在代码里面命名否则ce；

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
const int INF=0x3f3f3f3f;
#define mem(x,y) memset(x,y,sizeof(x))
#define SI(x) scanf("%d",&x)
#define PI(x) printf("%d",&x)
#define P_ printf(" ")
int ans[1010];
int nu[6];
struct Node{
	int cnt;
	int num;
	int hh,mm,dd;
	int sec;
	Node init(int x){
		cnt=x;
		scanf("%d",&num);
		nu[num]++;
		scanf("%d:%d:%d",&hh,&mm,&dd);
		sec=hh*60*60+mm*60+dd;
		ans[cnt]=50+10*num;
	}
	friend bool operator < (Node a,Node b){
		if(a.num!=b.num)return a.num<b.num;
		else return a.sec<b.sec;
	}
}dt[1010];
int main(){
	int N;
	while(SI(N),N!=-1){
		mem(nu,0);
		for(int i=0;i<N;i++){
			dt[i].init(i);
		}
		sort(dt,dt+N);
		for(int i=1;i<=4;i++){
			for(int j=0;j<N;j++){
				if(dt[j].num==i){
					if(nu[i]==1)ans[dt[j].cnt]+=5;
					else for(int k=j;k<j+nu[i]/2;k++){
						ans[dt[k].cnt]+=5;
					}
					break;
				}
			}
		}
		for(int i=0;i<N;i++){
			printf("%d\n",ans[i]);
		}
		puts("");
	}
	return 0;
}

转载于:https://www.cnblogs.com/handsomecui/p/5102432.html