杭电1718--Rank

Rank

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4411    Accepted Submission(s): 1707

Problem Description

Jackson wants to know his rank in the class. The professor has posted a list of student numbers and marks. Compute Jackson’s rank in class; that is, if he has the top mark(or is tied for the top mark) his rank is 1; if he has the second best mark(or is tied) his rank is 2, and so on.

 

 

Input

The input consist of several test cases. Each case begins with the student number of Jackson, an integer between 10000000 and 99999999. Following the student number are several lines, each containing a student number between 10000000 and 99999999 and a mark between 0 and 100. A line with a student number and mark of 0 terminates each test case. There are no more than 1000 students in the class, and each has a unique student number.

 

 

Output

For each test case, output a line giving Jackson’s rank in the class.

 

 

Sample Input

20070101

20070102 100

20070101 33

20070103 22

20070106 33

0 0

 

 

Sample Output

2

 

 

Source

2007省赛集训队练习赛（2）

 

 

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RE:  每个同学有学号、 根据学号找排名。 先把成绩单排一下， 整一下排名， 简单AC.

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm> 
using namespace std;

struct student
{
    int hao, mark, pre;
} num[1010];

bool cmp(student hao, student mark)
{
    return hao.mark > mark.mark; 
}
int main()
{
    int a;
    while(~scanf("%d", &a))
    {
        int k = 0;
        for(; ;)
        {
                
            int b, c;
            scanf("%d %d", &b, &c);
            if(b == 0 && c == 0)
                break;
            else{
                num[k].hao = b; num[k++].mark = c;
            }
        }
        sort(num, num + k, cmp);
        int Q = 1;                        //排名会有并列的。
        num[0].pre = Q; int temp = Q;
        for(int i = 1; i < k; i++)
        {
            if(num[i].mark == num[i-1].mark)
            {
                num[i].pre = temp;
                ++Q;
            }
            else
            {
                ++Q;
                num[i].pre = Q;
                temp = Q;
            }
            //++Q;
        }
        int i;
        for(i = 0; i < k; i++)
            if(num[i].hao == a)
                break;
        printf("%d
", num[i].pre);
    }
    return 0;
}