Rank
Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4160 Accepted Submission(s): 1616
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
第一次错是因为 只默认假定有可能两个人成绩相同。所以在排好序的分数中搜索那个孩子分数的时候，出现多次输出名次。其实找到后就应该break；那么不用测试肯定ac了
。
#include<iostream>
#include<algorithm>
using namespace std;
struct person
{
int n;
int b;
}per[1011];
bool cmp(person a,person c)
{
return a.b>c.b;
}
int main()
{
int x,i=0;
while(cin>>x)
{
i=0;
while(cin>>per[i].n>>per[i].b,per[i].n+per[i].b)
{i++;}
sort(per,per+i,cmp);
int M;
for(int j=0;j<i;j++)
{
if(per[j].n==x)
{
M=per[j].b;
for(int k=0;k<i;k++)
if(per[k].b==M)
{
cout<<k+1<<endl;
break;
}
}
}
}
return 0;
}