杭電 HDU ACM 1718 Rank



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 2007省賽集訓隊練習賽（2） 第一次錯是因為 只默認假定有可能兩個人成績相同。所以在排好序的分數中搜索那個孩子分數的時候，出現多次輸出名次。其實找到後就應該break；那麼不用測試肯定ac了 。

#include
#include
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