Online JudgeProblem SetAuthorsOnline ContestsUser
Web Board
Home Page
F.A.Qs
Statistical Charts
Problems
Submit Problem
Online Status
Prob.ID:
Register
Update your info
Authors ranklist
Current Contest
Past Contests
Scheduled Contests
Award Contest
User ID:
Password:
  Register
北京大学《ACM-ICPC竞赛训练》暑期课面向全球招生。容量有限,报名从速!
Language:
Unhappy Jinjin
Time Limit: 1000MSMemory Limit: 65536K
Total Submissions: 11486Accepted: 8409

Description

Jinjin is a junior school student. Besides the classes in school, Jinjin's mother also arranges some supplementary classes for her. However, if Jinjin studies for more than eight hours a day, she will be unhappy on that day. On any day she gets unhappy, the more time she studies, the unhappier she will be. Now we got Jinjin's class schedule for the next several days and your task is to find out whether she will be unhappy on these days; if she will be unhappy, on which day she will be the unhappiest.

Input

There may be several test cases. In the first line of each test case, there is an integer N (1 <= N <= 7), which represents the number of days you should analyze. Then there comes N lines, each contains two non-negative integers (each smaller than 10). The first integer represents how many hours Jinjin studies at school on the day, and the second represents how many hours she studies in the supplementary classes on the same day.

A case with N = 0 indicates the end of the input, and this case should not be processed.

Output

For each test case, output a line contains a single integer. If Jinjin will always be happy, the integer should be 0; otherwise, the integer should be a positive integer K, which means that Jinjin will be the unhappiest on the K-th day. If the unhappiest day is not unique, just output the earliest one among these unhappiest days.

Sample Input

7
5 3
6 2
7 2
5 3
5 4
0 4
0 6
1
4 4
0

Sample Output

3
0

Hint

Here is a sample solution of this problem using C language:
#include <stdio.h>

int main(){
while(1) {
int i, n;
int maxday, maxvalue = -1;
scanf("%d", &n);
if (n == 0) break;
for (i = 1; i <= n; i++) {
int a, b;
scanf("%d%d", &a, &b);
if (a + b > maxvalue) {
maxvalue = a + b;
maxday = i;
}
}
if (maxvalue <= 8) printf("0\n");
else printf("%d\n", maxday);
}
return 0;
}

Source

[Submit]   [Go Back]   [Status]   [Discuss]

Home Page   Go Back  To top


All Rights Reserved 2003-2013 Ying Fuchen,Xu Pengcheng,Xie Di
Any problem, Please Contact Administrator