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.

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.

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.

75 36 27 25 35 40 40 614 40

30

Here is a sample solution of this problem using C language: 

#include 
    
     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;}
    

Jinjin要学习N天，学习时间分为学校时间和补课时间，如果Jinjin某天学习了超过八小时（>= 8），则Jinjin会不开心，如果他有过不开心，那么求他学习时间最长的一天（从1开始），如果有几天学习时间都最长，则输出最早的一天；如果Jinjin学习时间小于等于八小时，则输出0。

模拟。

代码（1AC）：

#include 
    
     #include 
     
      #include 
      
       int main(void){    int i, days;    int max;    int first, second;    int flag;    while (scanf("%d", &days), days != 0){        max = -1;        flag = 0;        for (i = 0; i < days; i++){            scanf("%d%d", &first, &second);            if (first + second > 8 && max < first + second){                flag = i + 1;                max = first + second;            }        }        printf("%d\n", flag);    }    return 0;}