POJ 2656 Unhappy Jinjin

Unhappy Jinjin

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 11472 Accepted: 8396

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

75 36 27 25 35 40 40 614 40

Sample Output

30

超级大水题：

#include <iostream>#include <cstdio>using namespace std;int main(int argc, const char * argv[]) {    int n;    int a,b;    while(scanf("%d",&n)&&n)    {        int mx=0,src=0;        for(int i=1;i<=n;i++)        {            scanf("%d %d",&a,&b);            if(a+b>8&&a+b>mx)            {                mx=a+b;//将最不开心的那一天的学习时间存储下来，与接下来的输入进行判断                src=i;//存储最不开心的是哪一天            }        }        cout<<src<<endl;    }    return 0;}