这个题不难，但我是蒟蒻。。。

先看题目

题目：

Ever the maturing businessman, Farmer John realizes that he must manage his time effectively. He has N jobs conveniently numbered 1..N (1 <= N <= 1,000) to accomplish (like milking the cows, cleaning the barn, mending the fences, and so on).

To manage his time effectively, he has created a list of the jobs that must be finished. Job i requires a certain amount of time T_i (1 <= T_i <= 1,000) to complete and furthermore must be finished by time S_i (1 <= S_i <= 1,000,000). Farmer John starts his day at time t=0 and can only work on one job at a time until it is finished.

Even a maturing businessman likes to sleep late; help Farmer John determine the latest he can start working and still finish all the jobs on time.

作为一名忙碌的商人，约翰知道必须高效地安排他的时间.他有N工作要 做，比如给奶牛挤奶，清洗牛棚，修理栅栏之类的.

为了高效，列出了所有工作的清单.第i分工作需要T_i单位的时间来完成，而 且必须在S_i或之前完成.现在是0时刻.约翰做一份工作必须直到做完才能停 止.

所有的商人都喜欢睡懒觉.请帮约翰计算他最迟什么时候开始工作，可以让所有工作按时完成.（如果无法完成全部任务，输出-1）

输入格式

* Line 1: A single integer: N

* Lines 2..N+1: Line i+1 contains two space-separated integers: T_i and S_i

（翻译：* 第1行: 单个整数: n

* 第2行. n + 1行: 第 i + 1行包含两个空格分隔的整数: t _ i 和 s _ i）

输出格式

* Line 1: The latest time Farmer John can start working or -1 if Farmer John cannot finish all the jobs on time.

（翻译：* 第一行: 约翰可以开始工作的最迟时间，如果农民约翰不能按时完成所有工作，则为 -1）

输入输出样例

输入 #1复制

4

3 5

8 14

5 20

1 16

输出 #1复制

2

说明/提示

Farmer John has 4 jobs to do, which take 3, 8, 5, and 1 units of time, respectively, and must be completed by time 5, 14, 20, and 16, respectively.

Farmer John must start the first job at time 2. Then he can do the second, fourth, and third jobs in that order to finish on time.

（翻译：约翰有4个工作要做，分别需要3、8、5和1个时间单位，必须分别在5、14、20和16个时间段内完成。那么他就可以按照这个顺序做第二、第四、第三项工作，以便按时完成。）

思路：

只要倒着遍历模拟就行了，比较类似P1843奶牛晒衣服

代码：

（本蒟蒻用c++，python的可以走了）

#include<bits/stdc++.h>//万能头
using namespace std;
struct Work{int ut,lt;
}a[1010];
bool cmp(Work x,Work y){return x.lt<y.lt || x.lt==y.lt&&x.ut<y.ut;//比较大小
}
int main(){int n;cin>>n;for(int i=1;i<=n;i++){cin>>a[i].ut>>a[i].lt;}sort(a+1,a+n+1,cmp);int zwsj=a[n].lt-a[n].ut;//定义初始值for(int i=n-1;i>=1;i--){//模拟if(zwsj>a[i].lt){zwsj=a[i].lt;}zwsj-=a[i].ut;}if(zwsj>-1){cout<<zwsj;}else{cout<<-1;}
}