Farmer John went to cut some wood and left N (2 <= N <= 100,000) cows eating the grass, as usual. When he returned, he found to his horror that the cows were in his garden eating his beautiful flowers. Wanting to minimize the subsequent damage, FJ decided to take immediate action and transport the cows back to their barn. Each cow i is at a location that is Ti minutes (1 <= Ti <= 2,000,000) away from the barn. Furthermore, while waiting for transport, she destroys Di (1 <= Di <= 100) flowers per minute. No matter how hard he tries,FJ can only transport one cow at a time back to the barn. Moving cow i to the barn requires 2*Ti minutes (Ti to get there and Ti to return). Write a program to determine the order in which FJ should pick up the cows so that the total number of flowers destroyed is minimized.

* Line 1: A single integer

N * Lines 2..N+1: Each line contains two space-separated integers, Ti and Di, that describe a single cow's characteristics

第1行输入N，之后N行每行输入两个整数Ti和Di．

* Line 1: A single integer that is the minimum number of destroyed flowers

一个整数，表示最小数量的花朵被吞食．

约翰用6，2，3，4，1，5的顺序来运送他的奶牛．

Silver

考虑相邻的两只牛交换的条件是d[b]*t[a]<d[a]*t[b]。

按条件排序即可。

 #include<iostream>

 #include<cstdio>

 #include<cstring>

 #include<cstdlib>

 #include<cmath>

 #include<algorithm>

 #define LL long long

 using namespace std;

 LL n;

 struct data{

     LL t,d;

     bool operator <(const data tmp)const {

         return tmp.d*t<d*tmp.t;

     }

 }a[];

 int main(){

     scanf("%lld",&n);

     for(int i=;i<=n;i++) scanf("%lld%lld",&a[i].t,&a[i].d);

     sort(a+,a+n+);

     LL sum=,tim=;

     for(int i=;i<=n;i++){

         sum+=a[i].d*tim;

         tim+=a[i].t*;

     }

     printf("%lld\n",sum);

     return ;

 }