Description
Though ZCC has many Fans, ZCC himself is a crazy Fan of a coder, called "Memset137".
It was on Codefires(CF), an online competitive programming site, that ZCC knew Memset137, and immediately became his fan.
But why?
Because Memset137 can solve all problem in rounds, without unsuccessful submissions; his estimation of time to solve certain problem is so accurate, that he can surely get an Accepted the second he has predicted. He soon became IGM, the best title of Codefires. Besides, he is famous for his coding speed and the achievement in the field of Data Structures.
After become IGM, Memset137 has a new goal: He wants his score in CF rounds to be as large as possible.
What is score? In Codefires, every problem has 2 attributes, let's call them Ki and Bi(Ki, Bi＞0). if Memset137 solves the problem at Ti-th second, he gained Bi-Ki*Ti score. It's guaranteed Bi-Ki*Ti is always positive during the round time.
Now that Memset137 can solve every problem, in this problem, Bi is of no concern. Please write a program to calculate the minimal score he will lose.(that is, the sum of Ki*Ti).
Input
The first line contains an integer N(1≤N≤10^5), the number of problem in the round.
The second line contains N integers Ei(1≤Ei≤10^4), the time(second) to solve the i-th problem.
The last line contains N integers Ki(1≤Ki≤10^4), as was described.
Output
One integer L, the minimal score he will lose.
Sample Input
3
10 10 20
1 2 3
10 10 20
1 2 3
Sample Output
150
HINT
Memset137 takes the first 10 seconds to solve problem B, then solves problem C at the end of the 30th second. Memset137 gets AK at the end of the 40th second.
L = 10 * 2 + (10+20) * 3 + (10+20+10) * 1 = 150.
唉……一道水水的贪心还推了半天……毕竟我老了
按e/k排序完直接搞
假设我们当前有一个解。考虑交换相邻的两个i和i+1会不会比原来优
原来的答案是e[i]*k[i]+(e[i]+e[i+1])*k[i+1]
改完的答案是e[i+1]*k[i+1]+(e[i]+e[i+1])*k[i]
减一下，再约掉一下，得到e[i]*k[i+1]-e[i+1]*k[i]
只要这个式子小于0，那么交换完答案就会减少。
即e[i]*k[i+1]<e[i+1]*k[i]
即e[i]/k[i]<e[i+1]/k[i+1]
没开long long结果作死了一次
#include<cstdio>
#include<algorithm>
using namespace std;
int n,s[100010];
long long ans;
struct dat{int e,k;}a[100010];
bool operator<(dat a,dat b)
{return a.e*b.k<a.k*b.e;}
int main()
{
scanf("%d",&n);
for (int i=1;i<=n;i++)scanf("%d",&a[i].e);
for (int i=1;i<=n;i++)scanf("%d",&a[i].k);
sort(a+1,a+n+1);
for (int i=1;i<=n;i++)s[i]=s[i-1]+a[i].e;
for (int i=1;i<=n;i++)ans+=(long long)s[i]*a[i].k;
printf("%lld
",ans);
}