2014 (多校）1011 ZCC Loves Codefires，zcccodefires

2014 (多校）1011 ZCC Loves Codefires，zcccodefires

自從做了多校，整個人都不好了，老是被高中生就算了，題老是都不懂=-=原諒我是個菜鳥，原諒我智力不行。唯一的水題。

Problem 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

310 10 201 2 3

 
Sample Output

150HintMemset137 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. 

 

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int maxn=1e5+10;struct node{    int x,y;}a[maxn];int cmp(node l1,node l2){    return l1.x*l2.y<l2.x*l1.y;}int main(){    int n;    while(~scanf("%d",&n))    {        for(int i=0;i<n;i++)            scanf("%d",&a[i].x);        for(int i=0;i<n;i++)            scanf("%d",&a[i].y);        sort(a,a+n,cmp);        long long sum=0,s=0;        for(int i=0;i<n;i++)        {            sum+=a[i].x;            s+=sum*a[i].y;        }        printf("%I64d\n",s);    }    return 0;}