ZCC Loves Codefires Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 988 Accepted Submission(s): 500
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-KiTi score. It’s guaranteed Bi-KiTi 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
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.
贪心题。 很容易想到的是，为了让答案尽可能的小，最好使e小的和K大的尽可能早的完成。也就是说E和K的大小对结果都会有影响。
我们按E/K排个序。
注意冒泡会TLE。。。别问我怎么知道的T T
也别问我为什么不用Sort….因为我不会写CMP函数。。。
噗，其实很简单嘛
1
2
3 #include <iostream>
4 #include <algorithm>
5 #include <cstdio>
6 #include <cstring>
7
8 using namespace std;
9 const int N=100005;
10 int n;
11 struct Q
12 {
13 long long e,k;
14 }q[N];
15 bool cmp(Q a,Q b)
16 {
17 return a.e*b.k<a.k*b.e;
18 }
19
20 int main()
21 {
22 cin>>n;
23 for (int i=1;i<=n;i++)
24 scanf("%I64d",&q[i].e);
25 for (int i=1;i<=n;i++)
26 scanf("%I64d",&q[i].k);
27 sort(q+1,q+1+n,cmp);
28 long long ans=0,time=0;
29 for (int i=1;i<=n;i++)
30 {
31 time=time+q[i].e;
32 ans=ans+q[i].k*time;
33
34 }
35 printf("%I64d\n",ans);
36 return 0;
37 }