HDU 4882 ZCC Loves Codefires (Greedy), hdu4882

HDU 4882 ZCC Loves Codefires (Greedy), hdu4882
Zookeeper ZCC Loves Codefires

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission (s): 223 Accepted Submission (s): 123

Problem DescriptionThough ZCC has extends Fans, ZCC himself is a crazy Fan of a coder, called "memsetfan ".
It was on Codefires (CF), an online competitive programming site, that ZCC knew memsettings, and immediately became his fan.
But why?
Because Memset137 can solve all problem in rounds, without extends 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, memsetincluhas 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 memsetbench 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 ).
 

InputThe 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.

OutputOne 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.  

 

Author Zhenhai Middle School

Source2014 Multi-University Training Contest 2

The time required for n questions to complete question I is e [I]. The coefficient of question I is k [I].
You can complete the question competition in any order and the total time consumed from the start to the completion of the I question is t [I]
Then k [I] * t [I] points should be deducted for completing the I-th question. How many points should be deducted for completing all questions;

Consider any adjacent question I. j. Changing I, j. I, j. All questions before and after j have no effect on the result.
Only the deduction points for the I and j questions and the original (t + e [I]) * k [I] + (t + e [I] + e [j]) * k [j] becomes (t + e [I] + e [j]) * k [I] + (t + e [j]) * k [j]
We can see that only e [I] * k [j] is changed to e [j] * k [I]. Therefore, if e [I] * k [j] <e [j] * k [I] Then I should be completed before j;
Sort the data in this order;

#include<cstdio>#include<algorithm>using namespace std;typedef long long ll;const int N = 100050;ll k[N], o[N], t[N], e[N], n;bool cmp (int i, int j){    return (e[i] * k[j] < e[j] * k[i]);}int main(){    while (scanf ("%I64d", &n) != EOF)    {        for (int i = 1; i <= n; ++i)            scanf ("%I64d", &e[i]);        for (int i = 1; i <= n; ++i)        {            o[i] = i;            scanf ("%I64d", &k[i]);        }        sort (o + 1, o + n + 1, cmp);        ll ans=0;        for (int i = 1; i <= n; ++i)        {            int j=o[i];            t[j]=t[o[i-1]]+e[j];            ans += k[j] * t[j];        }        printf ("%I64d\n", ans);    }    return 0;}