Iiuc ONLINE CONTEST 2008/uva 11389 the bus Driver Problem: Greed

11389-the Bus Driver Problem

Time limit:1.000 seconds

Http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem &problem=2384

In a city there are n bus drivers. Also There are n morning bus routes & N afternoon bus routes with various lengths. Each driver is assigned one morning Route & one evening route. For any driver, if him total route length for a day exceeds D, he has to is paid overtime for every hour after the Hours at a flat R taka/hour. Your task is to assign one morning Route & one evening route to all bus driver so this total overtime amount The authority has to pay is minimized.

Input

The "a" of each test case has three integers n, D and R, as described above. In the second line, there are n space separated integers which the are of the lengths morning. Similarly the third line has n space separated integers denoting the evening route. The lengths are positive integers less than or equal to 10000. The end of the input was denoted by a case with three 0 s.

Output

For each test case, print the minimum possible overtime amount the authority must.

Constraints

-1≤n≤100

-1≤d≤10000

-1≤r≤5

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Sample Input

2 20 5

10 15

10 15

2 20 5

10 10

10 10

0 0 0

Output for Sample Input

50

0

An increase of one minus is necessarily optimal, which can be disproved by exchanging the position of two elements.

Complete code:

/*0.016s*/
    
#include <cstdio>  
#include <algorithm>  
#include <functional>  
using namespace Std;  
    
int a[105], b[105];  
    
int main ()  
{  
    int n, D, r, Sum, temp;  
    while (scanf ("%d%d%d", &n, &d, &r), N)  
    {for  
        (int i = 0; i < n; ++i)  
            scanf ("%d", &a[i]);  
        for (int i = 0; i < n; ++i)  
            scanf ("%d", &b[i]);  
        Sort (A, a + N);  
        Sort (b, B + N, greater<int> ());  
        sum = 0;  
        for (int i = 0; i < n; ++i)  
        {  
            temp = A[i] + b[i]-D;  
            if (temp > 0) sum + + temp;  
        }  
        printf ("%d\n", Sum * r);  
    }  
    return 0;  
}

Author: csdn Synapse7