学步园

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Problem E: The Bus Driver Problem

Input: standard input

Output: standard output

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 his
total route length for a day exceeds d, he has to be paid overtime for every hour after the first d hours at a flat r taka / hour. Your task is to assign one morning route & one evening route to each bus driver
so that the total overtime amount that the authority has to pay is minimized.

Input

The first line 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 are the lengths of the morning routes given
in meters. Similarly the third line has n space separated integers denoting the evening route lengths. The lengths are positive integers less than or equal to 10000. The end of input is denoted by a case with three 0 s.

Output

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

Constraints

-           1 ≤ n ≤ 100

-           1 ≤ d ≤ 10000

-           1 ≤ r ≤ 5

Sample Input

Output for Sample Input

2 20 5

10 15

10 15

2 20 5

10 10

10 10

0 0 0

题目很简单看代码就应该知道是什么意思了！！！！！

#include <stdio.h>
#include <math.h>
#include <algorithm>
#include <string.h>
#include <stdlib.h>
using namespace std;
int mlen[110];
int alen[110];
int main()
{
    int n,d,r;
    while(scanf("%d%d%d",&n,&d,&r)==3)
    {
        if(n==0&&d==0&&r==0)
        break;
        int i ;
        int sum=0;
        for(i = 0;i<n;++i)
        scanf("%d",&mlen[i]);
        for(i = 0;i<n;++i)
        scanf("%d",&alen[i]);
        sort(mlen,mlen+n);
        sort(alen,alen+n);
        for(i = 0;i<n;++i)
        if(mlen[i]+alen[n-1-i]>d)
        sum+=mlen[i]+alen[n-1-i]-d;
        printf("%d\n",sum*r);
    }
    return 0;
}

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