原题：
| In a city there are n bus drivers. Also there are n morning bus routes &nafternoon 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 exceedsd, he has to be paid overtime for every hour after the firstd 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.
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| Input | |
| The first line of each test case has three integers n, d and r, as described above. In the second line, there arenspace separated integers which are the lengths of the morning routes given in meters. Similarly the third line hasnspace 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.
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| Output | |
| For each test case, print the minimum possible overtime amount that the authority must pay.
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| Constraints | |
| - 1 ≤ n ≤ 100 - 1 ≤ d ≤ 10000 - 1 ≤ r ≤ 5
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| Sample Input | Output for Sample Input |
| 2 20 5 10 15 10 15 2 20 5 10 10 10 10 0 0 0 | 50 0 |
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分析：
简单贪心——排序
源码：
#include <stdio.h> #include<algorithm> int b[110]; int w[110]; using namespace std; int main() { int n,d,r; while(scanf("%d%d%d",&n,&d,&r)&&(n!=0||d!=0||r!=0)) { int i ; int sum=0; for(i = 0; i<n; ++i) { scanf("%d",&b[i]); } for(i = 0; i<n; ++i) { scanf("%d",&w[i]); } sort(b,b+n); sort(w,w+n); for(i = 0; i<n; ++i) { if(b[i]+w[n-1-i]>d) sum+=b[i]+w[n-1-i]-d; } printf("%d\n",sum*r); } return 0; }
