HDU ACMSteps 1.3.3:: Tian Ji ——The Horse Racing

#include <iostream>

#include <memory.h> 

#include <algorithm>

using namespace std;

/*

HDU题目:: Tian Ji -- The Horse Racing

Input

The input consists of up to 50 test cases. Each case starts with a positive integer n (n <= 1000) on the first line, which is 

the number of horses on each side. The next n integers on the second line are the speeds of Tian’s horses. Then the next n 

integers on the third line are the speeds of the king’s horses. The input ends with a line that has a single 0 after the last

test case.

Output

For each input case, output a line containing a single number, which is the maximum money Tian Ji will get, in silver dollars.

Sample Input

3

92 83 71

95 87 74

2

20 20

20 20

2

20 19

22 18

0

Sample Output

200

0

0

分析:

输入马匹规模n，输入田忌n匹马的速度，输入国王n匹马的速度.其思路是用最小的耗费赢取最大的胜利。对于快马的比较，能赢则赢。不能赢则考虑尽量用最小的代价打平不输钱（这里主要是代价观点）

*/

bool cmp(int a,int b)

{

    return a>b;

}

int main(){
int tian[1002];
int king[1002];
memset(tian,0,1002*sizeof(int));
memset(king,0,1002*sizeof(int));
int n,i,fast,fask,slot,slok,win_num;
while(true){
cin>>n;
if(n==0) break;
for(i=0;i<n;i++)//输入两人马匹速度 
cin>>tian[i]; 
for(i=0;i<n;i++)
cin>>king[i]; 
sort(tian,tian+n,cmp);
sort(king,king+n,cmp);
fast=0,fask=0,slot=slok=n-1,win_num=0;
while(fast<=slot&&fask<=slok){
if(tian[fast]>king[fask]){//快马比皇上快 
win_num++;
fast++,fask++;
}
else//不赢的时候主要考虑能否用以获取平局
if(tian[slot]>king[slok]){//田忌的慢马比皇上的慢马快 
win_num++;
slot--,slok--;
}
else
{//慢马无法赢，那么用来挡输或打平 
if(tian[slot]<king[fask])//用田忌的慢马去输皇上最快的马 
win_num--;
slot--,fask++;
}
}
cout<<(win_num<<1)*100<<endl;
memset(tian,0,n*sizeof(int));//清零
memset(king,0,n*sizeof(int));
}
return 0;

}