题意：田忌赛马，问你田忌最多能赢多少银子。
析：贪心，绝对贪心的题，贪心策略是：
1.如果田忌当前的最快的马能追上齐王的，那么就直接赢一局
2.如果田忌当前的最慢的马能追上齐王的，那么就直接赢一局
3.如果田忌当前的最慢的马不能超过齐王的，那么就输一局，并把齐王最快的干掉
通过以上策略，就是田忌赢的最多。
代码如下：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <unordered_map>
#include <unordered_set>
#define debug() puts("++++");
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e3 + 5;
const int mod = 2000;
const int dr[] = {-1, 1, 0, 0};
const int dc[] = {0, 0, 1, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
return r >= 0 && r < n && c >= 0 && c < m;
}
int a[maxn], b[maxn];
int main(){
while(scanf("%d", &n) == 1 && 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, greater<int>());
sort(b, b + n, greater<int>());
int ans = 0;
int fro1 = 0, fro2 = 0;
int rear1 = n-1, rear2 = n-1;
while(fro1 <= rear1){
if(a[fro1] > b[fro2]){
ans += 200;
++fro1;
++fro2;
}
else if(a[rear1] > b[rear2]){
ans += 200;
--rear1;
--rear2;
}
else if(a[rear1] <= b[rear2]){
if(a[rear1] < b[fro2]) ans -= 200;
--rear1;
++fro2;
}
}
printf("%d\n", ans);
}
return 0;
}
/*
3
15 12 9
16 13 9
5
7 8 9 12 15
7 8 9 13 16
4
1 2 4 5
2 3 3 4
*/
