Game of Credit Cards
After the fourth season Sherlock and Moriary have realized the whole foolishness of the battle between them and decided to continue their competitions in peaceful game of Credit Cards.
Rules of this game are simple: each player bring his favourite n-digit credit card. Then both players name the digits written on their cards one by one. If two digits are not equal, then the player, whose digit is smaller gets a flick (knock in the forehead usually made with a forefinger) from the other player. For example, if n = 3, Sherlock's card is 123 and Moriarty's card has number 321, first Sherlock names 1and Moriarty names 3 so Sherlock gets a flick. Then they both digit 2so no one gets a flick. Finally, Sherlock names 3, while Moriarty names 1 and gets a flick.
Of course, Sherlock will play honestly naming digits one by one in the order they are given, while Moriary, as a true villain, plans to cheat. He is going to name his digits in some other order (however, he is not going to change the overall number of occurences of each digit). For example, in case above Moriarty could name 1, 2, 3 and get no flicks at all, or he can name 2, 3 and 1 to give Sherlock two flicks.
Your goal is to find out the minimum possible number of flicks Moriarty will get (no one likes flicks) and the maximum possible number of flicks Sherlock can get from Moriarty. Note, that these two goals are different and the optimal result may be obtained by using different strategies.
Input
The first line of the input contains a single integer n (1 ≤ n ≤ 1000) — the number of digits in the cards Sherlock and Moriarty are going to use.
The second line contains n digits — Sherlock's credit card number.
The third line contains n digits — Moriarty's credit card number.
Output
First print the minimum possible number of flicks Moriarty will get. Then print the maximum possible number of flicks that Sherlock can get from Moriarty.
Examples
Input
3 123 321
Output
0 2
Input
2 88 00
Output
2 0
Note
First sample is elaborated in the problem statement. In the second sample, there is no way Moriarty can avoid getting two flicks.
题意：
两个人有n张卡片，每张卡片上有0-9中的一个数字。第一个人一张一张出牌，第二个人不遵守规则，可以改变自己的牌序。
每局中，谁的点数小，就要被弹一次。
问第二个人最少要被弹多少次，第一个人最多要被弹多少次。
思路：
最佳的贪心策略应该是很容易想到的，就是用小的值去比较小的值。但是值得注意的是，一个数只能用一次，所以得记录每一个数有多少个。。 先排序，排序之后 如果 a[i] <= b[j], 就计数器 ++，用的数 --, 最后 n - 计数器 就是第一个答案，同理第二个答案就是 计数器， 只不过是条件是 a[i] < b[j].
AC代码：
#include<bits/stdc++.h>
using namespace std;
int numb[10];
int main()
{
int n;
while(~scanf("%d",&n)){
memset(numb,0,sizeof(numb));
char a[1500],b[1500];
scanf("%s %s",a,b);
for(int i = 0;i < n;i ++)
numb[b[i] - '0'] ++;
int y[10];
memcpy(y,numb,sizeof(numb));
sort(a,a + n);
sort(b,b + n);
int ans = 0;
for(int i = 0;i < n;i ++){
int flag = 0;
for(int j = 0;j < n;j ++){
if(a[i] <= b[j]){
if(numb[b[j]-'0'] > 0){
flag = 1;
numb[b[j]-'0'] --;
break;
}
}
}
if(flag == 1) ans ++;
}
printf("%d\n",n - ans);
ans = 0;
for(int i = 0;i < n;i ++){
int flag = 0;
for(int j = 0;j < n;j ++){
if(a[i] < b[j]){
if(y[b[j]-'0'] > 0){
flag = 1;
y[b[j]-'0'] --;
break;
}
}
}
if(flag == 1) ans ++;
}
printf("%d\n",ans);
}
return 0;
}