CodeForces 518B Tanya and Postcard

开始想着找到YAY的次数，用第一个字符串的长度减去它，就是剩的另一个结果。呃，显然这想法是错的。

the letter in the given position has only the correct value but it is in the wrong case, then the girl says "WHOOPS".

这才是第二个数据的正确求法。

#include<stdio.h>
#include<iostream>
#include<math.h>
#include<string.h>
#include<iomanip>
#include<stdlib.h>
#include<ctype.h>
#include<algorithm>
#include<deque>
#include<functional>
#include<iterator>
#include<vector>
#include<list>
#include<map>
#include<queue>
#include<set>
#include<stack>
#define CPY(A, B) memcpy(A, B, sizeof(A))
typedef long long LL;
typedef unsigned long long uLL;
const int MOD = int (1e9) + 7;
const int INF = 0x3f3f3f3f;
const LL INFF = 0x3f3f3f3f3f3f3f3fLL;
const double EPS = 1e-9;
const double OO = 1e20;
const double PI = acos (-1.0);
const int dx[] = {-1, 0, 1, 0};
const int dy[] = {0, 1, 0, -1};
using namespace std;
const int maxn=200005;
char s[maxn],t[maxn];
int cs[52],ct[52];
int get (char ch) {
return (ch>='a'&&ch<='z') ?ch-'a':ch-'A'+26;
}
int main() {
while (~scanf ("%s %s",s,t) ) {
int ls=strlen (s),lt=strlen (t);
memset (cs,0,sizeof (cs) );
for (int i=0; i<ls; i++) {cs[get (s[i])]++;}
memset (ct,0,sizeof (ct) );
for (int i=0; i<lt; i++) {ct[get (t[i])]++;}//统计字母个数
int h=0;
for (int i=0; i<52; i++) {
int mm=min (cs[i],ct[i]);
cs[i]-=mm,ct[i]-=mm; h+=mm;//匹配的个数
}
int ku=0;
// the letter in the given position has only the correct value but it is in the wrong case, then the girl says "WHOOPS".
for (int i=0; i<26; i++) {
int kk=min (cs[i],ct[i+26]) +min (cs[i+26],ct[i]);//大写与小写，小写与大写
ku+=kk;
}
printf ("%d %d\n",h,ku);
}
return 0;
}