Since Judge Nicole Hosh moved to Egypt for her Computer Science Masters in AASTMT, in 2014, she has been training with coach Fegla and attending his camps in Egypt. She, also, set a number of problems for TCPC and JCPC and was a judge in LCPC and SCPC. Her best friend Noura was so proud of her so she was trying to convince her to start writing Codeforces Div. 2 round. After various attempts to convince her, Nicole finally agreed, and so, she started collecting some problems with different difficulties from her ex-contestant friends.
Judge Nicole collected 7 ideas for problems of different levels, she wants to create 5 problems for the next contest, one for each difficulty level, from A to E (difficulty 1 to 5). Given the difficulty level of the problems she currently has, she can merge the ideas of two problems, one of level x, and the other of level y to get a problem of level x + y.
For example, Judge Nicole can merge two problems of difficulties A and D, to get one problem of difficulty E (1 + 4 = 5).
Merging more than two problems into one will produce a problem with a long statement which is hard to explain, so she won’t do this (i.e., each problem is merged with another at most once). Also, she can’t merge a resultant problem again, and she can't use the same problem twice.
Input
The first line of input contains an integer T (1 ≤ T ≤ 330), the number of test cases.
Each test case will contain only one string S of length 7. Each letter of the string represents the difficulty level of a problem (from A to E), 'A' is the easiest and 'E' is the hardest.
Output
For each test case print "YES" if she can prepare a contest using the current problems, otherwise print "NO".
Examples
Input
3 EBEABDA CEDEACA BDAAEAA
Output
YES NO YES
题意：有ＡＢＣＤＥ五种难度的题，分别代表１２３４５，可以合并两个题目，合并题目的难度等于他们的难度和，如B＝Ａ＋Ａ，类似于２＝１＋１。现在给出一个七位长度的序列，问能否让这个序列种每种难度的题都至少有一个。每一种题只能合并一次，即不能有２个以上的题一起合并。
题解：暴力解法，把有情况都找出来看行不行。合并两个的有Ｃ(7,2)种情况，合并4个的有C(7,2) * C(5,2)种情况。
#include <bits/stdc++.h>
using namespace std;
int main()
{
int a[10];
char str[10];
int t,flag,cnt;
scanf("%d",&t);
while(t--)
{
flag = 0,cnt = 1;
memset(a,0,sizeof(a));
scanf("%s",str);
for(int i = 0;i < 7;i++)
{
a[str[i] - 'A' + 1]++;
}
for(int i = 1;i <= 5;i++)
{
if(!a[i]) cnt = 0;
}
if(cnt == 1) flag = 1;
if(!flag)
{
for(int i = 0;i < 7;i++)
{
for(int j = i + 1;j < 7;j++)
{
cnt = 1;
int x = str[i] - 'A' + 1,y = str[j] - 'A' + 1;
a[x]--,a[y]--,a[x + y]++;
for(int k = 1;k <= 5;k++)
{
if(!a[k]) cnt = 0;
}
a[x]++,a[y]++,a[x + y]--;
if(cnt == 1) flag = 1;
}
}
}
if(!flag)
{
for(int i = 0;i < 7;i++)
{
for(int ii = i + 1;ii < 7;ii++)
{
for(int j = 0;j < 7;j++)
{
for(int jj = j + 1;jj < 7;jj++)
{
cnt = 1;
if(i == j || ii == jj || i == jj || j == ii) continue;
int x = str[i] - 'A' + 1;
int y = str[ii] - 'A' + 1;
int r = str[j] - 'A' + 1;
int s = str[jj] - 'A' + 1;
a[x]--,a[y]--,a[x + y]++;
a[r]--,a[s]--,a[r + s]++;
for(int k = 1;k <= 5;k++)
{
if(!a[k]) cnt = 0;
}
a[x]++,a[y]++,a[x + y]--;
a[r]++,a[s]++,a[r + s]--;
if(cnt == 1) flag = 1;
}
}
}
}
}
if(flag == 1)
{
printf("YES\n");
}
else
{
printf("NO\n");
}
}
return 0;
}