UVA 620 - Cellular Structure

真心不知道怎么用动归，然后网上找到的做法：

I think this problem's tip is..
len=string length
Mutant : len is 1, and str[0]='B'
Simple : if len is 1, and str[0]='A'
Fully-Grown : str[len-1]='B', and str[len-2]='A'
Mutagenic : str[0]='B', and str[len-1]='A'
Else : Mutant

然后再写出自己的代码就行了。

#include<cstdio>
#include<cstring>
#include<cstdlib>

const char ans[4][20] = { "SIMPLE", "FULLY-GROWN", "MUTAGENIC", "MUTANT"};
int T, len, res;
char ch[1005];

int solve()
{
    if( len % 2 == 0) // !(len & 1)
        return 4;
    if( len == 1) {
        if( ch[0] == 'A')
            return 1;
        else return 4;
    }

    if( ch[len - 2] == 'A' && ch[len - 1] == 'B') // FULL-GROWN
        return 2;

    if( ch[0] == 'B' && ch[len - 1] == 'A') // MUTAGENIC
        return 3;

    return 4;
}

int main()
{
    scanf( "%d", &T);
    while( T --)
    {
        scanf( "%s", ch);
        len = strlen(ch);
        printf( "%s\n", ans[solve() - 1]);
    }
    return 0;
}