zoj 3171 - The Hidden 7's

题目：给你一个字符串，统计其中能找到多少个构成seven的子序列。

分析：dp。按单词的字母顺序DP，每个字母统计他前面字串的个数。

说明：（2011-09-19 14:03）

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define DataSize 10005

char         str[ DataSize ];
char          data[] = "0seven";

long long seven[ 6 ][ DataSize ];

int main()
{
    while ( scanf("%s",str) != EOF ) {
        int len = strlen( str );
        for ( int i = 0 ; i < len ; ++ i )
            if ( str[ i ] >= 'A' && str[ i ] <= 'Z' )
                str[ i ] = str[ i ] - 'A' + 'a';
        
        memset( seven, 0, sizeof( seven ) );
        for ( int i = 0 ; i < len ; ++ i )
            seven[ 0 ][ i ] = 1;
        seven[ 1 ][ 0 ] = (str[ 0 ]=='s');
            
        for ( int k = 1 ; k <= 5 ; ++ k )
        for ( int i = 1 ; i < len ; ++ i )
            if ( str[ i ] == data[ k ] )
                seven[ k ][ i ] = seven[ k ][ i-1 ] + seven[ k-1 ][ i-1 ];
            else
                seven[ k ][ i ] = seven[ k ][ i-1 ];

        printf("%lld\n",seven[ 5 ][ len-1 ]);
    }
    return 0;
}