poj3073

• 比赛状态堪忧，笑看自己找不着北..
• 把心态放好吧- -
• 反正窝从一開始就仅仅是为了多学习才上道的
• 至少已经从学习和智商上给窝带来了一些帮助
• 智商带不动，好辛苦～～～～（＞＿＜）～～～～　
• 说说这题吧…这题就是个SB题。考虑前i个字符能匹配的方案数。我们仅仅须要考虑它后几位是否能配上一组题目给出的字符就可以，于是有
  dp[i]=∑j=1ndp[j](if.字符[j,i]匹配上了某一组给定字符)

#include <cstdio>

#include <vector>

#include <string>

#include <cctype>

#include <cstring>

#include <iostream>

#include <algorithm>

using namespace std;

const int MAX = 128 << 2;

int dp[MAX];

char rp[MAX >> 2][6] =

{

    "4", "|3", "(", "|)", "3", "|=", "6", "#", "|",

    "_|", "|<", "|_", "|\\/|", "|\\|", "0", "|0", /*-P*/

    "(,)", "|?", "5", "7", "|_|", "\\/", "\\/\\/",

    "><", "-/", "2"

};

int main()

{

    char buffer[MAX];

    char s[MAX];

    while (cin >> buffer && buffer[0] != 'e')

    {

        s[0] = '\0';

        int len = strlen(buffer);

        for (int i = 0; i < len; ++i)

        {

            strcat(s, rp[buffer[i] - 'A']);

        }

        len = strlen(s);

        memset(dp, 0, sizeof(dp));

        for (int i = 0; i < len; ++i)

        {

            char ch = s[i + 1];

            s[i + 1] = '\0';

            for (int t = 0; t < 26; ++t)

            {

                if (strcmp(rp[t], s) == 0)

                {

                    ++dp[i];

                    break;

                }

            }

            for (int j = 1; j <= i; ++j)

            {

                for (int t = 0; t < 26; ++t)

                {

                    if (strcmp(rp[t], s + j) == 0)

                    {

                        dp[i] += dp[j - 1];

                    }

                }

            }

            s[i + 1] = ch;

        }

        cout << dp[len - 1] << endl;

    }

    return 0;

}