HDU - 1865 1string（大数）

题目链接：http://acm.hust.edu.cn/vjudge/contest/121397#problem/F

http://acm.hdu.edu.cn/showproblem.php?pid=1865

Description

You will be given a string which only contains ‘1’; You can merge two adjacent ‘1’ to be ‘2’, or leave the ‘1’ there. Surly, you may get many different results. For example, given 1111 , you can get 1111, 121, 112,211,22. Now, your work is to find the total number of result you can get. 

Input

The first line is a number n refers to the number of test cases. Then n lines follows, each line has a string made up of ‘1’ . The maximum length of the sequence is 200. 

Output

The output contain n lines, each line output the number of result you can get . 

 

Sample Input

Sample Output

AC代码:

 #include <stdio.h>
 #include <string.h>
 #define maxn 300
 int str1[maxn][maxn];
 int main()
 {
     int T, i, j;
     scanf("%d ",&T);

     while(T--)
     {
         char str[];
         gets(str);

         int len=strlen(str);
         str1[][]=;
         str1[][]=;

         for(i=;i<len;i++)
         {
             int c=,k;
             for(k=;k<maxn;k++)
             {
                 int s=str1[i-][k]+str1[i-][k]+c;
                 str1[i][k]=s%;
                 c=s/;
             }
             while(c)
             {
                 str1[i][k++]=c%;
                 c/=;
             }
         }

         for(i=;i>=;i--)
             if(str1[len-][i])
                 break;

         for(j=i;j>=;j--)
             printf("%d",str1[len-][j]);
             printf("\n");
     }
     return ;
 }

A完还A：

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

 #define N 210
  int f[][N];

  void init()
  {
      int i, j;
      f[][] = ;
        f[][] = ;

        for(i = ; i <= ; i++)
         for(j = ; j <= ; j++)
        {
            f[i][j] += f[i-][j] + f[i-][j];
            if(f[i][j] >= )
            {
                f[i][j] -= ;
                f[i][j+]++;
            }
        }
  }

 int main()
 {

    int i, j, t;
    char s[N];

    scanf("%d", &t);
    init();

    while(t--)
    {
        scanf("%s", s);

        int n = strlen(s);

         int k = ;
         for (; f[n][k] ==  ; k--);
         for (; k >=  ; k--)
             printf ("%d",f[n][k]);

             printf("\n");

    }

    return ;
 }