HDU

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
3
1
11
11111
Sample Output
1
2
8

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[300];
        gets(str);

        int len=strlen(str);
        str1[0][0]=1;
        str1[1][0]=2;

        for(i=2;i<len;i++)
        {
            int c=0,k;
            for(k=0;k<maxn;k++)
            {
                int s=str1[i-1][k]+str1[i-2][k]+c;
                str1[i][k]=s%10;
                c=s/10;
            }
            while(c)
            {
                str1[i][k++]=c%10;
                c/=10;
            }
        }

        for(i=299;i>=0;i--)
            if(str1[len-1][i])
                break;

        for(j=i;j>=0;j--)
            printf("%d",str1[len-1][j]);
            printf("
");
    }
    return 0;
}

A完还A：

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

#define N 210
 int f[1010][N];

 void init()
 {
     int i, j;
     f[0][1] = 1;
       f[1][1] = 1;

       for(i = 2; i <= 1000; i++)
        for(j = 1; j <= 200; j++)
       {
           f[i][j] += f[i-1][j] + f[i-2][j];
           if(f[i][j] >= 10)
           {
               f[i][j] -= 10;
               f[i][j+1]++;
           }
       }
 }

int main()
{

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

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


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

       int n = strlen(s);


        int k = 150;
        for (; f[n][k] == 0 ; k--);
        for (; k >= 1 ; k--)
            printf ("%d",f[n][k]);

            printf("
");

   }

   return 0;
}