HDU 1865（大数）

1sting

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 4020    Accepted Submission(s): 1493

Problem 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

//斐波那契数列加大数  正常的斐波那契秩只需一维数组，由于位数太多只能用二维数组来模拟
#include <stdio.h>
#include <string.h>
int str1[210][300]={{0}};
int main()
{
    int n;
    scanf("%d",&n);
    getchar();
    while(n--)
    {
        char str[300];
        gets(str);
        int len=strlen(str);
        str1[0][0]=1;
        str1[1][0]=2;
        for(int i=2;i<len;i++)
        {
            int c=0,k;
            for(k=0;k<300;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;
            }
        }
        int i;
        for(i=299;i>=0;i--)  //判断是否为零
            if(str1[len-1][i])
                break;
        for(int j=i;j>=0;j--)
            printf("%d",str1[len-1][j]);
        printf("\n");
    }
    return 0;
}