杭电1865--1sting

1sting

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4133    Accepted Submission(s): 1547

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

 

 

Author

z.jt

 

 

Source

2008杭电集训队选拔赛——热身赛

 

 

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//思路： 递推； 数中只存在1 和2 这两个数字 → → → f(n) = f(n-1) + f(n-1) ;

#include <stdio.h>
#include <string.h>
int fei[220][1001];
int main()
{
    int n;
    char str[220];
    int i, j;
    int temp, plus = 0;
    memset(fei, 0, sizeof(fei));
    fei[1][0] = 1;
    fei[2][0] = 2; 
    for(i=3; i<201; i++)  
    {
        for(j=0; j<1001; j++)     //二维数组下标表示数的位数；
        {
            temp = fei[i-1][j] + fei[i-2][j] + plus;
            fei[i][j] = temp%10;
            plus = temp/10;
        }
    }
    scanf("%d", &n);
    while(n--)
    {
        scanf("%s", str);
        int len = strlen(str);
        for(i=1000; i>=0; i--)
        if(fei[len][i])
        break;
        for(; i>=0; i--)
        printf("%d", fei[len][i]);
        printf("
");
    }
    return 0;
}