hdoj 1865 1sting



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

1sting

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3399    Accepted Submission(s): 1314

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

311111111

 

Sample Output

128

 

#include<stdio.h>#include<string.h>#define NUM 200#define MAX 100int a[NUM+10][MAX];int main(){int i,j,k;for(i=0;i<NUM+10;i++)memset(a[i],0,sizeof(a[i]));a[1][0]=1;a[2][0]=2;for(i=3;i<NUM+10;i++){//把200以内的斐波那契数列存到二维数组中 for(j=0;j<MAX;j++){a[i][j]+=(a[i-1][j]+a[i-2][j]);if(a[i][j]>=10){a[i][j+1]=a[i][j]/10;a[i][j]=a[i][j]%10;}}}int T,n;char s[NUM+10];scanf("%d",&T);while(T--){    scanf("%s",s);    n=strlen(s);    //printf("%d#\n",n);    for(i=MAX-1;i>0;i--)    if(a[n][i]!=0) break;    for(;i>=0;i--)    printf("%d",a[n][i]);    printf("\n");}return 0;}

只要能看出是斐波那契数列就可以了，然后用大数加法