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

  
  

   
   3
1
11
11111
  
  

 

Sample Output

  
  

   
   1
2
8
  
  

 

#include<stdio.h>
#include<string.h>
#define NUM 200
#define MAX 100
int 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