hduoj 1865 1sting

1sting

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7255    Accepted Submission(s): 2835

 

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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#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int sum=0;

int b[203][205]={0};

void pre()
{
	int i,j;
	b[0][0]=1;
	b[1][0]=2;
	b[2][0]=3;
	int f=1;
	int add=0;
	for(i=3;i<202;i++)
	{
		for(j=0;j<23;j++){     // 这个地方哇了10次 
			b[i][j]=(b[i-1][j]+b[i-2][j]+add);
			add=b[i][j]/10000;
			b[i][j]%=10000;
		
		}
		if(add>0)
		{
			b[i][j]=add;
			j++;
		}
		f=j;
	
	}
}
int main()
{string a;
	int n,i;
	pre();
	cin>>n;
	while(n--)
	{
		cin>>a;
		int n=a.length()-1;
	    if(n==0)
	    {
	    	cout<<1<<endl;
	    	continue;
		}
		i=204;
		while(b[n][i]==0)
		i--;
		printf("%d",b[n][i]);
		for(i--;i>=0;i--)
		printf("%04d",b[n][i]);
		printf("\n");
		a.clear();
	}
	return 0;
}