【杭电-oj】-1865-1sting（大数斐波那契数列）

1sting

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5021    Accepted Submission(s): 1864

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

 

这道题也是斐波那契数列，只是需要用字符串把1的个数转换为n而已。

#include
#include
int main()
{
	int t,l,k;
	char a[201];
	int f[1001][211]={0};			//全部初始化为0 
	f[1][1]=1;
	f[2][1]=2;
	for(int i=3;i<=1000;i++)
	{
		for(int j=1;j<=210;j++)			//二维数组，i表示第几个数，j表示这个数的第几位，
		{								// 然后对j进行类似大数相加的运算 
			f[i][j]+=f[i-1][j]+f[i-2][j];
			if(f[i][j]>9)
			{
				f[i][j]-=10;
				f[i][j+1]++;	
			}	
		}	
	}	
	scanf("%d",&t);
	while(t--)
	{
		scanf("%s",a);
		l=strlen(a);
		k=100;
		for( ;f[l][k]==0;k--);			//开始全部初始化为零，直到遇到一个f[n][k]!=0,此时跳出循环，开始输出 
		for( ;k>=1;k--)					//这种倒着输出不同于之前的倒着输出，此时不用考虑是否进位，只要不超出范围即可 
			printf("%d",f[l][k]);
		printf("\n");
	}
	return 0;
}