hduoj 1865 1string 【大数】【菲波那切数列】

1sting

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 5310    Accepted Submission(s): 2030

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

【思路】1 先求题意，我们发现是s[1]=1,s[2]=2, 其余的 n>=3 时s[n]=s[n-1]+s[n-2];                
       2 此题最长的列为200，所以轻轻松松就超出int ，long long的范围，我们要改用数组a[205][102],来存储每一位上的数字，(前者[205]记录输入的1的个数，后者[102]记录结果的每一位)。

AC代码：

#include<stdio.h>
#include<string.h>
int a[205][102];
//注意此处要比底下函数中的j的最大值开的大一点
void count()
{
	int i,j,p,q;
	memset(a,0,sizeof(a));//数组清零
	a[1][0]=1;a[2][0]=2;
	for(i=3;i<203;i++)
    //以下步骤模拟大数计算，初始化斐波那契数列
	{
		p=q=0;
		for(j=0;j<=100;j++)
		{
			p=a[i-1][j]+a[i-2][j]+q;
			a[i][j]=p%10;
			q=p/10;
		}printf("  %d  ",a[i][0]);
	}
}
int main()
{
	count();
	int n,i,j,len;
	char s[205];
	scanf("%d",&n);
	while(n--)
	{
		getchar();
		scanf("%s",s);
		len=strlen(s);
		for(i=100;i>=0;i--)//找到数值的最后一位
		    if(a[len][i]!=0)
		       break;
		for(j=i;j>=0;j--)
          // 注意上面的函数计算的值的数位是逆序的
		   printf("%d",a[len][j]);
		printf("
");
	}
	return 0;
}