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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
311111111
 


Sample Output
128

【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("\n");	}	return 0;}

hduoj 1865 1string 【大数】【菲波那切数列】