1sting
Time Limit : 5000/1000ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)
Total Submission(s) : Accepted Submission(s) :
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
这道题和大菲波数一样的,就多了些字符串中字符数统计,一个strlen函数就能解决,后面怎么办,自己回顾吧!
代码如下:
#include<stdio.h>
#include<string.h>
int main ()
{
int a[201][100];
int T,i,c,n,m,j;
memset(a,0,sizeof(a));
a[1][0]=1;a[2][0]=2;
for(c=0,i=3;i<201;i++)
for(j=0,m=0;j<=100;j++)
{
c=a[i-1][j]+a[i-2][j]+m;
a[i][j]=c%10;
m=c/10;
}
scanf("%d",&T);
getchar();
while(T--)
{
char b[201];
int l,k;
gets(b);
l=strlen(b);
for(i=99,k=0;i>=0;i--)
{
if(a[l][i]!=0)
k++;
if(k!=0)
printf("%d",a[l][i]);
}
printf("\n");
}
return 0;
}
原文:http://blog.csdn.net/hanhai768/article/details/23606181