HDU 1865 1sting

1sting

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4151    Accepted Submission(s): 1559

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

 /*采用递推的思想，求出f(n)=f(n-2)+f(n-1),因为数据量大，采用压位高精度方法，采用普通高精度超时了*/

代码如下

#include<iostream>#include<cstring>#include<cstdio>#define maxn 1000000using namespace std;int a[10000][300];/*一个数组单元存放8位数，总共可以存放2400位数*/char b[maxn];int main(){int i,j,len,t;a[1][0]=1;a[2][0]=2;for(i=3;i<10000;i++){for(j=0;j<300;j++) a[i][j]+=a[i-1][j]+a[i-2][j];for(j=0;j<300;j++){if(a[i][j]>=100000000) a[i][j+1]+=a[i][j]/100000000,a[i][j]%=100000000;}}cin>>t;while(t--){cin>>b;len=strlen(b);int max=299;while(!a[len][max]) max--;cout<<a[len][max];for(int i=max-1;i>=0;i--) printf("%08d",a[len][i]);cout<<endl;  } return 0;}