hdu1865 (high-precision, law-seeking)

Source: Internet
Author: User

1sting Time limit:5000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)
Total submission (s): 2599 Accepted Submission (s): 1032


Problem Description you'll be given a string which only contains ' 1 '; You can merge both 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 was to find the total number of result can get.

Input the first line was a number n refers to the number of test cases. Then n lines follows, each line has a string of made up of ' 1 '. The maximum length of the sequence is 200.

Output of the output contain n lines, each line output the number of the result can get.

Sample Input

3 1 11 11111
Sample Output
1 2 8

To find a recursive relationship, the end can be 1 or 2, 1 is the time F (x-1), is 2 of the time for F (x-2);

So f (x0=f (x-1) +f (x-2); Fibonacci sequence

#include <iostream>
#include <cstring>
#include <string>
using namespace std;
const int MAXN = $, MAXM = +;
int ANS[MAXN][MAXM];

void Init () {
    ans[1][0]=1;
    ans[2][0]=2;
    for (int i=3, i<maxn; i++) {for
        (int j=0; j<maxm; J + +) {
            ans[i][j]+=ans[i-1][j]+ans[i-2][j];
            if (ans[i][j]>9) {ans[i][j+1]+=ans[i][j]/10; ans[i][j]%=10;}

}} void print (int n) {
    int i = MAXM-1;
    while (ans[n][i]==0) i--;
    while (i!=-1) cout<<ans[n][i--];
    cout<<endl;
}

int main ()
{
    init ();
    int t;
    String str;
    cin>>t;
    while (t--) {
        cin>>str;
        Print (Str.size ());
    }
    return 0;
}



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