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本文介绍了一种使用动态规划解决密码组合问题的方法，通过给定的字符串还原可能的密码组合数量，每组数字范围为0到255且不包含前导零。

D Password

Accept:25	Submit:68
Time Limit:1000MS	Memory Limit:65536KB

Description

ykwd's password is a number sequence. Every number in this sequence is no less than 0 and no larger than 255, and without leading zeros. ykwd wrote his password down ,but he didn't add spaces between numbers. For example, if his password is 1 34 8 205,he will write down 1348205 . Now, he has forgot his password, he only has the string he has written down, he wants to know how many different possible passwords he can have?

Input

The first line has one interger t, denoting the number of cases.
Nest t lines, each line includes a string ykwd wrote down.
You can suppose that each string is no longer than 1000.

Output

For each case you should output one line. Because the answer may be very large, you need only ouput the answer mod 1234567.

Sample Input

2
1234
1024

Sample Output

7
5

submit Discuss

/*

简单的动态规划。。 

*/ 
#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
#define manx 1009

long long dp[manx];

int main(){
    int t;
    cin>>t;
    while(t--){
        memset(dp,0,sizeof(dp));
        string s;
        cin>>s;
        if(s.size()>=1)  dp[0]=1;
        if(s.size()>=2){
            int sum = s[1]-'0';
            sum += (s[0]-'0')*10;
            if(sum>=10) dp[1]=2; 
            else dp[1]=1;
        }    
        
        if(s.size()>=3){
            int sum = s[2]-'0';
            sum += (s[1]-'0')*10;
            dp[2]+=dp[1];
            if(sum>=10) dp[2]+=1;
            sum += (s[0]-'0')*100;
            if(sum >= 100 && sum<=255) dp[2]+=1;
        }
        
        for(int i=3;i<s.size();i++){
            int sum = s[i]-'0';
            dp[i]+=dp[i-1];  dp[i]%=1234567;
            
            sum = sum + (s[i-1]-'0')*10;
            if(sum >= 10) dp[i]+=dp[i-2];  dp[i]%=1234567;
            
            sum = sum + (s[i-2]-'0')*100;
            if(sum >= 100 && sum<=255) dp[i]+=dp[i-3];  dp[i]%=1234567;
        }        
        cout<<dp[s.size()-1]<<endl;
    }
}