HDU3555Bomb数位DP入门题目

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3555

Bomb

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 6448    Accepted Submission(s): 2242

Problem Description

The counter-terrorists found a time bomb in the dust. But this time the terrorists improve on the time bomb. The number sequence of the time bomb counts from 1 to N. If the current number sequence includes the sub-sequence "49", the power of the blast would add one point.
Now the counter-terrorist knows the number N. They want to know the final points of the power. Can you help them?

 

Input

The first line of input consists of an integer T (1 <= T <= 10000), indicating the number of test cases. For each test case, there will be an integer N (1 <= N <= 2^63-1) as the description.

The input terminates by end of file marker.

 

Output

For each test case, output an integer indicating the final points of the power.

 

Sample Input

   
   
   
   

    
    
    
    3
1
50
500
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    0
1
15


    
    
    
    
 
     
 
      Hint 
     
From 1 to 500, the numbers that include the sub-sequence "49" are "49","149","249","349","449","490","491","492","493","494","495","496","497","498","499", so the answer is 15. 
    
    
    
    
     
   
   
   
   

 

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;

long long  dp[33][3];
long long  DP(long long N)
{
    int bit[33];
            N++;
    long long sum=0;
    int len=0;
        while(N){
            bit[++len]=N%10;
            N/=10;
        }
        bit[len+1]=0;
        bool flag=0;
        for(int i=len;i>=1;i--){
            sum+=dp[i-1][2]*bit[i];
            if(!flag&&bit[i]>4)  sum+=dp[i-1][1];
            if(flag)    sum+=dp[i-1][0]*bit[i];
            if(bit[i+1]==4&&bit[i]==9)  flag=1;
        }
        return sum;
}
void init()
{
    memset(dp,0,sizeof(dp));dp[0][0]=1;
    for(int i=1;i<=20;i++){
        dp[i][0]=dp[i-1][0]*10-dp[i-1][1];
        dp[i][1]=dp[i-1][0];
        dp[i][2]=dp[i-1][1]+dp[i-1][2]*10;
    }
}
int main()
{
    init();
    int T;
    long long x;
    scanf("%d",&T);
    while(T--){
         cin>>x;
        cout<<DP(x)<<endl;
    }
    return 0;
}