HDU-3555-Bomb-数位dpBomb

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概述

Bomb

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 9396 Accepted Submission(s): 3314

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

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.

Author

fatboy_cw@WHU

Source

2010 ACM-ICPC Multi-University Training Contest（12）——Host by WHU

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数位dp ，和hdu-2089是一样的方法。 hdu-2089题解

需要注意的是一定要用__int64 , long long 会wa。

#include<iostream>
#include<cstdio>
#include<vector>
#include<cstring>
#include<algorithm>
#include<fstream>
using namespace std;

ifstream fin("cin.in");
ofstream fout("test.out");

__int64 n;
int num,  pre;
__int64 ans;
__int64 dp[25][2];

void solve()
{
    memset(dp, 0, sizeof(dp));
    ans=1;
    dp[0][0]=1;
    pre=0;
    __int64 t=n;
    for(int i=1;;i++){
        num=n%10;
         n/=10;
        if(num<=4&&num>0){

            if(num==4&&pre==9){

                ans=dp[i-2][0] * pre -  dp[i-2][1];

            }

            ans+=dp[i-1][0]*num;

        }

        if(num>4){

            ans+=dp[i-1][0]*num - dp[i-1][1];

        }

        if(n==0) break;

        dp[i][0]=dp[i-1][0]*10 - dp[i-1][1];
        dp[i][1]=dp[i-1][0];

        pre=num;
    }
    cout<<t+1-ans<<endl;
    //printf("%I64dn", t+1-ans);
    return ;
}
int main()
{

    int T;
    cin>>T;
    while(T--){
        cin>>n;
        solve();
    }
    return 0;
}