HDU 3555 Bomb （数位dp）

Bomb

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 15699    Accepted Submission(s): 5715

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.

Author

fatboy_cw@WHU

Source

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

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题解：把HDU 2089 改改就行了.......我还因为一个if判断条件吃了几发TLE....

AC代码：

//#include<bits/stdc++.h>
#include<iostream>
#include<stdio.h>
#include<cstring>
#include<string.h>
using namespace std;
typedef long long ll;
int dig[63];
ll dp[63][2][63];

ll dfs(int pos,bool have,int last,int flag)
{
    int i;
    if (pos < 0) return have;
    if (flag==0 && dp[pos][have][last]!=-1) return dp[pos][have][last];
    
    int n=flag ? dig[pos] : 9;
    ll ans=0;
    for(i=0;i<=n;i++)
    {
        if (last==4 && i==9) ans += dfs(pos-1,true,i,flag && (i==n));
        else ans+=dfs(pos-1,have,i,flag && (i==n));
    }
    if (flag==0)
    {
        dp[pos][have][last]=ans;
    }
    return ans;
}
ll solve(ll x)
{
    int len = 0; 
    while(x)
    {
        dig[len++] = x % 10;
        x /= 10;
    }
    return dfs(len-1,0,0,1);
}

int main()
{
    ll x;
    int t;
    scanf("%d",&t);
    while(t--)
    {
        memset(dp,-1,sizeof(dp));
        scanf("%I64d",&x);
        printf("%I64d\n",solve(x));
    }
    return 0;
}