【数位DP-板子题目】HDU-3555-Bomb- [只要49]

Bomb

Time Limit: / MS (Java/Others)    Memory Limit: / K (Java/Others)
Total Submission(s):     Accepted Submission(s): 

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  to N. If the current number sequence includes the sub-sequence "", 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 ( <= T <= ), indicating the number of test cases. For each test case, there will be an integer N ( <= N <= ^-) 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

Hint
From  to , the numbers that include the sub-sequence "" are "","","","","","","","","","","","","","","",
so the answer is .

题解：

#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
#include <queue>
#include <stack>
#include <vector>
#include <math.h>
#include <string.h>
#include<set>
using namespace std;
#define inf 0x3f3f3f3f
const double pi=acos(-1.0);    ///数位DP，只要49
#define ll long long
#define lson root<<1
#define rson root<<1|1
const ll mod = ;
#define ull unsigned long long
ll digit[];
ll dp[][];
    ///limit表示是否受限，最初进行的时候必定受限！
ll Cul(int len,bool if4,bool limit){

    if(len<)return ;
    if(!limit&&dp[len][if4]!=-)return dp[len][if4];
    int up_bound=(limit==)?digit[len]:;
    ll ans=;
    for(int i=;i<=up_bound;i++)
    {
        if(if4&&i==)
            continue;
        ans+=Cul(len-,i==,limit&&i==digit[len]);
    }
    dp[len][if4]=ans;
    return ans;
}

ll solve(ll n){
    int k=;
    while(n){
        digit[++k]=n%;
        n/=;
    }
    memset(dp,-,sizeof(dp));
    return Cul(k,false,true);
}

int main(){

    ll n,T;
    cin>>T;
    while(T--){
        scanf("%lld",&n);
        printf("%lld\n",n-solve(n)+);

    }

    return ;
}