hdu 3555 Bomb

Bomb

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)

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

数位DP

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define eps 1e-4
#define bug(x)  cout<<"bug"<<x<<endl;
const int N=2e2+10,M=1e6+10,inf=2147483647;
const ll INF=1e18+10,mod=2147493647;
int bit[N];
ll f[N][20];
ll dp(int pos,int pre,int flag)
{
    if(pos==0)return 1;
    if(flag&&f[pos][pre])return f[pos][pre];
    int x=flag?9:bit[pos];
    ll ans=0;
    for(int i=0;i<=x;i++)
    {
        if(pre!=4||(pre==4&&i!=9))
            ans+=dp(pos-1,i,flag||i<x);
    }
    return f[pos][pre]=ans;
}
ll getans(ll x)
{
    memset(f,0,sizeof(f));
    int len=0;
    while(x)
    {
        bit[++len]=x%10;
        x/=10;
    }
    return dp(len,0,0);
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        ll n;
        scanf("%lld",&n);
        printf("%lld
",n-getans(n)+1);
    }
    return 0;
}