Bomb

Posted 2020-08-12 mxzf0213

Bomb

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

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
分析：数位dp；
代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
#define mod 1000000007
#define inf 0x3f3f3f3f
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
#define pii pair<int,int>
#define Lson L, mid, ls[rt]
#define Rson mid+1, R, rs[rt]
const int maxn=1e5+10;
using namespace std;
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
inline ll read()
{
    ll x=0;int f=1;char ch=getchar();
    while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
    while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
    return x*f;
}
int m,k,t,d[20];
ll n,dp[20][3];
ll dfs(int now,int ca,int ok)
{
    if(now<0)return ca==2;
    if(ok&&dp[now][ca]!=-1)return dp[now][ca];
    int len=ok?9:d[now];
    ll ans=0;
    for(int i=0;i<=len;i++)
    {
        if(ca==0)
        {
            if(i==4)ans+=dfs(now-1,1,ok||i<len);
            else ans+=dfs(now-1,0,ok||i<len);
        }
        else if(ca==1)
        {
            if(i==9)ans+=dfs(now-1,2,ok||i<len);
            else if(i==4)ans+=dfs(now-1,1,ok||i<len);
            else ans+=dfs(now-1,0,ok||i<len);
        }
        else if(ca==2)
        {
            ans+=dfs(now-1,2,ok||i<len);
        }
    }
    if(ok)dp[now][ca]=ans;
    return ans;
}
ll solve(ll p)
{
    int num=0;
    memset(dp,-1,sizeof(dp));
    while(p)
    {
        d[num++]=p%10;
        p/=10;
    }
    return dfs(num-1,0,0);
}
int main()
{
    int i,j;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lld",&n);
        printf("%lld\n",solve(n));
    }
    //system("Pause");
    return 0;
}