HDU3555 Bomb[数位DP]

时间：2022-05-29 03:47:12

Bomb

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 16362    Accepted Submission(s): 5979

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.

和 不要62 类似
注意：
1. long long
2.if(a[i+1]==4&&a[i]>=9) flag=1;
不能+d[i-1][0]，因为9的特殊性不可能有比9大的个位数使后面可能出现49，天际线就是49了
#include <iostream>

#include <cstdio>

#include <algorithm>

#include <cstring>

#include <map>

using namespace std;

const int N=,INF=1e9+;

typedef long long ll;

inline ll read(){

    char c=getchar();ll x=,f=;

    while(c<''||c>''){if(c=='-')f=-;c=getchar();}

    while(c>=''&&c<=''){x=x*+c-'';c=getchar();}

    return x*f;

}

ll n;

ll d[N][];

void dp(){

    d[][]=;

    for(int i=;i<=;i++){

        d[i][]=*d[i-][]-d[i-][];//!

        d[i][]=d[i-][];

        d[i][]=*d[i-][]+d[i-][];

        //printf("d %d %d\n",d[i][0],d[i][2]);

    }

}

ll sol(ll n){

    int a[N],len=,flag=;

    ll ans=;

    while(n) a[++len]=n%,n/=;

    a[len+]=;

    for(int i=len;i>=;i--){

        ans+=d[i-][]*a[i];

        if(flag) ans+=a[i]*d[i-][];

        else if(a[i]>) ans+=d[i-][];//maybe 49

        if(a[i+]==&&a[i]==) flag=;//cannot +d[i-1][0],cause skyline and flag=1

        //printf("%d %d %d\n",i,ans,flag);

    }

    if(flag) ans++;

    return ans;

}

int main(){

    dp();

    int T=read();

    while(T--){

        n=read();

        printf("%lld\n",sol(n));

    }

}

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