HDU3555 Bomb[数位DP]

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=25,INF=1e9+5;
typedef long long ll;
inline ll read(){
    char c=getchar();ll x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
    return x*f;
}
ll n;
ll d[N][4];
void dp(){
    d[0][0]=1;
    for(int i=1;i<=20;i++){
        d[i][0]=10*d[i-1][0]-d[i-1][1];//!
        d[i][1]=d[i-1][0];
        d[i][2]=10*d[i-1][2]+d[i-1][1];
        //printf("d %d %d\n",d[i][0],d[i][2]);
    }
}
ll sol(ll n){
    int a[N],len=0,flag=0;
    ll ans=0;
    while(n) a[++len]=n%10,n/=10;
    a[len+1]=0;
    for(int i=len;i>=1;i--){
        ans+=d[i-1][2]*a[i];
        
        if(flag) ans+=a[i]*d[i-1][0];
        else if(a[i]>4) ans+=d[i-1][1];//maybe 49
        if(a[i+1]==4&&a[i]==9) flag=1;//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));
    }
}