Bomb HDU - 3555 （数位DP）

Posted 2021-02-06 smallhester

Bomb HDU - 3555 （数位DP）

 

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? 

InputThe 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. 
OutputFor 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.

题意：求1-n之间有几个数带有49
题解：一题数位DP，可以用记忆化搜索解决，感觉板子还是有点懂了，但不知道改了之后会不会了，具体的dfs的每一步的具体操作已经放在代码里了

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<sstream>
#include<cmath>
#include<stack>
#include<cstdlib>
#include <vector>
#include<queue>
using namespace std;

using namespace std;
#define ms(a,b) memset(a,b,sizeof(a))
#define lson rt*2,l,(l+r)/2
#define rson rt*2+1,(l+r)/2+1,r
typedef unsigned long long ull;
typedef long long ll;
const int MAXN=1e4+5;
const double EPS=1e-8;
const int INF=0x3f3f3f3f;
const int MOD = 1e9+7;

#define ll long long
#define llu unsigned long long
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
const int maxn =  1e5+5;
const int  mod = 1e9+7;

int bit[40];
ll f[40][4];
ll dp(int pos,int st,bool flag)
{
    //pos为位，st为状态，st=0：没有49，st=1：前一位为4，st=2，表示49都已经出现过了
    //flag表示高位与原数是否相同
    if(pos == 0)
        return st == 2;
    if(flag && f[pos][st] != -1)
        return f[pos][st];
    ll ans = 0;
    int x;
    if(flag)
        x = 9;
    else
        x = bit[pos];
    //cout<<x<<endl;
    for(int i = 0;i <= x; i++)
    {
        if((st == 2) || (st == 1 && i == 9))    //如果上一位已经有49（st==2）或者上一位为4（st==1）当前位为9
            ans += dp(pos-1,2,flag || i<x);     //只能加上上一个状态，且上一个状态一定为已经有49了
        else if(i == 4)
            ans += dp(pos-1,1,flag || i<x);     //当前为4和前一位也为4的状态是一样的
        else                                    //维持st=0的状态
            ans += dp(pos-1,0,flag || i<x);
    }
    if(flag)
        f[pos][st] = ans;   //记忆化
    return ans;
}

ll calc(ll x)
{
    int len = 0;
    while(x)
    {
        bit[++len] = x % 10;
        x /= 10;
    }
    //cout<<len<<endl;
    return dp(len,0,0);
}
int main()
{
    int t;
    scanf("%d",&t);
    memset(f,-1,sizeof f);
    while(t--)
    {
        ll n;
        scanf("%lld",&n);
        printf("%lld
",calc(n));
    }
}