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

Sample Output

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\n",calc(n));

    }

}

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Bomb HDU - 3555 （数位DP）

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