数位DP之找等凹数字

题目链接

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <time.h>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
using namespace std;
long long dp[20][20][10][2][2][2];
int num[20];
int s[20];
//第 i 位，长度为 len，上一位是什么，前面是否递增过，前面是否递减过，当前是否符合回文串的性质,flag就是普通的前一位是否有限制，q是盘是否前导0
long long rec(int i,int pre,int up,int down,int flag,int q,int len,int ispa)
{
    if(i<0)
        return up&&down&&ispa;
    if(dp[i][len][pre][up][down][ispa]!=-1&&!flag&&!q)
        return dp[i][len][pre][up][down][ispa];
    long long res=0;
    int o=s[i];
    for(int j=0;j<10;j++)
    {
        num[i]=j;
        if(j>o&&flag)
            break;
        if(q)
            res+=rec(i-1,j,0,0,j<o?0:flag,q&&j==0,len-(q&&j==0),ispa);
        else if(j==pre)
        {
            if(ispa&&i<len/2)
                res+=rec(i-1,j,up,down,j<o?0:flag,q&&j==0,len,j==num[len-i-1]);
            else
                res+=rec(i-1,j,up,down,j<o?0:flag,q&&j==0,len,ispa);
        }
        else if(j>pre)
        {
            if(!down)
                continue;
            if(ispa&&i<len/2)
                res+=rec(i-1,j,1,down,j<o?0:flag,q&&j==0,len,j==num[len-i-1]);
            else
                res+=rec(i-1,j,1,down,j<o?0:flag,q&&j==0,len,ispa);
        }
        else if(j<pre)
        {
            if(up)
                continue;
            if(ispa&&i<len/2)
                res+=rec(i-1,j,up,1,j<o?0:flag,q&&j==0,len,j==num[len-i-1]);
            else
                res+=rec(i-1,j,up,1,j<o?0:flag,q&&j==0,len,ispa);
        }
    }
    if(!flag&&!q)
        dp[i][len][pre][up][down][ispa]=res;
    return res;
}
long long cal(long long x)
{
    int len=0;
    while(x)
    {
        s[len++]=x%10;
        x/=10;
    }
    return rec(len-1,0,0,0,1,1,len,1);
}
int main()
{
    memset(dp,-1,sizeof(dp));
    long long l,r;
    int t;
    scanf("%d",&t);
    while(t--){
    scanf("%lld%lld",&l,&r);
    printf("%lld\n",cal(r)-cal(l-1));
    }
    return 0;
}

未检测，不能交题，但是应该是对的
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <time.h>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
using namespace std;
typedef long long ll;
int a[20];
int keep[20];
int dp[20][20][20][2][2][2];
ll dfs(int len,int pos,int pre,int up,int down,int limit,int flag0,int judge)
{
    if(pos<0)
        return up&&down&&judge;
    if(!flag0&&!limit&&dp[len][pos][pre][up][down][judge]!=-1)
        return dp[len][pos][pre][up][down][judge];
    ll sum=0;
    int tmp=a[pos];
    for(int i=0;i<=9;i++)
    {
        keep[pos]=i;
        if(limit&&i>tmp)
            break;
        if(flag0)
            sum+=dfs(len-(flag0&&i==0),pos-1,i,0,0,limit&&i==tmp,flag0&&i==0,judge);
        else if(i==pre)
        {
            if(judge&&pos<len/2)
                sum+=dfs(len,pos-1,i,up,down,limit&&i==tmp,flag0&&i==0,i==keep[len-pos-1]);
            else
                sum+=dfs(len,pos-1,i,up,down,limit&&i==tmp,flag0&&i==0,judge);
        }
        else if(i>pre)
        {
            if(!down)
                continue;
            if(judge&&pos<len/2)
                sum+=dfs(len,pos-1,i,1,down,limit&&i==tmp,flag0&&i==0,i==keep[len-pos-1]);
            else
                sum+=dfs(len,pos-1,i,1,down,limit&&i==tmp,flag0&&i==0,judge);
        }
        else if(i<pre)
        {
            if(up)
                continue;
            if(judge&&pos<len/2)
                sum+=dfs(len,pos-1,i,up,1,limit&&i==tmp,flag0&&i==0,i==keep[len-pos-1]);
            else
                sum+=dfs(len,pos-1,i,up,1,limit&&i==tmp,flag0&&i==0,judge);
        }
    }
    if(!flag0&&!limit)
        dp[len][pos][pre][up][down][judge]=sum;
    return sum;
}
ll solve(ll x)
{
    if(x<=100)
        return 0;
    int pos=0;
    while(x)
    {
        a[pos++]=x%10;
        x/=10;
    }
    return dfs(pos,pos-1,0,0,0,1,1,1);
}
int main()
{
    memset(dp,-1,sizeof(dp));
    ll l,r;
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lld%lld",&l,&r);
        printf("%lld\n",solve(r)-solve(l-1));
    }
    return 0;
}