BNUOJ 52325 Increasing or Decreasing 数位dp

传送门：BNUOJ 52325 Increasing or Decreasing题意：求[l,r]非递增和非递减序列的个数思路：数位dp，dp[pos][pre][status]

1. pos：处理到第几位
2. pre：前一位是什么
3. status：是否有前导零

递增递减差不多思路，不过他们计算的过程中像5555,444 这样的重复串会多算，所以要剪掉。个数是(pos-1)*9+digit[最高位]，比如一位重复子串是：1,2,3,4...9,9个，二位重复子串：11,22,33,44,...,99,9个；同理，其他类推；

不过这个题如果dp值每算完一个[l,r]就清零，会超时。那么我们这么分析，算[l1,r1]，[l2,r2]这两个区间时，dp是否真的有必要清零呢，答案是否定的，记忆化搜索的过程中记录的dp值如果计算过，那么当其他值算到他时，这个值是可以用的。具体的自己想想就好了

/**************************************************************
    Problem:BNUOJ 52325 Increasing or Decreasing
    User: youmi
    Language: C++
    Result: Accepted
    Time:    380 ms
    Memory:    1632 KB
****************************************************************/
//#pragma comment(linker, "/STACK:1024000000,1024000000")
//#include<bits/stdc++.h>
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <cmath>
#include <queue>
#include <deque>
#include <string>
#include <vector>
#define zeros(a) memset(a,0,sizeof(a))
#define ones(a) memset(a,-1,sizeof(a))
#define sc(a) scanf("%d",&a)
#define sc2(a,b) scanf("%d%d",&a,&b)
#define sc3(a,b,c) scanf("%d%d%d",&a,&b,&c)
#define scs(a) scanf("%s",a)
#define sclld(a) scanf("%lld",&a)
#define pt(a) printf("%d\n",a)
#define ptlld(a) printf("%lld\n",a)
#define rep(i,from,to) for(int i=from;i<=to;i++)
#define irep(i,to,from) for(int i=to;i>=from;i--)
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
#define lson (step<<1)
#define rson (lson+1)
#define eps 1e-6
#define oo 0x3fffffff
#define TEST cout<<"*************************"<<endl
const double pi=4*atan(1.0);

using namespace std;
typedef long long ll;
template <class T> inline void read(T &n)
{
    char c; int flag = 1;
    for (c = getchar(); !(c >= ‘0‘ && c <= ‘9‘ || c == ‘-‘); c = getchar()); if (c == ‘-‘) flag = -1, n = 0; else n = c - ‘0‘;
    for (c = getchar(); c >= ‘0‘ && c <= ‘9‘; c = getchar()) n = n * 10 + c - ‘0‘; n *= flag;
}
ll Pow(ll base, ll n, ll mo)
{
    ll res=1;
    while(n)
    {
        if(n&1)
            res=res*base%mo;
        n>>=1;
        base=base*base%mo;
    }
    return res;
}
//***************************

int n;
const int maxn=100000+10;
const ll mod=1000000007;
int digit[30];
ll dp0[20][20][2];
ll dp1[20][20][2];
int tot=0;
ll dfs0(int pos,int pre,int status,int limit)
{
    if(pos<0)
        return status;
    if(!limit&&dp0[pos][pre][status]!=-1)
        return dp0[pos][pre][status];
    int ed=limit?digit[pos]:9;
    ll res=0;
    if(status==0)
    {
        for(int i=0;i<=min(pre,ed);i++)
        {
            if(i==0)
                res+=dfs0(pos-1,10,0,limit&&(i==ed));
            else
                res+=dfs0(pos-1,i,1,limit&&(i==ed));
        }
    }
    else
    {
        for(int i=0;i<=min(ed,pre);i++)
            res+=dfs0(pos-1,i,status,limit&&(i==ed));
    }
    if(!limit)
        dp0[pos][pre][status]=res;
    return res;
}
ll dfs1(int pos,int pre,int status,int limit)
{
    if(pos<0)
        return status;
    if(!limit&&dp1[pos][pre][status]!=-1)
        return dp1[pos][pre][status];
    int ed=limit?digit[pos]:9;
    ll res=0;
    for(int i=pre;i<=ed;i++)
        res+=dfs1(pos-1,i,status||i,limit&&(i==ed));
    if(!limit)
        dp1[pos][pre][status]=res;
    return res;
}
void work(ll num)
{
    tot=0;
    while(num)
    {
        digit[tot++]=num%10;
        num/=10;
    }
}
ll solve(ll num)
{
    if(num==0)
        return 0;
    ll ans=(tot-1)*9+digit[tot-1];
    ll temp=0;
    int tt=0;
    while(tt<tot)
        temp=temp*10+digit[tot-1],tt++;
    if(temp>num)
        ans--;
    return ans;
}
int main()
{
    //freopen("in.txt","r",stdin);
    int T_T;
    scanf("%d",&T_T);
    ones(dp0);
    ones(dp1);
    for(int kase=1;kase<=T_T;kase++)
    {
        ll num;
        read(num);
        num--;
        work(num);
        ll temp0=dfs0(tot-1,10,0,1);
        temp0+=dfs1(tot-1,0,0,1);
        temp0-=solve(num);
        read(num);
        work(num);
        ll temp1=dfs0(tot-1,10,0,1);
        temp1+=dfs1(tot-1,0,0,1);
        temp1-=solve(num);
        ptlld(temp1-temp0);
    }
    return 0;
}