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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<<"*************************"<<endlconst 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;}

 

BNUOJ 52325 Increasing or Decreasing 数位dp