BNUOJ 52325 Increasing or Decreasing 数位dp

题目链接：

https://acm.bnu.edu.cn/v3/contest_show.php?cid=8506#problem/I

I. Increasing or Decreasing

Case Time Limit: 1000ms
Memory Limit: 524288KB

题意

求[l,r]上，数位满足非递增获非递减的数的个数。

题解

1、dp[i][j][k]表示低i位的第i位为j的，k1时表示满足非递减，k2时满足非递增，k==0时表示每一位都相等。

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<ctime>
#include<vector>
#include<cstdio>
#include<string>
#include<bitset>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)
#define scf scanf
#define prf printf

typedef long long LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;

const int INF=0x3f3f3f3f;
const LL INFL=0x3f3f3f3f3f3f3f3fLL;
const double eps=1e-8;
const double PI = acos(-1.0);

//start----------------------------------------------------------------------

const int maxn=22;

int arr[maxn],tot;
///type根据dp具体维数调整
LL dp[maxn][11][4];
///ismax标记表示前驱是否是边界值
///ser标记前驱是否是前导零
LL dfs(int len,int dig,int type,bool ismax,bool iszer) {
    if (len == 0) {
        ///递归边界，这说明前驱都合法了
        return 1LL;
    }
    if (!ismax&&dp[len][dig][type]>=0) return dp[len][dig][type];
    LL res = 0;
    int ed = ismax ? arr[len] : 9;

    ///这里插入递推公式
    for (int i = 0; i <= ed; i++) {
        if(i==0&&iszer){
            ///处理前导零
            res+=dfs(len-1,10,0,ismax&&i==ed,1);
        }else{
            if(type==0){
                if(i==dig||dig==10) res+=dfs(len-1,i,0,ismax&&i==ed,0);
                else if(i>dig) res+=dfs(len-1,i,1,ismax&&i==ed,0);
                else if(i<dig) res+=dfs(len-1,i,2,ismax&&i==ed,0);
            }else if(type==1){
                if(i>=dig){
                    res+=dfs(len-1,i,type,ismax&&i==ed,0);
                }
            }else if(type==2){
                if(i<=dig){
                    res+=dfs(len-1,i,type,ismax&&i==ed,0);
                }
            }
        }
    }
    return ismax ? res : dp[len][dig][type] = res;
}

LL solve(LL x) {
    tot = 0;
    while (x) { arr[++tot] = x % 10; x /= 10; }
    LL ret=0;
    return dfs(tot,10,0,true,true);
}

void init() {
    clr(dp,-1);
}

int main() {
    init();
    int n; scf("%d",&n);
    while(n--){
        LL l,r;
        scf("%lld%lld",&l,&r);
        prf("%lld
",solve(r)-solve(l-1));
    }
    return 0;
}

//end-----------------------------------------------------------------------

2、状态和上面差不多，不过是固定k，分别求出来，既ans=非递增+非递减-每个数位都相同。

#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<stack>
#include<ctime>
#include<vector>
#include<cstdio>
#include<string>
#include<bitset>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define sz() size()
#define pb(v) push_back(v)
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)
#define scf scanf
#define prf printf

typedef long long LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;

const int INF=0x3f3f3f3f;
const LL INFL=0x3f3f3f3f3f3f3f3fLL;
const double eps=1e-8;
const double PI = acos(-1.0);

//start----------------------------------------------------------------------

const int maxn=22;

int arr[maxn],tot;
///type根据dp具体维数调整
LL dp[maxn][11][4];
///ismax标记表示前驱是否是边界值
///ser标记前驱是否是前导零
LL dfs(int len,int dig,int type,bool ismax,bool iszer) {
    if (len == 0) {
        ///递归边界，这说明前驱都合法了
        return 1LL;
    }
    if (!ismax&&dp[len][dig][type]>=0) return dp[len][dig][type];
    LL res = 0;
    int ed = ismax ? arr[len] : 9;

    ///这里插入递推公式
    for (int i = 0; i <= ed; i++) {
        if(i==0&&iszer){
            ///处理前导零
            res+=dfs(len-1,10,type,ismax&&i==ed,true);
        }else{
            if(type==1){
                if(i>=dig||dig==10) res+=dfs(len-1,i,type,ismax&&i==ed,false);
            }else if(type==2){
                if(i<=dig||dig==10) res+=dfs(len-1,i,type,ismax&&i==ed,false);
            }else if(type==0){
                if(i==dig||dig==10) res+=dfs(len-1,i,type,ismax&&i==ed,false);
            }
        }
    }
    return ismax ? res : dp[len][dig][type] = res;
}

LL solve(LL x) {
    tot = 0;
    while (x) { arr[++tot] = x % 10; x /= 10; }
    LL ret=0;
    ret+=dfs(tot,10,1,true,true);
    ret+=dfs(tot,10,2,true,true);
    ret-=dfs(tot,10,0,true,true);
    return ret;
}

void init() {
    clr(dp,-1);
}

int main() {
    init();
    int n; scf("%d",&n);
    while(n--){
        LL l,r;
        scf("%lld%lld",&l,&r);
        prf("%lld
",solve(r)-solve(l-1));
    }
    return 0;
}

//end-----------------------------------------------------------------------