P2602 [ZJOI2010]数字计数

自己独立做出来的第一道数位dp(虽然一共做了3道qwq

#include <bits/stdc++.h>
#define inf 2333333333333333
#define N 1000010
#define p(a) putchar(a)
#define For(i,a,b) for(long long i=a;i<=b;++i)
//by war
//2020.7.14
using namespace std;
long long l,r,t;
long long f[20][20][20],ans[20];
vector<long long>dim,ki;

struct node{
    long long k[10];
}g[20][20][20];

void in(long long &x){
    long long y=1;char c=getchar();x=0;
    while(c<'0'||c>'9'){if(c=='-')y=-1;c=getchar();}
    while(c<='9'&&c>='0'){ x=(x<<1)+(x<<3)+c-'0';c=getchar();}
    x*=y;
}
void o(long long x){
    if(x<0){p('-');x=-x;}
    if(x>9)o(x/10);
    p(x%10+'0');
}

long long dfs(long long x,long long st,long long op){
    if(!x) return 1;
    if(~f[x][st][op]){
        For(i,0,9) ans[i]+=g[x][st][op].k[i];
        return f[x][st][op];
    }
    long long maxx=op?dim[x]:9,r=0,k=0;
    For(i,0,9) g[x][st][op].k[i]=ans[i];
    For(i,0,maxx){
        if(st==11 && i==0) dfs(x-1,11,op&(i==maxx));
        else{
            r+=dfs(x-1,i,op&(i==maxx));
            ans[i]+=t*r;
            k+=r;
            r=0;
        }
    }
    For(i,0,9) g[x][st][op].k[i]=ans[i]-g[x][st][op].k[i];
    return f[x][st][op]=k;
}

void deal(long long x){
    memset(f,-1,sizeof f);
    memset(g,0,sizeof g);
    dim.clear();
    dim.push_back(666);
    while(x){
        dim.push_back(x%10);
        x/=10;
    }
    dfs(dim.size()-1,11,1);
}

signed main(){
    in(l);in(r);
    t=1;
    deal(r);
    t=-1;
    deal(l-1);
    For(i,0,9) o(ans[i]),p(' ');
    return 0;
}
原文地址:https://www.cnblogs.com/war1111/p/13298074.html