HDU - 5787 K-wolf Number 数位dp

Posted 2020-11-12 cjlhy

HDU - 5787

直接数位dp就好了。

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include<bits/stdc++.h>
#define LL long long
#define LD long double
#define ull unsigned long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ALL(x) (x).begin(), (x).end()
#define fio ios::sync_with_stdio(false); cin.tie(0)                     ;

using namespace std;

const int N = 2e5 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = (int)1e9 + 7;
const double eps = 1e-8;
const double PI = acos(-1);

template<class T, class S> inline void add(T &a, S b) {a += b; if(a >= mod) a -= mod;}
template<class T, class S> inline void sub(T &a, S b) {a -= b; if(a < 0) a += mod;}
template<class T, class S> inline bool chkmax(T &a, S b) {return a < b ? a = b, true : false;}
template<class T, class S> inline bool chkmin(T &a, S b) {return a > b ? a = b, true : false;}

//mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

const int M = (11 * (11 * (11 * (10 * 11 + 10) + 10) + 10) + 10) + 1;

LL dp[20][M];
LL L, R, K;
int a[N], tot;
bool can[N][6];

LL dfs(int p, int mask, bool limit) {
    if(!can[mask][K]) return 0;
    if(p == -1) return 1;
    if(!limit && ~dp[p][mask]) return dp[p][mask];
    LL ans = 0;
    int dn = mask == 0;
    int up = limit ? a[p] : 9;
    for(int i = dn; i <= up; i++) {
        int nmask = 0;
        if(mask == M - 1 && !i) nmask = mask;
        else nmask = (mask * 11 + i) % M;
        ans += dfs(p - 1, nmask, limit && (i == up));
    }
    if(!limit) dp[p][mask] = ans;
    return ans;
}

LL solve(LL x) {
    tot = 0;
    for(LL i = x; i; i /= 10) {
        a[tot++] = i % 10;
    }
    return dfs(tot - 1, M - 1, 1);
}

int v[5];
bool vis[10];

bool check(int *v, int k) {
    memset(vis, 0, sizeof(vis));
    for(int i = 0; i + k <= 5; i++) {
        for(int j = i; j < i + k; j++) {
            if(v[j] == 10) continue;
            if(vis[v[j]]) return false;
            vis[v[j]] = true;
        }
        for(int j = i; j < i + k; j++) {
            if(v[j] == 10) continue;
            vis[v[j]] = false;
        }
    }
    return true;
}

int main() {
    for(v[0] = 0; v[0] <= 10; v[0]++) {
        for(v[1] = 0; v[1] <= 10; v[1]++) {
            for(v[2] = 0; v[2] <= 10; v[2]++) {
                for(v[3] = 0; v[3] <= 10; v[3]++) {
                    for(v[4] = 0; v[4] <= 10; v[4]++) {
                        int x = 0;
                        for(int i = 0; i < 5; i++) {
                            x = x * 11 + v[i];
                        }
                        for(int k = 2; k <= 5; k++) {
                            can[x][k] = check(v, k);
                            if(!can[x][k]) break;
                        }
                    }
                }
            }
        }
    }
    while(scanf("%lld%lld%lld", &L, &R, &K) != EOF) {
        memset(dp, -1, sizeof(dp));
        printf("%lld
", solve(R) - solve(L - 1));
    }
    return 0;
}

/*
*/