HDU 5787 K-wolf Number

5787

思路：

数位dp

#include<bits/stdc++.h>
using namespace std;
#define LL long long 
#define pb push_back
#define mem(a, b) memset(a, b, sizeof(a))

LL dp[25][12][12][12][12];
int a[25];
int cnt = 0, k;
LL dfs(int pos, int a1, int a2, int a3, int a4, bool limit, bool zero) {
    if(pos == -1) {
        return 1;
    }
    if(!limit) {
        if(k == 2) {
            if(~dp[pos][a4][0][0][0]) return dp[pos][a4][0][0][0];
        }
        else if(k == 3) {
            if(~dp[pos][a3][a4][0][0]) return dp[pos][a3][a4][0][0];
        }
        else if(k == 4) {
            if(~dp[pos][a2][a3][a4][0]) return dp[pos][a2][a3][a4][0];
        }
        else if(k == 5) {
            if(~dp[pos][a1][a2][a3][a4]) return dp[pos][a1][a2][a3][a4];
        }
    }
    int up = 9;
    if(limit) up = a[pos];
    LL ans = 0;
    for (int i = 0; i <= up; i++) {
        if(k == 2) {
            if(i != a4) {
                if(zero&&i==0)ans += dfs(pos-1, a2, a3, a4, 11, limit&&i==up, zero&&i==0);
                else ans += dfs(pos-1, a2, a3, a4, i, limit&&i==up, zero&&i==0);
            }
        }
        else if(k == 3) {
            if(i != a3 && i != a4) {
                if(zero&&i==0)ans += dfs(pos-1, a2, a3, a4, 11, limit&&i==up, zero&&i==0);
                else ans += dfs(pos-1, a2, a3, a4, i, limit&&i==up, zero&&i==0);
            }
        }
        else if(k == 4) {
            if(i != a2 && i != a3 && i != a4) {
                if(zero&&i==0)ans += dfs(pos-1, a2, a3, a4, 11, limit&&i==up, zero&&i==0);
                else ans += dfs(pos-1, a2, a3, a4, i, limit&&i==up, zero&&i==0);
            }
        }
        else if(k == 5) {
            if(i != a1 && i != a2 && i != a3 && i != a4) {
                if(zero&&i==0)ans += dfs(pos-1, a2, a3, a4, 11, limit&&i==up, zero&&i==0);
                else ans += dfs(pos-1, a2, a3, a4, i, limit&&i==up, zero&&i==0);
            }
        } 
    }
    if(!limit) {
        if(k == 2) {
            dp[pos][a4][0][0][0] = ans;
        }
        else if(k == 3) {
            dp[pos][a3][a4][0][0] = ans;
        }
        else if(k == 4) {
            dp[pos][a2][a3][a4][0] = ans;
        }
        else if(k == 5) {
            dp[pos][a1][a2][a3][a4] = ans;
        }
    }
    return ans;
} 
LL solve(LL n) {
    if(n == 0) return 1;
    cnt = 0;
    mem(dp, -1);
    while(n) a[cnt++] = n%10, n /= 10;
    return dfs(cnt-1, 11, 11, 11, 11, 1, 1); 
}
int main(){
    LL L, R;
    while(~ scanf("%lld %lld %d", &L, &R, &k)){
        printf("%lld\n", solve(R) - solve(L-1));
    }
    return 0;
}