hdu 5787 K-wolf Number (数位dp)

K-wolf Number

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 737    Accepted Submission(s): 265

Problem Description

Alice thinks an integer x is a K-wolf number, if every K adjacent digits in decimal representation of x is pairwised different.
Given (L,R,K), please count how many K-wolf numbers in range of [L,R].

 

Input

The input contains multiple test cases. There are about 10 test cases.

Each test case contains three integers L, R and K.

1≤L≤R≤1e18
2≤K≤5

 

Output

For each test case output a line contains an integer.

 

Sample Input

1 1 2

20 100 5

 

Sample Output

1

72

 

建立一个pre[]数组来传递一个数之前四个数的值，传递时检查一下前面k - 1个值是否符合条件，注意dfs时去除前导0的影响就好了。

 

#include <bits/stdc++.h>
using namespace std;
#define ll long long
int dig[20];
ll dp[20][11][11][11][11];
ll l, r, k;

ll dfs(int pos, int pre[], int lim, int zero) {
    if(pos == -1) {
        //for(int i = 1; i <= 4; i++) printf("%d ", pre[i]); puts("");
        return 1;
    }
    ll& tmp = dp[pos][pre[1]][pre[2]][pre[3]][pre[4]];
    if(!lim && tmp != -1) return tmp;
    int End = lim ? dig[pos] : 9;
    ll ret = 0;
    for(int i = 0; i <= End; i++) {
        int flag = 0;
        int tmpPre[5];
        for(int j = 1; j <= k - 1; j++) if(i == pre[j]) {
            flag = 1;
            break;
        }
        if(flag) continue;
        for(int j = 2; j <= 4; j++) tmpPre[j] = pre[j - 1];
        if(zero && !i) tmpPre[1] = 10;
        else tmpPre[1] = i;
        if(zero && !i) ret += dfs(pos - 1, tmpPre, lim && (i == End), 1);
        else ret += dfs(pos - 1, tmpPre, lim && (i == End), 0);
    }
    if(!lim) tmp = ret;
    return ret;
}

ll func(ll num) {
    int n = 0;
    while(num) {
        dig[n++] = num % 10;
        num /= 10;
    }
    int pre[5];
    for(int i = 1; i <= 4; i++) pre[i] = 10;
    return dfs(n - 1, pre, 1, 1);
}

int main() {
    while(~scanf("%I64d %I64d %I64d", &l, &r, &k)) {
        memset(dp, -1, sizeof(dp));
        printf("%I64d
", func(r) - func(l - 1));
    }
}