BZOJ 1026 windy数

1026

思路：

数位dp

代码：

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

int a[15];
int dp[15][15];
int dfs(int pos, int pre, bool limit, bool zero) {
    if (pos == -1) return 1;
    if(!limit && !zero && ~dp[pos][pre]) return dp[pos][pre];
    int top = limit ? a[pos] : 9;
    int ans = 0;
    for (int i = 0; i <= top; i++) {
        if(zero){
            ans += dfs(pos - 1, i, limit&&i==a[pos], zero&&i==0);
        }
        else {
            if(abs(pre - i) <= 1) continue;
            ans += dfs(pos - 1, i, limit&&i==a[pos], zero&&i==0);
        }
    } 
    if(!limit && !zero) dp[pos][pre] = ans;
    return ans;
}
int solve(int n) {
    int cnt = 0;
    mem(a, 0); 
    while (n) {
        a[cnt++] = n % 10;
        n /= 10;
    }
    return dfs(9, 0, 1, 1);
} 
int main() {
    int a, b;
    scanf("%d%d", &a, &b);
    mem(dp, -1);
    printf("%d\n",solve(b) - solve(a - 1));
    return 0;
}