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解题思路

数位 \(\text{DP}\)
设状态 \(dp[now][las][0/1][0/1]\) 表示当前 \(\text{DP}\) 到第 \(i\) 位，前一个数是 \(las\)，有没有顶到上界，有没有前导零的答案。
转移十分显然。

细节注意事项

• 咕咕咕

参考代码

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cctype>
#include <cmath>
#include <ctime>
#include <vector>
#define rg register
using namespace std;

template < typename T > inline void read(T& s) {
    s = 0; int f = 0; char c = getchar();
    while (!isdigit(c)) f |= c == '-', c = getchar();
    while (isdigit(c)) s = s * 10 + c - 48, c = getchar();
    s = f ? -s : s;
}

const int _ = 11;

int a[_], dp[_][_];

inline int dfs(int now, int las, int lim, int zero) {
    if (now == 0) return 1;
    if (!lim && !zero && dp[now][las] != -1) return dp[now][las];
    int res = 0, tp = lim ? a[now] : 9;
    for (rg int j = 0; j <= tp; ++j)
        if (abs(j - las) >= 2) {
            int _lim = lim && j == tp;
            int _zero = zero && j == 0;
            int _las = _zero ? -2 : j;
            int _now = now - 1;
            res += dfs(_now, _las, _lim, _zero);
        }
    if (!lim && !zero) dp[now][las] = res;
    return res;
}

inline int solve(int x) {
    int n = 0;
    for (rg int i = x; i; i /= 10) a[++n] = i % 10;
    memset(dp, -1, sizeof dp);
    return dfs(n, -2, 1, 1);
}

int main() {
#ifndef ONLINE_JUDGE
    freopen("in.in", "r", stdin);
#endif
    int l, r;
    read(l), read(r);
    printf("%d\n", solve(r) - solve(l - 1));
    return 0;
}

完结撒花 \(qwq\)

「SCOI2009」windy数