传送门
Luogu
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 \(qwq\)
原文地址:https://www.cnblogs.com/zsbzsb/p/11746554.html
时间: 2024-10-07 18:37:48