[Bzoj1026][SCOI2009]windy数(数位dp)

题目链接:https://www.lydsy.com/JudgeOnline/problem.php?id=1026

比较明显的数位dp,dp[pos][num]表示pos位和上一位数差为num的个数。

初始将num设为-2,然后记忆化开始搜

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 const double eps = 1e-8;
 5 const ll INF = 9e18 + 7;
 6 const int maxn = 5e5 + 10;
 7 inline ll read() {
 8     ll n = 0, f = 1; char ch = getchar();
 9     while (ch<'0' || ch>'9') { f = -1, ch = getchar(); }
10     while (ch >= '0'&&ch <= '9') { n = n * 10 + ch - '0', ch = getchar(); }
11     return n * f;
12 }
13 int dp[20][12];
14 int p[20];
15 int dfs(int pos, int num, int limit) {
16     if (pos == -1)
17         return 1;
18     if (dp[pos][num] != -1 && !limit&&num >= 0)
19         return dp[pos][num];
20     int up = limit ? p[pos] : 9;
21     int ans = 0;
22     for (int i = 0; i <= up; i++) {
23         if (abs(i - num) < 2)continue;
24         int w = i;
25         if (num == -2 && i == 0)
26             w = num;
27         ans += dfs(pos - 1, w, limit && (i == up));
28     }
29     if (!limit && !num >= 0)
30         dp[pos][num] = ans;
31     return ans;
32 }
33 int solve(int x) {
34     int pos = 0;
35     while (x) {
36         p[pos++] = x % 10;
37         x /= 10;
38     }
39     return dfs(pos - 1, -2, 1);
40 }
41 int main() {
42     int l, r;
43     l = read(), r = read();
44     memset(dp, -1, sizeof(dp));
45     printf("%d\n", solve(r) - solve(l - 1));
46     //system("pause");
47 }

 

posted @ 2019-07-01 20:13  祈梦生  阅读(...)  评论(...编辑  收藏