UESTC 250 windy数 数位dp

题目链接

#include<bits/stdc++.h>
using namespace std;
#define mem1(a) memset(a, -1, sizeof(a))
#define ll long long
int dp[20][20], digit[20], len;
ll dfs(int len, int pre, bool fp, bool first) {     //first表示前面的数是否全部为0， pre是前一个数
    if(!len)
        return 1;
    if(!first&&!fp && dp[len][pre]!=-1)
        return dp[len][pre];
    int ret = 0, maxx = fp?digit[len]:9;
    for(int i = 0; i<=maxx; i++) {
        if(!first)
            if(abs(i-pre)<2)
                continue;
        ret += dfs(len-1, i, fp&&i == maxx, first&&i==0);
    }
    if(!fp&&!first)
        return dp[len][pre] = ret;
    return ret;
}
int cal(int n) {
    len = 0;
    while(n) {
        digit[++len] = n%10;
        n/=10;
    }
    return dfs(len, 0, true, true);
}
int main()
{
    int a, b;
    mem1(dp);
    while(~scanf("%d%d", &a, &b)) {
        printf("%d\n", cal(b)-cal(a-1));
    }
}