数位DP GYM 100827 E Hill Number

 

题目链接

题意：判断小于n的数字中，数位从高到低成上升再下降的趋势的数字的个数

分析：简单的数位DP，保存前一位的数字，注意临界点的处理，都是套路。

#include <bits/stdc++.h>

typedef long long ll;
const int N = 70 + 5;
char str[N];
ll dp[N][10][2];
int len;

ll DFS(int pos, int pre, int up, int limit) {
    if (pos == len) {
        return 1;
    }
    ll &now = dp[pos][pre][up];
    if (!limit && now != -1) {
        return now;
    }
    ll ret = 0;
    int d = limit ? str[pos] - '0' : 9;
    for (int i=0; i<=d; ++i) {
        if (up) {
            if (i >= pre) {
                ret += DFS (pos + 1, i, up, limit && i == d);
            } else {
                ret += DFS (pos + 1, i, 0, limit && i == d);
            }
        } else if (i <= pre) {
            ret += DFS (pos + 1, i, 0, limit && i == d);
        }
    }
    if (!limit) {
        now = ret;
    }
    return ret;
}

int main() {
    int T; scanf ("%d", &T);
    while (T--) {
        scanf ("%s", str);
        len = strlen (str);
        int flag = -1;
        for (int i=1; i<len; ++i) {
            if (str[i] < str[i-1]) {
                flag = 0;
            }
            if (flag == 0 && str[i] > str[i-1]) {
                flag = 1;
                break;
            }
        }
        if (flag == 1) {
            puts ("-1");
        } else {
            memset (dp, -1, sizeof (dp));
            printf ("%I64d\n", DFS (0, 0, 1, 1) - 1);
        }
    }
    return 0;
}