数位DP BZOJ 1026 [SCOI2009]windy数

 

题目链接

前面全是0的情况特判

#include <bits/stdc++.h>

int dp[10][10];
int digit[10];

int DFS(int pos, int val, int zero, bool limit) {
    if (pos == -1) {
        return 1;
    }
    int &now = dp[pos][val];
    if (!limit && zero && now != -1) {
        return now;
    }
    int d = limit ? digit[pos] : 9;
    int ret = 0;
    if (zero == 0) {
        for (int i=0; i<=d; ++i) {
            ret += DFS (pos - 1, i, i, limit && i == d);
        }
    } else {
        for (int i=0; i<=d; ++i) {
            if ((i-val) * (i - val) < 4) {
                continue;
            }
            ret += DFS (pos - 1, i, 1, limit && i == d);
        }
    }
    if (!limit && zero) {
        now = ret;
    }
    return ret;
}

int solve(int x) {
    int n = 0;
    while (x) {
        digit[n++] = x % 10;
        x /= 10;
    }
    int ret = 0;
    return DFS (n - 1, 0, 0, true);
}

int main() {
    memset (dp, -1, sizeof (dp));
    int a, b;
    while (scanf ("%d%d", &a, &b) == 2) {
        printf ("%d\n", solve (b) - solve (a - 1));
    }
    return 0;
}

　　

转载于:https://www.cnblogs.com/Running-Time/p/5503302.html