hdu2089 简单dp

/**********************jibancanyang**************************
*Author        :jibancanyang
*Created Time  : 二  4/12 23:15:53 2016
*File Name     : hdu2089.cpp
*题目类型:简单dp
*感悟:dp[i]为小于等于i的满足条件数的个数,可以o(n)的更新,但是这个题的标准做法是数位dp.
***********************1599664856@qq.com**********************/

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <stack>
using namespace std;
typedef pair<int, int> pii;
typedef long long ll;
typedef unsigned long long ull;
vector<int> vi;
#define pr(x) cout << #x << ": " << x << "  "
#define pl(x) cout << #x << ": " << x << endl;
#define xx first
#define yy second
#define sa(n) scanf("%d", &(n))
#define rep(i, a, n) for (int i = a; i < n; i++)
#define vep(c) for(decltype((c).begin() ) it = (c).begin(); it != (c).end(); it++)
const int mod = int(1e9) + 7, INF = 0x3fffffff, maxn = 1e5 + 12;
int dp[maxn * 10];
bool ok[maxn * 10];
bool judge(int x) {
if(ok[x / 10]) {
if ( !(x / 10 % 10 == 6 && x % 10 == 2) && x % 10 != 4) {
return ok[x] = true;
} else return ok[x] = false;
}
return ok[x] = false;
}
int main(void)
{
#ifdef LOCAL
//freopen("in.txt", "r", stdin);
//freopen("out.txt", "w", stdout);
#endif
cin.sync_with_stdio(false);
dp[0] = 1, ok[0] = 1;
for (int i = 1; i <= maxn * 10; i++) {
judge(i);
dp[i] = dp[i - 1] + ok[i];
}
int n, m;
while (cin >> n >> m, n *n + m *m) {
cout << dp[m] - (n == 0 ? 0 : dp[n - 1]) << endl;
}
return 0;
}