数位dp、

练习题衔接：传送门

入门题

HDU2089

尝试了用两种方法解，感觉记忆化搜索特别方便。。。

#include <bits/stdc++.h>

using namespace std;
#define LL long long
const int qq = 1e6 + 10;
/*
void Init(){
	dp[0][0] = 1;
	for(int i = 1; i <= 7; ++i)
		for(int j = 0; j < 10; ++j)
			for(int k = 0; k < 10; ++k)
				if(j != 4 && !(j == 6 && k == 2))
					dp[i][j] += dp[i - 1][k];
}

int Solve(int n){
	int len = 0;
	while(n > 0){
		d[++len] = n % 10;
		n /= 10;
	}
	d[len + 1] = 0;
	int ans = 0, t;
	for(int i = len; i > 0; --i){
		for(int j = 0; j < d[i]; ++j)
			if(d[i + 1] != 6 || j != 2)
				ans += dp[i][j];
		if(d[i] == 4 || (d[i + 1] == 6 && d[i] == 2))	break;
	}
	return ans;
}	
*/
int dp[10][10], d[10];
int Dfs(int len, bool state, bool fp){
	if(!len)	return 1;
	if(fp && dp[len][state] != -1){
		return dp[len][state];
	}
	int t = fp ? 9 : d[len];
	int ret = 0;
	for(int i = 0; i <= t; ++i){
		if(i == 4 || (state && i == 2))	continue;
		ret += Dfs(len - 1, i == 6, fp || i != t);
	}
	return fp ? dp[len][state] = ret : ret;
}

int Solve(int n){
	int len = 0;
	while(n > 0){
		d[++len] = n % 10;
		n /= 10;
	}
	return Dfs(len, false, false);
}
int main(){
//	Init();
	int a, b;
	while(scanf("%d%d", &a, &b) != EOF){
		if(a == 0 && b == 0)	break;
		memset(dp, -1, sizeof dp);
		printf("%d\n", Solve(b) - Solve(a - 1));
	}
	return 0;
}