网易互娱2017实习生招聘游戏研发工程师在线笔试第二场 C

偶尔碰到这题，简单数位DP题，然而我已生疏了……

这次算是重新想到的，看来对DP的理解有增进了……

dp[i][j][k]，表示前i为，mod为j，是否出现2、3、5的剩下的数位可组成的数字。答案就是dp[len][0][0]

 

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>

using namespace std;

#define LL long long 

LL dp[20][7][2];
int num[20];


LL dfs(int len, int mod, bool one, int show, bool flag){
	if(len == 0){
		if(show == 1 && mod == 0){
			dp[len][mod][show] = 1;
		}
		else dp[len][mod][show] = 0;
		return dp[len][mod][show];
	}
	if(!flag && dp[len][mod][show] != -1) return dp[len][mod][show];
	
	int up = flag ? num[len]:9;
	LL ans = 0, tmp = show;
	
	for(int i = 0; i <= up ; i++){
		if(one && i == 8) continue;
		if(i == 2 || i == 3 || i == 5) tmp = (show | 1);
	//	cout <<" tmp = " << tmp << endl;
		ans += dfs(len - 1, (mod * 10 + i)%7, i == 1?true:false, tmp, (flag && i == up) ? true: false);
		tmp = show;
	//	printf("%I64d====%d\n", ans, i);
		
	}
	dp[len][mod][show] = ans;
	return ans;
	
}



long long slove(LL n){
	
	int len = 0;
	LL tmp = n;
	while(tmp){
		num[++len] = tmp % 10;
		tmp /= 10;
	}
	
//	printf("%I64d\n", n);
	
	dfs(len, 0, false, 0, true);
	
	return dp[len][0][0];
	
	
}

int main(){
	long long l, r;
	int T;
	scanf("%d", &T);
	while(T--){
		memset(dp, -1, sizeof(dp));
		scanf("%I64d%I64d", &l, &r);
		printf("%I64d\n", slove(r) - slove(l - 1));
	}
	return 0;
}