POJ3208 Apocalypse Someday(二分 数位DP)

数位DP加二分

//数位dp,dfs记忆化搜索
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
typedef long long LL;
#define N 20

LL dp[N][3];//dp[i][j]表示长度为i，前面有j个6时不含666的数的个数
int num[N];

//c6表示前面6的个数
LL dfs(int len, int c6, bool ismax){
	if(len == 0){
		return 1;
	}
	if(!ismax && dp[len][c6] >= 0){
		return dp[len][c6];
	}
	int max = ismax? num[len]:9;
	LL cnt = 0;
	for(int i = 0; i <= max; i++){
		if(c6 == 2 && i == 6){
			continue;
		}
		cnt += dfs(len-1, i == 6? c6 + 1: 0, ismax && i == max);
	}
	return ismax?cnt:dp[len][c6]=cnt;
}

LL solve(LL n){
	LL t = n;
	int len = 0;
	while(n){
		num[++len]=n%10;
		n/=10;
	}
	return t + 1 - dfs(len, 0, true);
}


int main(){
    memset(dp,-1,sizeof(dp));
    int t;
    cin>>t;
    while(t--){
    	LL n;
    	scanf("%I64d", &n);
    	LL l = 666, r = 50000000666ll;
    	while(l < r){
    		LL m = (l + r)>>1;
    		if(solve(m) < n){
    			l = m + 1;
    		}else{
    			r = m;
    		}
    	}
    	printf("%I64d
", l);
    }

    return 0;
}