poj3208 Apocalypse Someday （数位dp + 二分）

The number 666 is considered to be the occult “number of the beast” and is a well used number in all major apocalypse themed blockbuster movies. However the number 666 can’t always be used in the script so numbers such as 1666 are used instead. Let us call the numbers containing at least three contiguous sixes beastly numbers. The first few beastly numbers are 666, 1666, 2666, 3666, 4666, 5666…

Given a 1-based index n, your program should return the nth beastly number.

Input

The first line contains the number of test cases T (T ≤ 1,000).

Each of the following T lines contains an integer n (1 ≤ n ≤ 50,000,000) as a test case.

Output

For each test case, your program should output the nth beastly number.

Sample Input

Sample Output

1666
2666
66666

Sponsor

一道数位dp+二分

很久没写数位dp了借此题重新复习一下

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#define ll long long
int a[20];
ll dp[20][2][2][2];
ll n, m;
using namespace std;ll dfs(int pos, int pre1, int pre2, int k, bool limit){if(pos == 0)return k;if(!limit && dp[pos][pre1][pre2][k] != -1)return dp[pos][pre1][pre2][k];int up = limit? a[pos] : 9;ll cnt = 0;for(int i = 0; i <= up; i++){if(k == 1)cnt += dfs(pos - 1, i == 6,pre1, 1, i == a[pos] && limit);elsecnt += dfs(pos - 1, i == 6,pre1, (i == 6) && pre1 && pre2, i == a[pos] && limit);}if(!limit)dp[pos][pre1][pre2][k] = cnt;return cnt;
}ll solve(ll x){int sum = 0;while(x){a[++sum] = x % 10;x = x / 10; }return dfs(sum, 0, 0, 0, 1);
}int main(){memset(dp, -1, sizeof(dp));int t;cin >> t;while(t--){cin >> n;ll l = 1, r = 0x3f3f3f3f3f3f3f3f, mid;while(l <= r){mid = (l + r) / 2;if(solve(mid) < n)l = mid + 1;elser = mid - 1;}cout << l << endl;}}

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