Educational Codeforces Round 70 (Rated for Div. 2) D. Print a 1337-string...

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分类专栏： cf 文章标签： codeforce 构造

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本文详细解析了Codeforces平台上的D级难题，探讨了如何构造特定的数字序列以匹配预定的子序列数量，通过预处理和策略性插入元素，有效地解决了问题。

D. Print a 1337-string…

题目链接: https://codeforces.com/contest/1202/problem/D

Problem Description

The subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements.
You are given an integer $n$ .
You have to find a sequence $s$ consisting of digits ${1,3,7\}$ such that it has exactly $n$ subsequences equal to $1337$ .
For example, sequence $337133377$ has $6$ subsequences equal to $1337$ :

$1.\ 337\underline{1}3\underline{3}\underline{3}7\underline{7}$ (you can remove the second and fifth characters);
$2.\ 337\underline{1}\underline{3}3\underline{3}7\underline{7}$ (you can remove the third and fifth characters);
$3.\ 337\underline{1}\underline{3}\underline{3}37\underline{7}$ (you can remove the fourth and fifth characters);
$4.\ 337\underline{1}3\underline{3}\underline{3}\underline{7}7$ (you can remove the second and sixth characters);
$5.\ 337\underline{1}\underline{3}3\underline{3}\underline{7}7$ (you can remove the third and sixth characters);
$6.\ 337\underline{1}\underline{3}\underline{3}3\underline{7}7$ (you can remove the fourth and sixth characters).

Input

The first line contains one integer $t (1 \leq t \leq 10)$ — the number of queries.
Next $t$ lines contains a description of queries: the $i$ -th line contains one integer $n_i (1≤n_i≤10^9).$

Output

For the $i$ -th query print one string $s_i (1≤|s_i|≤10^5)$ consisting of digits ${1,3,7\}.$ String $s_i$ must have exactly $n_i$ subsequences $1337$ . If there are multiple such strings, print any of them.

Sample Input

2
6
1

Sample Output

113337
1337

题意

有 $t$ 次询问，每次询问给定一个 $n_i$ .
要求构造一个只包含 $1, 3, 7$ 的字符串 $s_i(1≤|s_i|≤10^5)$ ，要使得 $s_i$ 中有 $n_i$ 个 $1337$ 的子序列.

思路

因为对构造的字符串 $s_i$ 的长度有限制，所以考虑尽量的多放一些 $3$ 来缩短所构造出的字符串。
先预处理出 $C_i^2$ 的值， $i$ 从 $2$ 到 $44721$ ，因为 $C_{44722}^2$ 就已经超过了 $n_i$ 的上限，就不用处理了。
接下来就在各个 $3$ 之间尝试能不能插 $1$ 来使得 $1337$ 子序列的个数更接近 $n_i$ 啦。

代码

#include <bits/stdc++.h>
#define pi acos(-1.0)
#define ll long long
#define ull unsigned long long
#define esp 1e-9
#define inf 0x3f3f3f3f
#define inff 0x3f3f3f3f3f3f3f3f
#define Pair pair<ll, ll>
#define It list<node>::iterator
   
using namespace std;
 
const ll N = 5e4+5, mod = 1e9+7;
ll t, n, arr[N], cnt[N];
 
int main(){
	scanf("%lld", &t);
	//预处理
	for (ll i = 2; i <= 44721; i++){
		arr[i] = i*(i-1)/2;
	}
	while (t--){
		scanf("%lld", &n);
		memset(cnt, 0, sizeof(cnt));
		//寻找最多的3记录在m中
		ll l = 2, r = 44721, m;
		while (l <= r){
			ll mid = (l+r)/2;
			if (arr[mid] <= n){
				m = mid; l = mid+1;
			}
			else{
				r = mid-1;
			}
		}
		cnt[m]++;//在第一位放一个1
		n -= arr[m];//更新一下还差多少个子序列
		while (n){
			//在尽量靠前的位置插入1来快速缩小与n的差距
			ll l = 2, r = 44721, idx;
			while (l <= r){
				ll mid = (l+r)/2;
				if (arr[mid] <= n){
					idx = mid; l = mid+1;
				}
				else{
					r = mid-1;
				}
			}
			cnt[idx]++;//记录插入位置
			n -= arr[idx];//更新差距
		}
		//输出构造的字符串
		for (ll i = 1; i <= m; i++){
			while (cnt[m-i+1]){
				putchar('1');
				cnt[m-i+1]--;
			}
			putchar('3');
		}
		printf("7\n");
	}
 
	return 0;
}