这题也好

- 给定一个数n

- 求一个串含有n个子序列{1337}, 并输出这个串

- 串长小于1e5

>> face <<

tutorial:问题可以简化成 $n=cn{t}_1\ast C_{cn{t}_3}^2\ast n_3$,且$cn{t}_1+cn{t}_3+cn{t}_7\le 10^5$, 怎么办, 且注意到$cn{t}_3$这个数占权最大, 肯定越多越好, 那$cn{t}_3$最多是多少呢, 考虑$C_x^2\ge 10^5$,$x$大约是1e4数量级的, 现在就最大化$cn{t}_3$就好了, 余下的部分用$cn{t}_7$补在第一个$133$后面, 然后接333333…337;

#include <bits/stdc++.h> using namespace std; #define _rep(i, a, b) for (int i = (a); i <= (b); ++i) #define _rev(i, a, b) for (int i = (a); i >= (b); --i) #define _for(i, a, b) for (int i = (a); i < (b); ++i) #define _rof(i, a, b) for (int i = (a); i > (b); --i) #define oo 0x3f3f3f3f #define ll long long #define db double #define eps 1e-8 #define bin(x) cout << bitset<10>(x) << endl; #define what_is(x) cerr << #x << " is " << x << "s" << endl; #define met(a, b) memset(a, b, sizeof(a)) #define all(x) x.begin(), x.end() #define pii pair<int, int> int nxt() { int ret; scanf("%d", &ret); return ret; } string s; int main() { int t = nxt(); _rep(i, 1, t) { ll n = nxt(); if(n == 1){ cout << "1337" << endl; continue; } n *= 2; ll n3 = sqrt(n); while(n3 * (n3 - 1) <= n)n3++; n3--; ll left = (n - n3 * (n3 - 1)) / 2; s = "133"; _rep(i, 1, left) s.append("7"); _rep(i, 1, n3 - 2) s.append("3"); s.append("7"); cout << s << endl; //cout << s.size() << endl; } }