Codeforces Round #491 (Div. 2) F. Concise and clear

F. Concise and clear

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Vasya is a regular participant at programming contests and is already experienced in finding important sentences in long statements. Of course, numbers constraints are important — factorization of a number less than 1000000 is easier than of a number less than 1000000000. However, sometimes it's hard to understand the number at the first glance. Could it be shortened? For example, instead of 1000000 you could write ${\mathrm{106106,\; instead\; of 1000000000 —109109,\; instead\; of 1000000007 — 109+7109+7.}}^{}$

Vasya decided that, to be concise, the notation should follow several rules:

• the notation should only consist of numbers, operations of addition ("+"), multiplication ("*") and exponentiation ("^"), in particular, the use of braces is forbidden;
• the use of several exponentiation operations in a row is forbidden, for example, writing "2^3^4" is unacceptable;
• the value of the resulting expression equals to the initial number;
• the notation should consist of the minimal amount of symbols.

Given $\mathrm{nn,\; find\; the\; equivalent\; concise\; notation\; for\; it.}$

Input

The only line contains a single integer $\mathrm{nn (1\le n\le 100000000001\le n\le 10000000000).}$

Output

Output a concise notation of the number $\mathrm{nn.\; If\; there\; are\; several\; concise\; notations,\; output\; any\; of\; them.}$

Examples

input

Copy

2018

output

Copy

2018

input

Copy

1000000007

output

Copy

10^9+7

input

Copy

10000000000

output

Copy

100^5

input

Copy

2000000000

output

Copy

2*10^9

Note

The third sample allows the answer 10^10 also of the length $55.$

 思路：10^10特判一下，之后剩的位数不超过10位，这样通过改字符来缩减长度最多出现4个字符，又单独的乘法和加法不会缩减长度，所以一定有^，于是我们将n表示成a^b*c+d的形式，a从2枚举到sqrt(n)，b从2枚举到loga(n)，贪心找最大的c，然后唯一确定一个d，这样就得到了一个sqrt(n)*log(n)复杂度的假算法，会WA9。此算法假在c或d也可能表示成e^f的形式，于是我们通过一个map预处理出所有能表示成这个形式的数，查询的时候和map里的比一下取长度最短就行了。预处理和枚举复杂度都为O(sqrt(n)*log2(n))。

#include <iostream>
#include <fstream>
#include <sstream>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <queue>
#include <stack>
#include <vector>
#include <set>
#include <map>
#include <list>
#include <iomanip>
#include <cctype>
#include <cassert>
#include <bitset>
#include <ctime>

using namespace std;

#define pau system("pause")
#define ll long long
#define pii pair<int, int>
#define pb push_back
#define mp make_pair
#define clr(a, x) memset(a, x, sizeof(a))

const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-9;

/*
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace __gnu_pbds;
tree<pli, null_type, greater<pli>, rb_tree_tag, tree_order_statistics_node_update> T;
*/

ll n;
string ans;
void add(string &S, ll t) {
    char s[29];
    int l = 0;
    while (t) {
        s[++l] = t % 10;
        t /= 10;
    }
    while (l) {
        S += s[l] + '0';
        --l;
    }
}
map<ll, string> mmp;
void pre() {
    for (ll a = 2; a <= sqrt(n + 0.5); ++a) {
        ll tt = a;
        for (int b = 2; ; ++b) {
            tt *= a;
            if (tt > n) break;
            string t2 = "";
            add(t2, a);
            t2 += "^";
            add(t2, b);
            if (mmp.count(tt)) {
                string t1 = mmp[tt];
                if (t2.length() < t1.length()) {
                    mmp[tt] = t2;
                }
            } else {
                mmp[tt] = t2;
            }
        }
    }
}
int main() {
    scanf("%lld", &n);
    if (n < 1000) {
        printf("%lld
", n);
    } else {
        pre();
        add(ans, n);
        for (ll a = 2; a <= sqrt(n + 0.5); ++a) {
            ll tt = a;
            for (ll b = 2; ; ++b) {
                tt *= a;
                if (tt > n) break;
                string res = "";
                add(res, a);
                res += "^";
                add(res, b);
                ll c = n / tt;
                for (ll tc = c; tc >= max(1ll, c - 0); --tc) {
                    ll d = n - tt * tc;
                    string res2 = res;
                    if (1 != tc) {
                        res2 += "*";
                        string res4 = "";
                        add(res4, tc);
                        if (mmp.count(tc)) {
                            if (res4.length() < mmp[tc].length()) {
                                res2 += res4;
                            } else {
                                res2 += mmp[tc];
                            }
                        } else {
                            res2 += res4;
                        }
                    }
                    if (d) {
                        res2 += "+";
                        string res3 = "";
                        add(res3, d);
                        if (mmp.count(d)) {
                            if (res3.length() < mmp[d].length()) {
                                res2 += res3;
                            } else {
                                res2 += mmp[d];
                            }
                        } else {
                            res2 += res3;
                        }
                    }
                    if (res2.length() < ans.length()) {
                        ans = res2;
                    }
                }
            }
        }
        cout << ans << endl;
    }
    return 0;
}