ural 1222. Chernobyl’ Eagles

1222. Chernobyl’ Eagles

Time limit: 1.0 second
Memory limit: 64 MB

A Chernobyl’ eagle has several heads (for example, the eagle on the Russian National Emblem is a very typical one, having two heads; there exist Chernobyl’ eagles having twenty-six, one and even zero heads). As all eagles, Chernobyl’ eagles are very intelligent. Moreover, IQ of a Chernobyl’ eagle is exactly equal to the number of its heads. These eagles can also enormously enlarge their IQ, when they form a group for a brainstorm. IQ of a group of Chernobyl’ eagles equals to the product of IQ’s of eagles in the group. So for example, the IQ of a group, consisting of two 4-headed eagles and one 7-headed is 4*4*7=112. The question is, how large can be an IQ of a group of eagles with a given total amount of heads.

Input

There is one positive integer N in the input, N ≤ 3000 — the total number of heads of Chernobyl’ eagles in a group.

Output

Your program should output a single number — a maximal IQ, which could have a group of Chernobyl’ eagles, with the total amount of heads equal to N.

Sample

input	output
5	6

Problem Author: folklore, proposed by Leonid Volkov
Problem Source: The Seventh Ural State University collegiate programming contest

Tags: none  (hide tags for unsolved problems)

Difficulty: 141

 

题意：和为n，拆成若干个数，使得乘积最大

分析：显然，先拆成尽量多的3，在拆成2，这样乘积最大，因为肯定不能拆成1，然后之后所有数都可以用2，3构成，根据归纳法可以证明。

注意，4要拆成两个2

 #include <cstdio>
 #include <cstring>
 #include <cstdlib>
 #include <cmath>
 #include <deque>
 #include <vector>
 #include <queue>
 #include <iostream>
 #include <algorithm>
 #include <map>
 #include <set>
 #include <ctime>
 using namespace std;
 typedef long long LL;
 typedef double DB;
 #define For(i, s, t) for(int i = (s); i <= (t); i++)
 #define Ford(i, s, t) for(int i = (s); i >= (t); i--)
 #define Rep(i, t) for(int i = (0); i < (t); i++)
 #define Repn(i, t) for(int i = ((t)-1); i >= (0); i--)
 #define rep(i, x, t) for(int i = (x); i < (t); i++)
 #define MIT (2147483647)
 #define INF (1000000001)
 #define MLL (1000000000000000001LL)
 #define sz(x) ((int) (x).size())
 #define clr(x, y) memset(x, y, sizeof(x))
 #define puf push_front
 #define pub push_back
 #define pof pop_front
 #define pob pop_back
 #define ft first
 #define sd second
 #define mk make_pair
 inline void SetIO(string Name) {
     string Input = Name+".in",
     Output = Name+".out";
     freopen(Input.c_str(), "r", stdin),
     freopen(Output.c_str(), "w", stdout);
 }

 inline int Getint() {
     int Ret = ;
     char Ch = ' ';
     while(!(Ch >= '' && Ch <= '')) Ch = getchar();
     while(Ch >= '' && Ch <= '') {
         Ret = Ret*+Ch-'';
         Ch = getchar();
     }
     return Ret;
 }

 const int N = , Mod = ;
 int Arr[N], Len;
 int n;

 inline void Input() {
     scanf("%d", &n);
 }

 inline void Mul(int x) {
     For(i, , Len) Arr[i] *= x;
     For(i, , Len) {
         Arr[i+] += Arr[i]/Mod;
         Arr[i] %= Mod;
     }
     while(Arr[Len+]) {
         Len++;
         Arr[Len+] += Arr[Len]/Mod;
         Arr[Len] %= Mod;
     }
 }

 inline void Solve() {
     Arr[] = , Len = ;
     while(n > ) Mul(), n -= ;
     if(n == ) Mul(), n -= ;
     if(n == ) Mul(), n -= ;
     if(n == ) Mul(), n -= ;

     printf("%d", Arr[Len]);
     Ford(i, Len-, ) printf("%04d", Arr[i]);
     puts("");
 }

 int main() {
     #ifndef ONLINE_JUDGE
     SetIO("F");
     #endif
     Input();
     Solve();
     return ;
 }