Codeforces Round #382 (Div. 2) D. Taxes 歌德巴赫猜想
2 seconds
256 megabytes
standard input
standard output
Mr. Funt now lives in a country with a very specific tax laws. The total income of mr. Funt during this year is equal to n (n ≥ 2) burles and the amount of tax he has to pay is calculated as the maximum divisor of n (not equal to n, of course). For example, if n = 6 then Funt has to pay 3 burles, while for n = 25 he needs to pay 5 and if n = 2 he pays only 1 burle.
As mr. Funt is a very opportunistic person he wants to cheat a bit. In particular, he wants to split the initial n in several parts n1 + n2 + ... + nk = n (here k is arbitrary, even k = 1 is allowed) and pay the taxes for each part separately. He can't make some part equal to 1 because it will reveal him. So, the condition ni ≥ 2 should hold for all i from 1 to k.
Ostap Bender wonders, how many money Funt has to pay (i.e. minimal) if he chooses and optimal way to split n in parts.
The first line of the input contains a single integer n (2 ≤ n ≤ 2·109) — the total year income of mr. Funt.
Print one integer — minimum possible number of burles that mr. Funt has to pay as a tax.
4
2
27
3
题意:一个n可以拆成任意多个数,每个数都不为1,f(n)=最大的因子(除了其本身);使得拆的和最小;
思路:显然拆成素数会使得解更优,相当于问最少拆成几个素数;根据歌德巴赫猜想;详见代码;
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define mod 1000000007
#define esp 0.00000000001
const int N=1e5+,M=1e6+,inf=1e9;
const ll INF=1e18+;
int prime(int n)
{
if(n<=)
return ;
if(n==)
return ;
if(n%==)
return ;
int k, upperBound=n/;
for(k=; k<=upperBound; k+=)
{
upperBound=n/k;
if(n%k==)
return ;
}
return ;
}
int main()
{
int x;
scanf("%d",&x);
if(prime(x))
return puts("");
if(x%==)
return puts("");
if(prime(x-))
return puts("");
puts("");
return ;
}
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