Taxes CodeForces
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Mr. Funt now lives in a country with a very specific tax laws. The total income of mr. Funt during this year is equal ton (n ≥ 2) burles and the amount of tax he has to pay is calculated as the maximum divisor ofn (not equal to n, of course). For example, ifn = 6 then Funt has to pay 3 burles, while for n = 25 he needs to pay5 and if n = 2 he pays only1 burle.
As mr. Funt is a very opportunistic person he wants to cheat a bit. In particular, he wants to split the initialn in several parts n1 + n2 + ... + nk = n (herek is arbitrary, even k = 1 is allowed) and pay the taxes for each part separately. He can't make some part equal to1 because it will reveal him. So, the condition ni ≥ 2 should hold for alli from 1 to k.
Ostap Bender wonders, how many money Funt has to pay (i.e. minimal) if he chooses and optimal way to splitn 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:给出一个n,这个n可以分解成 n = n1 + n2 + …… + nk,其中k可以取任意数。要使得分解以后所有的n的最大因子(不包括自己本身)的和最小,问最小的和是多少。
思路:比赛的时候想到全部拆成素数是最好的,但是不知道怎么拆,看别人跑的特别快,就知道是数论题,绝望之下试了两发暴力,都是TLE了,GG。早上起来才知道有“哥德巴赫猜想”这个东西。
内容大概是如下两点:
1、所有大于2的偶数可以被分解成两个素数。
2、所有大于7的奇数可以被分解成三个素数。(n-3)为偶数,3是一个素数,所以是三个。
所以知道这个猜想之后就变得简单了:
1、偶数:n为2,答案是1,否则答案是2.
2、奇数:首先,n最少可以拆成三个素数,还有两种情况要考虑:n本身是一个素数的话答案就是1,n-2是一个素数答案就是2(一个奇数可以拆成一个偶数+一个奇数,偶数只有2是素数)。
#include <cstdio>#include<vector>#include<cmath>#include<algorithm>#include<cstring>using namespace std;typedef long long ll;int is_prime(int x){ for(int i=2;i<=(int)sqrt(x);i++){ if(x%i==0) return 0; } return 1;}int main(){ int n; scanf("%d",&n); int ans; if(n==2||n==3||n==5||n==7) ans=1; else if(n%2==0) ans=2; else if(n&1){ if(is_prime(n)) ans=1; else if(is_prime(n-2)) ans=2; else ans=3; } printf("%d\n",ans);return 0;}
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