【CodeForces735D】【哥德巴赫猜想】Taxes 题解

最新推荐文章于 2024-08-13 01:28:56 发布

MaverickFW

最新推荐文章于 2024-08-13 01:28:56 发布

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文章标签： codeforces 哥德巴赫猜想

本文链接：https://blog.youkuaiyun.com/MaverickFW/article/details/55005438

本文探讨了一种特殊的税收制度，其中个人需要支付其总收入的最大除数作为税款。为了减少税负，文章提出了一种将收入拆分为多个部分并分别纳税的方法，并通过算法给出了最小可能税额的解决方案。

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Taxes

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.

Input
The first line of the input contains a single integer n (2 ≤ n ≤ 2·109) — the total year income of mr. Funt.

Output
Print one integer — minimum possible number of burles that mr. Funt has to pay as a tax.

Example
Input
4
Output
2
Input
27
Output
3

哥德巴赫
偷税，就是把一个数字分解成质数，最少分几个。（当然中央的决定权也是很重要的）（划去）
拆成素数的话，答案就是1
素数当然是坠吼的

判断:
1.判断是否为偶数，偶数中如果是2输出1，否则输出2。不为偶数则继续
2.判断这个数字n是不是质数，是质数输出1，否则继续
3.判断n-2是不是质数，是质数输出2，否则输出3

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <string>
#include <iomanip>
#include <ctime>
#include <climits>
#include <cctype>
#include <algorithm>
#define clr(x) memset(x,0,sizeof(x))
#define LL long long
#ifdef WIN32
#define AUTO "%I64d"
#else
#define AUTO "%lld"
#endif

using namespace std;

LL n;

bool check(int x){
    for(int i = 2; i*i <= x; i++)
        if(x%i == 0)return false;
    return true;
}

int main()
{
    scanf(AUTO,&n);
    if(n>2&&n%2==0)
        printf("2\n");
    else if(n==2)
        printf("1\n");
    else{
        if(check(n))cout<<"1"<<endl;
        else if(check(n-2))cout<<"2"<<endl;
        else cout<<"3"<<endl;
    }
}