Codeforces Round #382 (Div. 1) B. Taxes
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题意:funt的年收入是n元,交税的费用=n的最大因数(n自身除外),现在他想逃税,把n拆成任意份,但是每一份不能是1,问他最少要交多少钱的税收。
思路:其实这个我楞是没想出来,上网翻了一下题解,原来一个大于2偶数必定能拆成两个素数相加,那么这道题就很简单了,只有几种情况,如果n是2那么答案就是1,如果n是偶数但大于2,那么答案就是2,如果n是奇数且n是素数答案就是1,如果n是奇数但n不是素数n-2是素数的话,答案就是2,如果n是奇数但n不是素数也不是素数的话,答案就是3。下面给代码:
#include<cstdio>#include<algorithm>#include<cstring>#include<iostream>#include<cmath>#include<queue>#include<string>#include<functional>typedef long long LL;using namespace std;#define maxn 105#define ll l,mid,now<<1#define rr mid+1,r,now<<1|1#define lson l1,mid,l2,r2,now<<1#define rson mid+1,r1,l2,r2,now<<1|1#define pi acos(-1.0)#define inf 0x3f3f3f3fconst int mod = 1e9 + 7;int main(){LL n;while (~scanf("%lld", &n)){if (n == 2){printf("1\n");}else if (n % 2){bool jud1 = false, jud2 = false;for (int i = 2; i <= sqrt(n); i++){if (!(n%i)){jud1 = true;break;}}n -= 2;for (int i = 2; i <= sqrt(n); i++){if (!(n%i)){jud2 = true;break;}}if (!jud1)printf("1\n");else if (jud1&&jud2)printf("3\n");elseprintf("2\n");}elseprintf("2\n");}}
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