PAT 甲级 A1059 Consecutive Factors

自己写的

#include 
#include 
const int maxn1 = 100010;
typedef long long ll;
struct Piki{
	ll pi;
	int ki;
}s[maxn1];

bool isPrime(ll n) {
	if(n == 1) return false;
	ll sqr = (ll)sqrt(1.0 * n); // 开根号 
	for(int i = 2; i <= sqr; i++) {
		if(n % i == 0) return false;
	}
	return true;
}

int much(ll a, ll b) {
	int ans = 1;
	if(b != 0 && b % a == 0 && isPrime(a)) {
		ans += much(a, b / a); 
	}
	return ans;
}

int main() {
	ll n;
	int ans = 0, j = 0;
	scanf("%lld", &n);
	ll sum = n;
	ll maxn = (ll)sqrt(1.0 * n);
	for(int i = 2; i <= maxn; i++) {
		ans = much(i, n) - 1;
		if(ans != 0) {
			s[j].pi = i;
			s[j].ki = ans;
			j++;
			n = n / (ll)pow(i, ans * 1.0);
		}
	}
	if(j == 0) {
		printf("%d=%d", sum, sum);
	} else {
		printf("%d=", sum);
		for(int i = 0; i < j; i++) {
	  		if(s[i].ki > 1) {
	  			printf("%d^%d", s[i].pi, s[i].ki);
			} else {
				printf("%d", s[i].pi);
			}
			if(i < j - 1) printf("*");
    	}
	}
	printf("\n");
	return 0;
}

算法笔记中的

#include 
#include 
const int maxn = 100010;
bool is_prime(int n) {
	if(n <= 1) return false;
	int sqr = (int)sqrt(1.0 * n);
	for(int i = 2; i <= sqr; i++) {
		if(n % i == 0) return false;
	}
	return true;
} 
// 求素数表
int prime[maxn], pNum = 0;
void Find_Prime() {
	for(int i = 1; i < maxn; i++) {
		if(is_prime(i) == true) {
			prime[pNum++] = i;
		}
	}
} 
struct factor {
	int x, cnt;
}fac[10];

int main() {
	Find_Prime(); // 注意写出来 
	int n, num = 0;
	scanf("%d", &n);
	if(n == 1) printf("1=1");
	else {
		printf("%d=", n);
		int sqr = (int)sqrt(1.0 * n); // 乘以1.0使其变为浮点型 
		for(int i = 0; i < pNum && prime[i] <= sqr; i++) {
		if(n % prime[i] == 0) {
			fac[num].x = prime[i];
			fac[num].cnt = 0;
			while(n % prime[i] == 0) {
				fac[num].cnt++;
				n /= prime[i];
			}
			num++;
		}
		if(n == 1) break;
	}
	if(n != 1) {
		fac[num].x = n;
		fac[num++].cnt = 1;
	}
	for(int i = 0; i < num; i++) {
		if(i > 0) printf("*");
		printf("%d", fac[i].x);
		if(fac[i].cnt > 1) {
			printf("^%d", fac[i].cnt);
		}
	}
}
	return 0;
}