贪心法（服务器最少跳转次数）

找约数的个数O(n^0.5)

#include<iostream>
#include<cmath>
#include<cstdio>
using namespace std;

int exe(int n) {
	int ans = 0;
	int bound = sqrt(n) + 1;
	for (int i = 1; i <= bound; i++) {
		if (n%i == 0) {
			++ans;
			if (n / i > bound) ++ans;//一对儿，在另一半也有一个
		}
	}
	return ans;
}
int main()
{
	//data
	int n = 0;
	int x = 0;
	//input
	cin >> n;
	for (size_t i = 0; i <n; i++)
	{
		cin >> x;
		cout<< exe(x)<<endl;
	}
	//output
	//system("pause");
	return 0;
}

代理服务器最少跳转几次

#include<stdio.h>
#include<string.h>
#include<stdlib.h>

int main() {
	//data
	int n,m;
	char proxy[1000][16];
	char server[2000][16];
	//input
	scanf("%d", &n);
	for (int i = 0; i < n; i++) {
		scanf("%s", proxy[i]);
	}
	scanf("%d", &m);
	for (int i = 0; i < m; i++) {
		scanf("%s", server[i]);
	}
	//贪心法
	int index = 0, count = 0, flag = 1;
	while (index < m) {
		int max = 0;
		for (int i = 0; i < n; i++) {
			//遍历每一个代理服务器，看其从当前位置最多可以访问到多少服务器max
			int j = index;
			while (strcmp(proxy[i], server[j]) && j < m)
				j++;
			if (j - index > max) max = j - index;
		}
		if (max == 0) {
			flag = 0;
			break;
		}
		count++;
		index += max;
	}
	//output
	if (flag)
		printf("%d\n", count - 1);  //因为第一次不算转换
	else 
		printf("-1\n");

	return 0;
}

质数的分解：120=2×2×2×3×5

#include <iostream>
#include <cmath>
using namespace std;
int main() {
	//这题的关键：
	//1、是sqrt,可以极大减少复杂度，若是到方根N仍大于1，则必还有且只还有1个质因数
	//2、每次瞬间整除都可帮助减少遍历范围
	long M = 100;
	while (cin >> M)
	{
		long count = 0;
		for (long j = 2; j <= sqrt(M); j++)
		{
			while (M%j == 0)
			{
				M = M / j;
				count++;
			}
			if (M <= 1)break;
		}
		if (M > 1)count++;
		cout << count << endl;
	}
	return 0;
}