2017_7台风套题

Posted 2020-10-02 蒟蒻zht的博客

传送门

 

A.最大的数

筛质数。

#include <cstdio>
#include <iostream>
#include <algorithm>
#define N 20001

int prime[N];
bool not_prime[N];
int n, cnt, ans, num;

inline int read()
{
	int x = 0, f = 1;
	char ch = getchar();
	for(; !isdigit(ch); ch = getchar()) if(ch == ‘-‘) f = -1;
	for(; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + ch - ‘0‘;
	return x * f;
}

inline void init()
{
	int i, j;
	for(i = 2; i < N; i++)
		if(!not_prime[i])
		{
			prime[++cnt] = i;
			for(j = i + i; j < N; j += i) not_prime[j] = 1;
		}
}

int main()
{
	int i, j, a, x;
	init();
	n = read();
	for(i = 1; i <= n; i++)
	{
		a = read();
		x = std::upper_bound(prime + 1, prime + cnt + 1, a) - prime;
		for(j = x - 1; j >= 1; j--)
			if(a % prime[j] == 0)
			{
				if(prime[j] > num || (prime[j] == num && a > ans))
				{
					num = prime[j];
					ans = a;
				}
				break;
			}
	}
	printf("%d\n", ans);
	return 0;
}

 

B.小红数列

找找规律。

#include <cstdio>
#include <iostream>

int n;

inline int read()
{
	int x = 0, f = 1;
	char ch = getchar();
	for(; !isdigit(ch); ch = getchar()) if(ch == ‘-‘) f = -1;
	for(; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + ch - ‘0‘;
	return x * f;
}

int main()
{
	n = read();
	printf("%d\n", n / 3 * 2 + (n % 3 == 2));
	return 0;
}