Gcc_Gdb_7_8_1 @ 2024-11-16 16:05:12
https://www.luogu.com.cn/problem/U505175
juruo 的代码:
#include <bits/stdc++.h>
using namespace std;
#define LL long long
#define Pii pair<int, int>
#define ULL unsigned long long
namespace gdb7
{
bool goodnum[10000010], is_prime[10000010], cnt;
int prime[1000010];
int main() {
int n;
cin >> n;
is_prime[0] = is_prime[1] = true;
for (int i = 2; i <= n; ++i) {
if (!is_prime[i]) {
// cout << " i" << i << endl;
prime[++cnt] = i;
for (LL j = 1; j <= n; ++j) {
goodnum[j] = true;
}
}
for (int j = 1; j <= cnt && i * prime[j] <= n; ++j) {
is_prime[i * prime[j]] = true;
if (i % prime[j] == 0) {
break;
}
}
}
for (int i = 1; i <= n; ++i) {
cout << "("<< is_prime[i] << ", " << goodnum[i] << "), ";
}
cout << "\ncnt:" << cnt << endl;
for (int i = 1; i <= cnt; ++i) {
cout << prime[i] << " ";
}
cout << endl;
for (int i = n; i >= 1; --i) {
if (goodnum[i]) {
cout << i << endl;
break;
}
}
return 0;
}
};
int main()
{
return gdb7::main();
}
by Gcc_Gdb_7_8_1 @ 2024-11-16 16:06:30
肯定 TLE
by TeaQiLang @ 2024-11-16 16:08:33
#include<bits/stdc++.h>
using namespace std;
bool qwq(int num){
for (int i=2;i<=sqrt(num);++i){
int temp=num;
while(temp%i==0){
temp/=i;
}
if(temp==1) return true;
}
return false;
}
int main() {
int n;
cin>>n;
for(int i=n;i>= 1;--i){
if(qwq(i)) {
cout<<i<<endl;
break;
}
}
return 0;
}
by TeaQiLang @ 2024-11-16 16:09:07
@Gcc_Gdb_7_8_1
by hjb13357896690 @ 2024-11-16 16:09:47
重打换思路,不要惋惜O(∩_∩)O~(既然你是蒟蒻,那就服从大佬超级蒟蒻的命令吧)
by TeaQiLang @ 2024-11-16 16:10:34
@hjb13357896690 采菊东篱
by hjb13357896690 @ 2024-11-16 16:11:32
@Gcc_Gdb_7_8_1 问一下是哪个团里的
by TeaQiLang @ 2024-11-16 16:13:23
@hjb13357896690 都说了是站外题
by MLE_Automaton @ 2024-11-16 16:14:20
@Gcc_Gdb_7_8_1 看不懂你的代码,我的思路:枚举log n往下的数,二分得到n的i次根向下取整,再幂i,取最大,总时间复杂度是log^3 n
by Gcc_Gdb_7_8_1 @ 2024-11-16 16:14:20
@hjb13357896690 某内部 OJ
by Gcc_Gdb_7_8_1 @ 2024-11-16 16:15:21
哦,对了,其中有调试代码