#include<bits/stdc++.h> using namespace std; long ikun(long x)//劈荆成王，伴kun远航 {long n=1,s=0,y=0; while(s<=x) { s=s+pow(n,2); if(s>x) break; else {y++;n++;} } return y; } int main() { long a; cin>>a; cout<<ikun(a); }

a=int(input())
b=0
c=1
flag=True
while flag:
    b+=c**2
    if b<=a:
        c+=1
    else:
        c-=1
        flag=False

print(c)

我是213

a=int(input())
b=0
c=1
flag=True
while flag:
    b+=c**2
    if b<=a:
        c+=1
    else:
        c-=1
        flag=False

print(c)
错了牟翔宇吃两吨

#include <iostream>
#include <math.h>
using namespace std;

int main()
{
    int n,x,i,m=0;
    cin >>x;
    for(i=1;i>0;i++){
        m+=pow(i,2);
        if(m>x){
            n=i-1;
            break;
        }
    }
    cout <<n;
}

#include <bits/stdc++.h>
using namespace std;
int bif(int a)
{
	int z=0,i,b=0,c[1000]={0};
	for(i=1;i<a;i++)
	{
		c[i]=i;
		b+=c[i]*c[i];
		z++;
		if(b>a)
			break;
	
	}
	return z-1;
}

int main() 
{
	int k,f;
	cin>>k;
	cout<<bif(k)<<endl;
return 0;
}

本题最优时间复杂度为O(logN)

$\Sigma_{i=1} {i^2}$ = n(n+1)(2n+1)/6

结合二分可达到log级别复杂度

AC代码

#coding: utf-8
n:int = int(input())
def judge(x: int):
    return x*(x + 1)*(2 * x + 1) <= 6 * n
left = 0
right = n+1
while (right - left > 1):
    mid = int((left + right) / 2)
    if (judge(mid)):
        left = mid
    else:
        right = mid
print("%d\n" % (left))

a=int(input())
b=0
c=1
flag=True
while flag:
    b+=c**2
    if b<=a:
        c+=1
    else:
        c-=1
        flag=False

print(c)
错了何格吃一吨

#include <iostream>
#include <cmath>
using namespace std;
int main() 
{
	long long x;
	cin >> x;
	long long n = 1;
	long long sum = 0;
	while (sum + n * n <= x) 
    {
		sum += n * n;
		n++;
	}
	cout << n - 1 << endl;
	return 0;
}

zZ~

#include <iostream>
#include<cmath>
using namespace std;
int n,x,m;
int main() {
    cin>>x;
    for(n=1;n<=x;n++){
        m=m+pow(n,2);
        if(m>x){
            cout<<n-1;
            break;
        }
    }
return 0;
}

c++

#include <iostream>
#include <cmath>
using namespace std;
int k=0;int x;
int f(int n)
{

    k+=pow(n,2);
    if(k>=x)
    {
        return n;
    }
    n++;
    return f(n);
}
int main()
{


   cin>>x;
   cout<<f(0)-1;
    return 0;
}

a=int(input())
n=1
m=0
while m<=a:
    m+=n**2
    n+=1
print(n-2)

python

x=int(input())

for i in range(1,x):

if i*(i+1)(2i+1)/6<=x and (i+1)(i+2)(2*i+3)/6>x:

print(i)