#include<iostream>
#include<cmath>
using namespace std;
int main()
{
int i,j,k;
cin>>k;
i=j=0;
long sum=0;
while (i<k)
{
i+=j+1;
j++;
sum+=j*j;
}
if (i>k) sum-=(i-k)*j;
cout<<sum;
return 0;
}

var
k,n:longint;
procedure main;
var
i,t,s:longint;
begin
readln(k);
t:=1;
s:=0;
for i:= 1 to k do
begin
if s<t
then
begin
n:=n+t;
inc(s);
end
else
begin
s:=1;
inc(t);
n:=n+t;
end;
end;
writeln(n);
end;
begin
main;
readln
end.

var
k,n:longint;
procedure main;
var
i,t,s:longint;
begin
readln(k);
t:=1;
s:=0;
for i:= 1 to k do
begin
if s<t
then
begin
n:=n+t;
inc(s);
end
else
begin
s:=1;
inc(t);
n:=n+t;
end;
end;
writeln(n);
end;
begin
main;
readln
end.

program coin;
var
k,n:longint;
procedure main;
var
i,t,s:longint;
begin
readln(k);
t:=1;
s:=0;
for i:= 1 to k do
begin
if s<t
then
begin
n:=n+t;
inc(s);
end
else
begin
s:=1;
inc(t);
n:=n+t;
end;
end;
writeln(n);
end;
begin
main;
readln
end.

O(1)算法
那些用O(n)算法的就不要发题解了，一点思维都没有

补充一个公式：1^2+2^2+…+n^2=n*(n+1)*(2*n+1)/6

#include<cmath>
#include<cstdio>
int n,m,ans;
int main()
{
scanf("%d",&n);
m=(-1+(int)sqrt(1+8*n))>>1;
ans=m*(m+1);
ans=ans*((m<<1)+1)/6+(n-(ans>>1))*(m+1);
printf("%d",ans);
return 0;
}

水题啊 那年的~

var
n,i:integer; s:longint;
begin
readln(n);
s:=0;
for i:=1 to n do
begin
n:=n-i;
s:=i*i+s;
if n<=0 then begin s:=s+i*n; writeln(s); halt; end;
end;
writeln(s);
end.

普及组什么时候这么水了。这个连刚学的都会做呀。。。。
var
x,ans,n,i,m:longint;
begin
readln(n);
x:=0; i:=0; m:=1;
while x<n do
begin
ans:=ans+m;
inc(x);
inc(i);
if i=m then
begin
inc(m);
i:=0;
end;
end;
writeln(ans);
end.

program lx;
var sum,s,a,d,k:longint;
begin
read(k);
d:=0;s:=0;sum:=0;
repeat
inc(d);
s:=s+d;
if s<=k then sum:=sum+d*d
else begin
s:=s-d;
sum:=sum+(k-s)*d;
s:=s+d;
end;
until s>=k;
write(sum);
end.

超水，这里就不发答案了

var
k,n,s:longint;
begin
readln(k);
n:=trunc((sqrt(8*k+1)-1)/2);
s:=n*(n+1)*(2*n+1) div 6 + (k-n*(n+1) div 2)*(n+1);
writeln(s);
readln;
end.

#include<iostream>
using namespace std;
int main()
{
int k,j,temp;
long long sum=0;
cin>>k;
for(int i=1,j=1,temp=1;i<=k;i++)
{
sum+=j;
temp--;
if(temp==0)temp=++j;
}
cout<<sum;
return 0;
}

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

include<iostream> include<cstdio>

using namespace std;
int main()
{
long long int a,b,c,e,f,g;
int d=0;
cin>>a;
while(1)
{
d++;
c=(d(d+1))/2;
if(c>=a)
{break; }
}
d=d-1;
e=(2a-d(d+1))/2(d+1);
f=d(d+1)(2*d+1)/6;
g=f+e;
cout<<g<<endl;
return 0;
}

program p1974;
var k,t,j,i,s:longint;
begin
read(k);
t:=0;
i:=0;
while t<k do
begin
inc(i);
t:=t+i;
end;
s:=0;
for j:=1 to i do
begin
s:=s+jj;
end;
s:=s-i(t-k);
write(s);
end.

#include<algorithm>
#include<cstring>
#include<fstream>

using namespace std;
ifstream fin("coin.in");
ofstream fout("coin.out");
int n,i=0,ct,net;

int main()
{
fin>>n;
while(n>0)
{
i++;
ct+=i;
n=n-i;
net=i;
}
//fout<<ct;
if(n<0)
{
ct=ct-(0-n,i);
}
fout<<ct;

return 0;
}

var n,k,i,j,sum,z:longint;
begin
readln(k);
j:=0;
repeat
j:=j+1;
i:=i+j;
until i>=k;
z:=i;
n:=z-k;
for i:=1 to j do sum:=sum+i*i;
sum:=sum-n*j;
writeln(sum);
end.

编译成功

Free Pascal Compiler version 2.6.4 [2014/03/06] for i386
Copyright (c) 1993-2014 by Florian Klaempfl and others
Target OS: Win32 for i386
Compiling foo.pas
Linking foo.exe
19 lines compiled, 0.0 sec , 27840 bytes code, 1628 bytes data

测试数据 #0: Accepted, time = 0 ms, mem = 436 KiB, score = 10

测试数据 #1: Accepted, time = 0 ms, mem = 436 KiB, score = 10

测试数据 #2: Accepted, time = 0 ms, mem = 436 KiB, score = 10

测试数据 #3: Accepted, time = 0 ms, mem = 436 KiB, score = 10

测试数据 #4: Accepted, time = 0 ms, mem = 436 KiB, score = 10

测试数据 #5: Accepted, time = 0 ms, mem = 440 KiB, score = 10

测试数据 #6: Accepted, time = 0 ms, mem = 436 KiB, score = 10

测试数据 #7: Accepted, time = 0 ms, mem = 436 KiB, score = 10

测试数据 #8: Accepted, time = 0 ms, mem = 436 KiB, score = 10

测试数据 #9: Accepted, time = 0 ms, mem = 436 KiB, score = 10

Accepted, time = 0 ms, mem = 440 KiB, score = 100

代码
program p1974;
var k,t,j,i,s:longint;
begin
read(k);
t:=0;
i:=0;
while t<k do
begin
inc(i);
t:=t+i;
end;
s:=0;
for j:=1 to i do
begin
s:=s+j*j;
end;
s:=s-i*(t-k);
write(s);
end.

奇葩
​

program p1974;
var k,i,j,s,t:longint;
begin
read(k);
t:=0;
i:=0;
s:=0;
while t<=k do
begin
i:=i+1;
t:=t+i;
end;
for j:=1 to i do
begin
s:=s+j*j;
end;
s:=s-i*(t-k);

write(s);
end.

var j,k,n,m,s,t:longint;
a:string;
b:array[1..100000000]of longint;

begin

read(n);
s:=1;
for j:=1 to n do
begin
t:=j;
while t<>0 do
begin
b[s]:=j; s:=s+1;
t:=t-1;
end;
end;
for j:=1 to n do m:=m+b[j];
write(m);

end.