#include<bits/stdc++.h>
using namespace std;
int main()
{
long long a=0,b=0,c=0,k,i,j;
cin>>k;
for(i=1;i<=100001;i++)
for(j=1;j<=i;j++)
{
a++;
b+=i;
while(a==k)
{
cout<<b<<endl;
return 0;

}

}
}

水的不能再水

#include <bits/stdc++.h>
using namespace std;
int main()
{
    long long n,tot=0;
    long long l=1,k=1;
    
    cin>>n;
    for (long long i=1;i<=n;i++)
    {
        if (l>k) l=1,k++;
        l++;
        tot += k;
    }
    cout<<tot;
}

#include <iostream>
using namespace std;
int main()
{
int k,a,b=0;
int c=0;
cin>>k;
for(a=1;b<=k;a++)
{
b=b+a;
c=c+a*a;
}
a=a-1;
c=c-(b-k)*a;
cout<<c;
return 0;
}

//大佬都在用数学 萌新就直接模拟了lol
#include <cstdio>
#include <iostream>

using namespace std;

int main(){
    int k;
    cin >> k;
    int sum = 0;
    int cur = 1;
    int remain = 1;
    for (int i = 1; i <= k; i++){
        if (remain == 0) remain = ++cur;
        sum += cur;
        remain--;
    }
    cout << sum << endl;
    return 0;
}

3q22321321321

asdasdasdasdas

pascal代码，只用了一重循环，大概是O（n）吧！！
var
n,ans,k,day:longint;
begin
readln(n);
ans:=1;
repeat
k:=k+ans*ans;
day:=day+ans;
ans:=ans+1;
until day>=n;
k:=k-(day-n)*(ans-1);
writeln(k);
end.

先自己加加，找规律就做了。没多想纯数学

#include<iostream>
#include<cmath>
using namespace std;
int main(){
int m,k,i,a=0;
cin>>k;
for(i=1;(0.5*i+0.5*i*i)<=k;i++){
a+=i*i;
m=i;
}
if(k==(0.5*i+0.5*i*i)){
cout<<a;
return 0;
}
else {
a+=(k-(0.5*m+0.5*m*m))*(m+1);
}
cout<<a;
return 0;
}

高中数列方法，需要很高的逻辑能力……你们看不懂算了
program P1974;
var
i,n,j:integer;
a:array[1..10000] of longint;
ans:longint;
begin
read(n);
for i:=1 to n do
a[i]:=(i*i+i) div 2;
i:=0;
repeat
i:=i+1;
until n<=a[i];
ans:=0;
j:=0;
repeat
j:=j+1;
ans:=ans+j*j;
until j=i-1;
ans:=ans+(n-a[i-1])*i;
writeln(ans);
end.

楼下等人代码好长，膜拜……
c++
#include<iostream>
using namespace std;
int main()
{
int n;
cin>>n;
int i=0,ans=0;
while(++i&&i<=n)
{
n-=i;
ans=ans+i*i;
}
ans=ans+i*n;
cout<<ans;
return 0;
}


AC超简易
#include<bits/stdc++.h>
using namespace std;
int a,ans;
int coin(int x)
{
int t,i;
if(a==0)
{
cout<<ans<<endl;
return 0;
}
x++;
t=x;
if(a>=t)
{
a=a-t;
for(i=1;i<=t;i++)
ans=ans+x;
coin(x);
}
else if(t>a)
{
for(i=1;i<=a;i++)
ans=ans+x;
a=0;
coin(x);
}
}
int main()
{
cin>>a;
coin(0);
}

var k,i,s,d:longint;
begin
readln(k);
s:=0;d:=0;i:=0;
repeat inc(i);
d:=d+i;
s:=s+i*i;
until d>=k;
d:=d-k;
s:=s-d*i;
writeln(s);
end.

#include <cstdio>
#include <cstdlib>

#define a

#ifndef a
int main() {
long long k, n = 1, tot = 0, ans = 0;
scanf("%lld", &k);
while ((tot + n) <= k) {
ans += n*n, tot += n++;
}
ans += (k - tot)*n;
printf("%lld", ans);
return 0;
}
#else
int main(){
long long k,n=1,tot=0,ans=0;
scanf("%lld",&k);
while((tot+n)<=k){
for(int i =1;i<=n*n;i++,ans++);
for(int i =1;i<=n;i++,tot++);
n++;
}
for(int i =1;i<=(k-tot)*n;i++,ans++);
printf("%lld",ans);
return 0;
}
#endif
上面的程序
测试数据 #0: Accepted, time = 0 ms, mem = 508 KiB, score = 10
测试数据 #1: Accepted, time = 0 ms, mem = 508 KiB, score = 10
测试数据 #2: Accepted, time = 0 ms, mem = 504 KiB, score = 10
测试数据 #3: Accepted, time = 0 ms, mem = 504 KiB, score = 10
测试数据 #4: Accepted, time = 0 ms, mem = 508 KiB, score = 10
测试数据 #5: Accepted, time = 0 ms, mem = 508 KiB, score = 10
测试数据 #6: Accepted, time = 15 ms, mem = 508 KiB, score = 10
测试数据 #7: Accepted, time = 0 ms, mem = 508 KiB, score = 10
测试数据 #8: Accepted, time = 0 ms, mem = 504 KiB, score = 10
测试数据 #9: Accepted, time = 0 ms, mem = 508 KiB, score = 10
Accepted, time = 15 ms, mem = 508 KiB, score = 100
下面的程序
测试数据 #0: Accepted, time = 0 ms, mem = 504 KiB, score = 10
测试数据 #1: Accepted, time = 0 ms, mem = 504 KiB, score = 10
测试数据 #2: Accepted, time = 0 ms, mem = 504 KiB, score = 10
测试数据 #3: Accepted, time = 0 ms, mem = 508 KiB, score = 10
测试数据 #4: Accepted, time = 0 ms, mem = 504 KiB, score = 10
测试数据 #5: Accepted, time = 0 ms, mem = 508 KiB, score = 10
测试数据 #6: Accepted, time = 0 ms, mem = 508 KiB, score = 10
测试数据 #7: Accepted, time = 0 ms, mem = 508 KiB, score = 10
测试数据 #8: Accepted, time = 0 ms, mem = 508 KiB, score = 10
测试数据 #9: Accepted, time = 0 ms, mem = 504 KiB, score = 10
Accepted, time = 0 ms, mem = 508 KiB, score = 100

#include<cstdio>
int main()
{
    int k,t,i=1,j=1,ans=0;
    scanf("%d",&k);
    for(t=1;t<=k;t++)
    {
        ans+=j;
        i--;
        if(i==0)
        {
            j++;
            i=j;
        }
    }
    printf("%d",ans);
    return 0;
}

这道题就是用来放松心情的
```c++
#include<iostream>
#include<cstdio>
using namespace std;

const int maxn = 10000 + 10;

int f[maxn];
void init () {
int cur = 0;
for (int i = 1; cur < maxn; i++) {
for (int j = 0; j < i; j++) {
if (++cur >= maxn) break;
f[cur] = f[cur-1]+i;
}
}
}

int main () {
init();
int k; cin >> k;
cout << f[k];
}
```

#include<stdio.h>
int main()
{
int k;
scanf("%d",&k);
int sum = 0;
int tmp = 1;
int tmp_ = 1;
for(int i = 1;i <= k; i++)
{
for(int j = 1;j <= tmp; j++)
{
if(tmp_ <= k)
{
sum += tmp;
tmp_++;
}
}
tmp++;
if(tmp_ > k)
break;
}
printf("%d",sum);
}

好水
pascal
var
a,i,j,n,t:longint;
begin
readln(n);
for i:=1 to n do
for j:=1 to i do
begin
inc(a,i);
inc(t);
if t=n then
begin
writeln(a);
halt;
end;
end;
end.


var
ans:longint;
i,k:longint;
begin
read(k);
ans:=0;
i:=1;
while i<=k do
begin
ans:=ans+i*i;
k:=k-i;
inc(i);
if i>k then ans:=ans+k*i;
end;
write(ans);
end.

var
a,i,j,n,t:longint;
begin
readln(n);
for i:=1 to n do
for j:=1 to i do
begin
a:=a+i;
t:=t+1;
if t=n then
begin
writeln(a);
break;
end;
end;
end.