最简单方法 求化简0.0

DarkMoon_Dragon @ 2016-08-16 17:24:11

#include<iostream>
using namespace std;
int main()
{
    int n,sum=0,m=1,k=1,tian=1,i;
    cin>>n;
    for(i=1;i<=n;i++)
    {   sum=sum+m;
        if(tian==k)m++,k++,tian=0;
        tian++;
    }
    cout<<sum;
}
一个循环搞定 (*￣▽)u┌┐ｄ(▽￣*)

by jiangzhy @ 2016-08-20 10:52:18

沙发

by arklight @ 2016-08-21 17:56:47

#include<iostream>
using namespace std;
int main()
{
    int k,j,i,sum1=0,sum2=0;cin>>k;
    for(i=1;sum1<=k-i;++i)sum1+=i;
    for(j=1;j<i;j++)
    {
        sum2+=j*j;
    }
    cout<<sum2+(k-sum1)*(j);
}

by MakiseKurisu_ @ 2016-09-07 20:36:59

Pascal【求翻译！】

by 丁镜脯 @ 2016-09-15 17:16:41

var
  n,i,j,s:longint;
begin
  readln(n);
  while n>0 do begin
    i:=i+1;
    if n>=i then begin
      for j:=1 to i do s:=s+i;
      n:=n-i;
    end else begin
      for j:=1 to n do s:=s+i;
      n:=0;
    end;
  end;
  writeln(s);
end.

by Alextokc @ 2016-10-21 19:47:16

可以的，不错。

by GQL666 @ 2016-10-24 18:55:40

66666666666666666666

我NOIP正好做的这道题，。、。

程序完美AC，但用PASCAL开了1000000000，

爆了。

wuwuwuwu

by Reaepita @ 2016-10-28 21:15:41

#include<iostream>
using namespace std;
int main()
{
    int a,s=0,i,tian=0;
    cin>>a;
    for(i=1;(i+1)*i/2<=a;i++)
    {
        s=s+i*i;
    }i-=1;
    s=s+(a-i*(i+1)/2)*(i+1);
    cout<<s;
}

by DarkMoon_Dragon @ 2016-10-28 21:16:08

@包航 智障

by Kelin @ 2016-11-27 08:08:18

你告诉我这是最简单做法kidding me ??