Pizza Cutting

　　　　　　　　　　　　　　Pizza Cutting

When someone calls Ivan lazy, he claims that it is his intelligence that helps him to be so. If his
intelligence allows him to do something at less physical effort, why should he exert more? He also
claims that he always uses his brain and tries to do some work at less effort; this is not his laziness,
rather this is his intellectual smartness.
Once Ivan was asked to cut a pizza into seven pieces to distribute it
among his friends. (Size of the pieces may not be the same. In fact, his
piece will be larger than the others.) He thought a bit, and came to the
conclusion that he can cut it into seven pieces by only three straight cuts
through the pizza with a pizza knife. Accordingly, he cut the pizza in the
following way (guess which one is Ivan’s piece):
One of his friends, who never believed in Ivan’s smartness, was startled
at this intelligence. He thought, if Ivan can do it, why can’t my computer?
So he tried to do a similar (but not exactly as Ivan’s, for Ivan will criticize
him for stealing his idea) job with his computer. He wrote a program that
took the number of straight cuts one makes through the pizza, and output a number representing the
maximum number of pizza pieces it will produce.
Your job here is to write a similar program. It is ensured that Ivan’s friend won’t criticize you for
doing the same job he did.
Input
The input file will contain a single integer N (0 ≤ N ≤ 210000000) in each line representing the number
of straight line cuts one makes through the pizza. A negative number terminates the input.
Output
Output the maximum number of pizza pieces the given number of cuts can produce. Each line should
contain only one output integer without any leading or trailing space.

Sample Input
5
10
-100

Sample Output

16
56

题意：就是给你一个数，然后用n条线最多可把一个披萨（平面）分成几块

tip：怒送好几次WA啊，首先0条线-》一块。。。

　　然后long long 有木有啊，有木有啊。。。

#include<iostream>
#define ll long long

using namespace std;

ll a[210000005];

int main()
{
    a[0]=1;
    a[1]=2;
    for(int i=2;i<=210000005;i++)
        a[i]=a[i-1]+i;

    int n;
    while(cin>>n)
    {
        if(n>=0)
            cout<<a[n]<<endl;
        else
            break;
    }
    return 0;
}