省赛AK练习赛之Pizza Cutting

原题：

Description

      

Problem  E

Pizza  Cutting

Input:standard input

Output: standard output

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

分析：

就是最少的刀数，切最多的饼～～～～记住公式——p（最少的刀数，切最多的饼的块数）=n×（n+1）/2+1

原码：

#include<stdio.h>
int main()
{
    long long n;
    while(scanf("%lld",&n) && n>=0)
    {
        printf("%lld\n",n*(n+1)/2+1);
    }
    return 0;
}