ACM1001：Sum Problem

problem description

in this problem, your task is to calculate sum(n) = 1 + 2 + 3 + ... + n.

 

input

the input will consist of a series of integers n, one integer per line.

 

output

for each case, output sum(n) in one line, followed by a blank line. you may assume the result will be in
the range of 32-bit signed integer.

 

sample input

1

100

 

sample output

1

 

5050

 ------------------------------------------------------------------------------------------------------------------

sn = n*(n+1)/2
把两个相同的自然数列逆序相加
2sn=1+n + 2+(n-1) + 3+(n-2) + ... n+1
=n+1 +n+1 + ... +n+1
=n*(n+1)
sn=n*(n+1)/2
另，
m到n的自然数之和:smn=(n-m+1)/2*(m+n)
(n>m)
smn=sn-s(m-1)
=n*(n+1)/2 -(m-1)*(m-1+1)/2
={n*(n+1) - m(m-1)}/2
={n*(n+1) - mn + m(1-m) + mn }/2
={n*(n-m+1)+ m(1+ n-m)}/2
=(n+m)(n-m+1)/2

注意：虽然题目说sum不会大于32位，但是n*(n+1)会大于32位。

 

#include <stdio.h>
#include <stdlib.h>

int main()
{
    long long n;
    //freopen("f:\\input.txt", "r", stdin);
    while (scanf("%lld", &n) != eof)
    {
        printf("%lld\n\n", n * (n + 1) / 2);
    }
    //freopen("con", "r", stdin);
    //system("pause");
    return 0;
}