Sum Problem
Time Limit: 1000/500 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 338086 Accepted Submission(s): 85117
Problem Description
Hey, welcome to HDOJ(Hangzhou Dianzi University Online Judge).
In this problem, your task is to calculate SUM(n) = 1 + 2 + 3 + ... + n.
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
Author
DOOM III
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这题其实没有想象的那么简单，要注意先除后乘这样不会爆掉，所以还要考虑奇偶，不过还有种办法是用double存，最后强转回int 也可以
#include<queue>
#include<math.h>
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
#define N 115
int main()
{
int n;
__int64 w;
while(~scanf("%d",&n))
{
if(n%2==0) w=n/2*(n+1);
else w=(n+1)/2*n;
printf("%I64d ",w);
}
return 0;
}
这题其实没有想象的那么简单，要注意先除后乘这样不会爆掉，所以还要考虑奇偶，不过还有种办法是用double存，最后强转回int 也可以
#include<queue>
#include<math.h>
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
#define N 115
int main()
{
int n;
__int64 w;
while(~scanf("%d",&n))
{
if(n%2==0) w=n/2*(n+1);
else w=(n+1)/2*n;
printf("%I64d ",w);
}
return 0;
}
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