HDU 1001 Sum Problem

Sum Problem

Time Limit: 1000/500 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 454929    Accepted Submission(s): 114527

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.

 

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

 

新手的水题，本来没什么难度，但对于像我这样初上路的新手，还是会遇到很多问题，写下来促进自己提升。

首先可以直接用叠加

#include<stdio.h>

int main() {

int n , sum = 0;
while (scanf("%d", &n) != EOF) {

for (int i = 0; i <= n; i++) {
sum += i;
}
printf("%d\n\n", sum);
}
return 0;
}

看起来没什么问题，但是提交上去却是Wrong Answer

仔细看一下可以发现sum每完成一个SUM(n)后没有清零，把值带入了下一个SUM(n)的计算中，正确的代码应该为：

#include<stdio.h>

int main() {

int n;
while (scanf("%d", &n) != EOF) {
int sum = 0; // sum应该在这里定义
for (int i = 0; i <= n; i++) {
sum += i;
}
printf("%d\n\n", sum);
}
return 0;
}

如果用公式做sum = (n+1)*n/2 则需要注意另一个问题。题目说结果不超过32bit,但是如果用公式做的话中间值(n+1)*n可能超过32bit，

提交如下代码

#include<stdio.h>

int main() {

int n, sum = 0;

while (scanf("%d", &n) != EOF) {

sum = (n+1)*n/2;
printf("%d\n\n", sum);
}

return 0;
}

果然是Wrong Answer

正确的代码

#include<stdio.h>

int main() {

    int n, sum = 0;

    while (scanf("%d", &n) != EOF) {

        if(n%2==0)
           sum=n/2*(1+n);
        else
           sum=(n+1)/2*n;

        printf("%d\n\n", sum);
    }

    return 0;
}

新手入门，处处是坑。