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。