Sum PRoblem

1100 Sample Output
15050 新手的水題，本來(lái)沒(méi)什么難度，但對(duì)于像我這樣初上路的新手，還是會(huì)遇到很多問(wèn)題，寫下來(lái)促進(jìn)自己提升。首先可以直接用疊加#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;}看起來(lái)沒(méi)什么問(wèn)題，但是提交上去卻是Wrong Answer仔細(xì)看一下可以發(fā)現(xiàn)sum每完成一個(gè)SUM(n)后沒(méi)有清零，把值帶入了下一個(gè)SUM(n)的計(jì)算中，正確的代碼應(yīng)該為：#include<stdio.h>int main() {int n;while (scanf("%d", &n) != EOF) {int sum = 0; // sum應(yīng)該在這里定義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 則需要注意另一個(gè)問(wèn)題。題目說(shuō)結(jié)果不超過(guò)32bit,但是如果用公式做的話中間值(n+1)*n可能超過(guò)32bit，提交如下代碼#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正確的代碼
#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;}新手入門，處處是坑。