HDU 1001 Sum Problem

2019-11-14 10:10:14

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Sum PRoblem

Time Limit: 1000/500 MS (java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 454929 Accepted Submission(s): 114527Problem DescriptionHey, welcome to HDOJ(Hangzhou Dianzi University Online Judge).In this problem, your task is to calculate SUM(n) = 1 + 2 + 3 + ... + n. InputThe input will consist of a series of integers n, one integer per line. OutputFor 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

1100 Sample Output15050 新手的水題，本來沒什么難度，但對于像我這樣初上路的新手，還是會(huì)遇到很多問題，寫下來促進(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;}看起來沒什么問題，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)問題。題目說結(jié)果不超過32bit,但是如果用公式做的話中間值(n+1)*n可能超過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。