3185 - 求1-n以内所有奇数的和
时间限制 : 1 秒
内存限制 : 128 MB
<span style="font-family:宋体;font-size:12pt;">【题目描述】</span><span style="font-size:12pt;"></span>
<span style="font-size:12pt;"><span> </span><span> </span></span><span style="font-family:宋体;font-size:12pt;">给定一个正整数</span><span style="font-size:12pt;">N</span><span style="font-family:宋体;font-size:12pt;">,输出大于等于</span><span style="font-size:12pt;">1</span><span style="font-family:宋体;font-size:12pt;">且小于等于</span><span style="font-size:12pt;">N</span><span style="font-family:宋体;font-size:12pt;">的所有奇数的和</span><span style="font-size:12pt;"></span>
<span style="font-family:宋体;font-size:12pt;">【输入格式】</span><span style="font-size:12pt;"></span>
<span style="font-size:12pt;"><span> </span><span> </span></span><span style="font-family:宋体;font-size:12pt;">一个正整数,即</span><span style="font-size:12pt;">N</span>
<span style="font-family:宋体;font-size:12pt;">【输出格式】</span><span style="font-size:12pt;"></span>
<span style="font-size:12pt;"><span> </span><span> </span></span><span style="font-family:宋体;font-size:12pt;">仅包含一行,即答案</span><span style="font-size:12pt;"></span>
<span style="font-family:宋体;font-size:12pt;">【样例输入】</span><span style="font-size:12pt;"></span>
<span style="font-size:12pt;"><span> </span><span> </span>10</span>
<span style="font-family:宋体;font-size:12pt;">【样例输出】</span><span style="font-size:12pt;"></span>
<span style="font-size:12pt;">25</span>
<span style="font-family:宋体;font-size:12pt;">【数据规模】</span><span style="font-size:12pt;"></span>
<span style="font-size:12pt;">100%</span><span style="font-family:宋体;font-size:12pt;">的数据,</span><span style="font-family:宋体;font-size:12pt;">n</span><span style="font-family:宋体;font-size:12pt;">≤<span>10000</span></span>
<span style="font-family:宋体;font-size:12pt;">【样例解释】</span><span style="font-size:12pt;"></span>
<span style="font-size:12pt;"><span> </span><span> </span>1+3+5+7+9=25</span>
<span style="font-family:宋体;font-size:12pt;"></span>
题目输入
题目输出
输入/输出样例
输入格式
输出格式
C语言解答
#include <stdio.h> int n, i, t; int main(void) { scanf("%d", &n); for (i = 1; i <= n; i++) { if (i % 2) { t += i; } } printf("%d\n", t); return 0; }
C++解答
#include<iostream> using namespace std; int main() { int n,s; cin>>n; s=0; for (int i=1;i<=n;i++) { if (i%2==1) s=s+i; } cout<<s<<endl; return 0; }
Java解答
import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner cin = new Scanner (System.in ); while (cin. hasNext ()){ int a = 0; int n =cin. nextInt (); for (int i = 1; i <= n ; i++) { if (i % 2 == 0) { continue; } a += i; } System.out.println(a); } } }
Python解答
# coding=utf-8 n = int(input()) t = 0 for i in range(1, n + 1): if i % 2 : t = t + i print(t)