(Problem 10)Summation of primes
The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.
Find the sum of all the primes below two million.
#include<stdio.h>
#include<math.h>
#include<stdbool.h> #define N 2000000 bool prim(int n)
{
int i;
for(i=; i*i<=n; i++)
{
if(n%i==)
return false;
}
return true;
} int main()
{
int i;
long long sum=;
for(i=; i<=N; i=i+)
{
if(prim(i))
{
sum+=i;
}
}
printf("%lld\n",sum); return ;
}
|
Answer:
|
142913828922 |
(Problem 10)Summation of primes的更多相关文章
- (Problem 47)Distinct primes factors
The first two consecutive numbers to have two distinct prime factors are: 14 = 2 7 15 = 3 5 The fi ...
- (Problem 37)Truncatable primes
The number 3797 has an interesting property. Being prime itself, it is possible to continuously remo ...
- (Problem 35)Circular primes
The number, 197, is called a circular prime because all rotations of the digits: 197, 971, and 719, ...
- (Problem 42)Coded triangle numbers
The nth term of the sequence of triangle numbers is given by, tn = ½n(n+1); so the first ten triangl ...
- (Problem 70)Totient permutation
Euler's Totient function, φ(n) [sometimes called the phi function], is used to determine the number ...
- (Problem 53)Combinatoric selections
There are exactly ten ways of selecting three from five, 12345: 123, 124, 125, 134, 135, 145, 234, 2 ...
- (Problem 49)Prime permutations
The arithmetic sequence, 1487, 4817, 8147, in which each of the terms increases by 3330, is unusual ...
- (Problem 36)Double-base palindromes
The decimal number, 585 = 10010010012(binary), is palindromic in both bases. Find the sum of all num ...
- (Problem 29)Distinct powers
Consider all integer combinations ofabfor 2a5 and 2b5: 22=4, 23=8, 24=16, 25=32 32=9, 33=27, 34=81, ...
随机推荐
- ORA-20000: ORU-10027: buffer overflow, limit of 10000 bytes
要用dbms_output.put_line来输出语句,遇到以下错误: ERROR 位于第 1 行: ORA-20000: ORU-10027: buffer overflow, limit ...
- Http报文格式学习及Get和Post主要区别总结
HTTP(HyperText Transport Protocol,超文本传送协议) http请求数据包的格式:头部(request line + header)+ 数据(data) 头部和数据包体 ...
- HTTP使用BASIC认证的原理及实现方法 (转载)
转自:http://blog.itpub.net/23071790/viewspace-709367 一. BASIC认证概述 在HTTP协议进行通信的过程中,HTTP协议定义了基本认证过程以允许 ...
- 在windows下配置对github的操作--基本操作
一.下载安装 git for widows软件 git for widows 是专门用来在windows下操作 github的软件,提供bash(命令行) 和 gui两种方式. 在bash下,其实就是 ...
- Scrambled Polygon(凸多边形,斜率)
Scrambled Polygon Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 7805 Accepted: 3712 ...
- Objective-c 方法的调用
在书写了类的声明和实现后,应用程序如何去调用它呢? 在Objective-c中,调用方法的简单格式如下: 1⃣ [实例 方法]; 如: [person setAge:32]; 其中 pe ...
- adb shell 命令
adb 概述 SDK的Tools文件夹下包含着Android模拟器操作的重要命令adb,adb的全称为(Android Debug Bridge就是调试桥的作用.通过adb我们可以在Eclipse中方 ...
- DontDestroyOnLoad(Unity3D开发之五)
Unity中我们从A场景切换到B场景的时候,A场景全部对象都会销毁,但有时候我不须要销毁某些东西. 比方一个简单的游戏的背景音乐,我不须要多次反复创建,多个场景播放这一个即可了.这个时候就须要用到Do ...
- 解決 centos中-bash: vim: command not found
用centos 的主机的時候, 用 vim 时出现 -bash: vim: command not found. 只能使用 vi. 那么如何安裝 vim 呢? 输入 rpm -qa|grep vim ...
- gcc代码反汇编查看内存分布[2]: arm-linux-gcc
arm-none-linux-gnueabi-gcc -v gcc version 4.4.1 (Sourcery G++ Lite 2010q1-202) 重点: 代码中的内存分配, 地址从低到高: ...