欧拉计划（1-5）
If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.
Find the sum of all the multiples of 3 or 5 below 1000.
10以下的自然数中，属于3和5的倍数的有3,5,6和9，它们之和是23.
找出1000以下的自然数中，属于3和5的倍数的数字之和。
Use C：
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
int main(int argc, const char * argv[])
{
clock_t start, end;
start = clock();
int sum=0;
for(int i = 1;i<1000;i++)
{
if(i%3==0 || i%5==0)
{
sum=sum+i;
}
}
end = clock();
printf("Answer : %d,The best response time : %f s",sum,
(double)(end - start) / (double)CLOCKS_PER_SEC);
return 0;
}
使用穷举法可以很容易得出结论，总耗费时间0.000012s，让我们来试试另外一种方法。
Use C:
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
int main(int argc, const char * argv[])
{
// insert code here...
#define n 3
#define m 5
clock_t start, end;
start = clock();
int num1 = (int)(1000-1)/n*n;
int num2 = (int)(1000-1)/m*m;
int num3 = (int)(1000-1)/(n*m)*n*m;
int sum = (num1/n)*(n+num1)/2 + (num2/m)*(m+num2)/2 - (num3/(n*m))*(n*m+num3)/2;
end = clock();
printf("The number is : %d,spent time : %f s",sum,
(double)(end - start) / (double)CLOCKS_PER_SEC);
return 0;
}
Answer: 233168 The best response time: 0.000001s
可以看出代码要比之前的复杂了许多，但是使用这种等差数列求和的方法
的运算时间只有0.000001s,总共减少了900多次的循环，运行速度提高了12倍，而且如果运算量更大，提高的效率越多。
以后每周会在这里分享五道题，以及自己的算法，如果大家有时间推荐先自己去
编译一下再拿过来对比，如果有更效率的算法，欢迎提出评论进行探讨。
Problem2:
Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be:
1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...
By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.
斐波那契数列中的每一项被定义为前两项之和。从1和2开始，斐波那契数列的前十项为：
1, 2, 3, 5, 8, 13, 21, 34, 55, 89, ...
考虑斐波那契数列中数值不超过4百万的项，找出这些项中值为偶数的项之和。
Use C:
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
int main(int argc, const char * argv[])
{
// insert code here...
clock_t start, end;
start = clock();
int n1 = 1;
int n2 = 2;
int sum = 0;
while (n1<4000000&&n2<4000000) {
if(n1 % 2 == 0)
sum = sum+n1;
n1 = n1 + n2;
if(n2 % 2 == 0)
sum = sum+n2;
n2 = n1 + n2;
}
end = clock();
printf("Answer : %d,The best response time: %f s",sum,
(double)(end - start) / (double)CLOCKS_PER_SEC);
return 0;
}
Answer : 4613732,The best response time: 0.000001s
Problem3:
The prime factors of 13195 are 5, 7, 13 and 29.
What is the largest prime factor of the number 600851475143 ?
13195的质数因子有5,7,13和29.
600851475143的最大质数因子是多少？
Use C:
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
int main(int argc, const char * argv[])
{
// insert code here...
clock_t start, end;
start = clock();
long int sum = 600851475143;
for(int i=3; i<sum; )
{
if(sum%i==0)
sum/=i;
else i+=2;
}
end = clock();
printf("Answer : %ld,The best response time: %f s",sum,
(double)(end - start) / (double)CLOCKS_PER_SEC);
return 0;
}
Answer : 6857,The best response time: 0.000042 s
Problem4:
A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 99.
Find the largest palindrome made from the product of two 3-digit numbers.
一个回文数指的是从左向右和从右向左读都一样的数字。最大的由两个两位数乘积构成的回文数是9009 = 91 * 99.
找出最大的有由两个三位数乘积构成的回文数。
Use C:
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <time.h>
int main(int argc, const char * argv[])
{
clock_t start, end;
start = clock();
int mul = 0;
int max = 0;
char a[7];
for(int i = 999 ; i > 99 ; i--)
{
for(int j = 990 ; j > 99 ; j-=11)
/*
我们把回文数的构成可以拆分成这样:
100000a + 10000b + 1000c + 100c + 10b + a
=100001a + 10010b + 1100c
=11(9091a + 910b + 100c)
所以可以看出回文数是一定能够被11整除的
*/
{
mul = i * j;
sprintf(a, "%d", mul);
if(a[0]==a[5]&&a[1]==a[4]&&a[2]==a[3])
{
if(mul > max)
{
max = mul;
}
}
}
}
end = clock();
printf("Answer : %d,The best response time: %f s",max,
(double)(end - start) / (double)CLOCKS_PER_SEC);
}
Answer : 906609,The best response time: 0.011134 s
Problem5:
2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.
What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?
2520是最小的能被1-10中每个数字整除的正整数。
最小的能被1-20中每个数整除的正整数是多少？
Use C:
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <time.h>
int main(int argc, const char * argv[])
{
clock_t start, end;
start = clock();
long int val = 2520;//因为1-20肯定是1-10的倍数，所以这里直接采用能够正好被1-10整除的最小正整数2520做为单位
long int x;
long int y;
for(x = val ; ; x += val)
{
for(y = 19;y > 10 ; y--)
{
if(x % y != 0)
{
break;
}
}
if(y == 10)
{
end = clock();
printf("Answer : %ld,The best response time: %f s",x,
(double)(end - start) / (double)CLOCKS_PER_SEC);
break;
}
}
return 0;
}
Answer : 232792560,The best response time: 0.001288 s