Project Euler 1-5
今天发现一个很好玩的网站：projecteuler.net/，都是我喜欢的数论题
处理数字，我最喜欢！
Problem 1
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.
#include<stdio.h>
int main()
{
int i,ans;
ans = 0;
for(i = 3;i < 1000;i++)
if(i % 3 == 0 || i % 5 == 0)
ans += i;
printf("%d\n",ans);
return 0;
}
Problem 2
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, ...
Find the sum of all the even-valued terms in the sequence which do not exceed four million.
#include<stdio.h>
int main()
{
long long n,t,tmp,ans;
t = 1;
n = 1;
ans = 0;
while(n <= 4000000)
{
tmp = n;
n = n + t;
t = tmp;
if(n % 2 == 0)
ans += n;
}
printf("%lld\n",ans);
return 0;
}
Problem 3
The prime factors of 13195 are 5, 7, 13 and 29.
What is the largest prime factor of the number 600851475143 ?
#include<stdio.h>
int main()
{
long long num = 600851475143ll;
int i = 2;
while (num > 1)
{
while(num % i == 0) num /= i;
i++;
}
i--;
printf("%d\n",i);
return 0;
}
Problem 4
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.
#include<stdio.h>
int check(int n)
{
if(back(n) == n)
return 1;
return 0;
}
int back(int n)
{
int t;
t = 0;
while(n != 0)
{
t = t * 10 + n % 10;
n /= 10;
}
return t;
}
int main()
{
int n,m,ans,max;
max = -1;
for(n = 100;n <= 999;n++)
for(m = n;m <= 999;m++)
{
ans = m * n;
if(check(ans) && ans > max)
max = ans;
}
printf("%d\n",max);
return 0;
}
Problem 5
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 number that is evenly divisible by all of the numbers from 1 to 20?
#include<stdio.h>
int main()
{
int ans;
ans = 1 * 16 * 9 * 5 * 7 * 11 * 13 * 17 * 19;
printf("%d\n",ans);
return 0;
}
2022年8月24日 21:44
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2022年9月24日 00:06
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2023年7月07日 02:53
Your code implementation seems to be correct. By using a loop and checking if each number is a multiple of 3 or 5, you successfully calculated the sum of all the multiples below 1000. digital marketing company in cochin The answer of 233168 matches the expected result. Well done on providing a working solution!