1. Preface
Official Site: https://pe-cn.github.io/
2. 1~100
2.1. Multiples of 3 and 5
/* Multiples of 3 and 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. 3的倍数和5的倍数 如果我们列出10以内所有3或5的倍数，我们将得到3、5、6和9，这些数的和是23。 求1000以内所有3或5的倍数的和。 */ #include<stdio.h> int main(){ int sum=0; for(int i=0;i<1000;i++){ if(i%3 == 0 || i%5 == 0){ sum+=i; } } printf("1000以内所有3或5的倍数的和：%d\n",sum); return 0; }
2.2. Even Fibonacci numbers
/* * Even Fibonacci numbers * 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. * * 偶斐波那契数 * 斐波那契数列中的每一项都是前两项的和。由1和2开始生成的斐波那契数列前10项为： * * 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, … * 考虑该斐波那契数列中不超过四百万的项，求其中为偶数的项之和。 * */ #include<stdio.h> #include<stdlib.h> int fun_recursive(int n); int main(){ int n; printf("请输入要输出项（自然数）斐波那契数列："); scanf("%d",&n); int *a = (int *)malloc((n+1)*sizeof(int));//如需存储，使用动态内存分配n+1个空间进行存储 int i,sum=0; for(i=1;i<n+1;i++){ int number=fun_recursive(i); printf("%d,",number); if(i!=0 && i%5==0){ printf("\n"); } if(number%2 == 0){ //求偶数项之和 sum+=number; } } printf("偶数项之和：%d\n",sum); return 0; } int fun_recursive(int n){ if(n<=1){ return n; }else{ return fun_recursive(n-1)+fun_recursive(n-2); } }
2.3. Largest prime factor
/* Largest prime factor * The prime factors of 13195 are 5, 7, 13 and 29. * * What is the largest prime factor of the number 600851475143 ? * * 最大质因数 * 13195的所有质因数为5、7、13和29。 * * 600851475143最大的质因数是多少？ * */ #include<stdio.h> int main(){ long int n,i; printf("请输入任意正整数："); scanf("%ld",&n); for(i=2;i<n;i++){ while(n!=i){ if(n%i == 0){ printf("%ld*",i); n/=i; }else{ break; } } } printf("%ld\n",n); return 0; }
2.4. Largest palindrome product
/* Largest palindrome product * 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. * * 最大回文乘积 * 回文数就是从前往后和从后往前读都一样的数。由两个2位数相乘得到的最大回文乘积是 9009 = 91 × 99。 * * 找出由两个3位数相乘得到的最大回文乘积。 * * */ #include<stdio.h> char *reverseStr(char *str,size_t len) int main(){ int max = 0; for(int i=100;i<=999;i++){ for(int j=100;j<=999;j++){ int palindrome = i*j; //字符串反转 *reverseStr() } } } char *reverseStr(char *str, size_t len){ char *start = str; char *end = str+len-1; char ch; if(str != NULL){ while(start<end){ ch = *start; *start++=*end; *end-- =ch; } } return str; }
# coding:utf-8
# Largest palindrome product
# 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.
#
# 最大回文乘积
# 回文数就是从前往后和从后往前读都一样的数。由两个2位数相乘得到的最大回文乘积是 9009 = 91 × 99。
#
# 找出由两个3位数相乘得到的最大回文乘积。
#
#
# 定义字符反转方法
def reverse(num):
strnum = str(num)[::-1]
return int(strnum)
max = None
for a in range(100,1000):
for b in range(100,1000):
rs=a*b
if(rs == reverse(rs) and rs > max):
max=rs;
print '由两个3位数相乘得到的最大回文乘积是：',max
2.5. Smallest multiple
/* Smallest multiple * 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? * * 最小倍数 * 2520是最小的能够被1到10整除的数。 * * 最小的能够被1到20整除的正数是多少？ * */ #include<stdio.h>