project euler

/*
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：%dn",sum);
        return 0;
}

/*
 * 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：%dn",sum);
        return 0;
}
int fun_recursive(int n){
        if(n<=1){
                return n;
        }else{
                return fun_recursive(n-1)+fun_recursive(n-2);
        }
}

/* 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("%ldn",n);
        return 0;
}

/* 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

/* 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>