欧拉计划1-5题
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.
题目大意：
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。
#include <stdio.h> #include <string.h> #include <ctype.h> void solve() { int sum,i; sum=0; for(i=3; i<1000; i++) { if(i%3==0 || i%5==0) { sum+=i; } } printf("%d\n",sum); } int main() { solve(); return 0; }
Completed on Sun, 31 Mar 2013, 14:35
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.
题目大意：
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。
#include <stdio.h> #include <string.h> #include <ctype.h> #include <math.h> #define N 4000000 int a[1001]; void solve() { int a,b,c,n,count=2; a=1,c=0,b=2; n=3; while(c<=N) { c=a+b; if(n%2!=0) { a=c; } else { b=c; } n++; if(c%2==0) { count+=c; } } printf("%d",count); } int main() { solve(); return 0; }
The prime factors of 13195 are 5, 7, 13 and 29.
What is the largest prime factor of the number 600851475143 ?
题目大意：
13195的质数因子有5,7,13和29.
600851475143的最大质数因子是多少？
#include<stdio.h> #include<math.h> #include<stdbool.h> #define N 600851475143 bool prim(int n) { int i; for(i=2; i*i<=n; i++) { if(n%i==0) return false; } return true; } int main() { long long s=sqrt(N); while(s--) { if(s%2!=0 && prim(s) && (N%s==0)) { printf("%lld\n",s); break; } } return 0; }
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.
题目大意：
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.
找出最大的有由个三位数乘积构成的回文数。
#include<stdio.h> #include<math.h> #include<string.h> #include<ctype.h> #include<stdlib.h> #include<stdbool.h> bool palindromic(int n) //判断一个整数是否为回文数 { char s[10]; sprintf(s,"%d",n); //将整数n保存在字符数组s中 int i,len; len=strlen(s); for(i=0; i<len/2; i++) { if(s[i]!=s[len-i-1]) return false; } return true; } bool have_the_factor(int n) //判断是否含有两个3位数的因数 { int s=999; int r,b; while(s>100) { if((n%s)==0 && ((n/s)>100 && (n/s)<1000)) return true; s--; } return false; } int main() { int i=1000000; while(i>0) { if(palindromic(i) && have_the_factor(i)) { printf("%d\n",i); break; } i--; } return 0; }
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?
题目大意：
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？
#include <stdio.h> #include <string.h> #include <ctype.h> #include <math.h> #define N 20 int gcd(int a, int b) { if(b==0) return a; else return gcd(b,a%b); } int lcm(int a, int b) { return a/(gcd(a,b))*b; } void solve() { int i,s=2; for(i=3; i<=N; i++) { s=lcm(s,i); } printf("%d\n",s); } int main() { solve(); return 0; }
