Project Euler begin

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.

 

對於1000一下的正整數中，3的倍數+5的倍數-gcd(3,5)的倍數

簡單的容斥原理：

#include<stdio.h>int cal(int a, int m) {if (a > m) {return 0;}int n, d;d = a;n = (m - a) / d + 1;if ((m - a) % d == 0) {n--;}return a * n + (n - 1) * n / 2 * d;}int main() {int res, n = 1000;res = cal(3, n) + cal(5, n) - cal(15, n);printf("%d\n", res);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, ...

By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.

 

#include<stdio.h>#define NMAX 4000000int num[NMAX];int main() {int i, res;num[1] = 1, num[2] = 2;for (i = 3, res = 2; i < NMAX; i++) {num[i] = num[i - 1] + num[i - 2];if (num[i] > NMAX) {break;}if (!(num[i] & 1)) {res += num[i];}}printf("%d %d\n", i, res);return 0;}

 用中文理解之：

對於上述數列，數列元素值小於4，000，000的所有的些項的值是偶數的總和，是多少？