学步园

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.

 

Answer:

233168

对于1000一下的正整数中，3的倍数+5的倍数-gcd(3,5)的倍数

简单的容斥原理：

#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.

 

Answer:

4613732

 

#include<stdio.h>
#define NMAX 4000000
int 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：

对于上述数列，数列元素值小于4，000，000的所有的些项的值是偶数的总和，是多少？