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.
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.
#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？
