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: