Paradise_Violet的博客#include <stdio.h> int main(void) { int i; int sum = 0; //sum 表示求和 for (i = 1; i <= 4; ++i) sum = sum + i; printf("sum = %d\n", sum); return 0; }
stitchshaw的博客for 但此法的缺陷是当n很大时,会循环很多次,影响运行时间。时间复杂度为O(n/2)=O(n) #include <iostream> using namespace std;... for (int i=1; i<=n; i++) if (i%2 == 0) sum += i; cout