上海汉得公司的面试题,有兴趣的看看
用编程语言实现以下各题,
1,求自然数10000以内所有11的倍数的和?
2,1+2+4+8+.......+2^100 = ?
3,有若干只鸡和兔,共30个头,100条腿,求有几只鸡几只兔?
求代码???
[解决办法]
#include <stdio.h>#define NUMBER 10000#define DEVIDE 11void main(){ int end = NUMBER / DEVIDE; long sum = 0; //sum = (1 + end) / 2 * end * DEVIDE; //sum = (1 + end) * end * DEVIDE / 2; sum = ((1 + end) * end * DEVIDE) >> 1; printf("sum = %ld\n", sum);}