【发布时间】:2015-09-29 05:33:47
【问题描述】:
问题1的简单解决方案是
static unsigned int solutionInefficient(unsigned int n){
unsigned int sum = 0;
for (unsigned int i = 0; i < n; i++){
if (i % 3 == 0 || i % 5 == 0) {
sum += i;
}
}
return sum;
}
我决定尝试使用 n = 2147483647 的不同测试用例,并在 12 秒内计算出最终结果。所以,我想出了另一个解决方案,它给了我相同的答案,只花了 2 秒:
static unsigned int solutionEfficient(unsigned int n){
unsigned int sum = 0;
unsigned int sum3 = 0;
unsigned int sum5 = 0;
unsigned int sum15 = 0;
for (unsigned int i = 3; i < n; i += 3){
sum3 += i;
}
for (unsigned int i = 5; i < n; i += 5){
sum5 += i;
}
for (unsigned int i = 15; i < n; i += 15){
sum15 += i;
}
return sum3 + sum5 - sum15;
}
我最后一次尝试进行更快的实现涉及一些谷歌搜索并使用算术求和公式,最后一段代码如下所示:
static unsigned int solutionSuperEfficient(unsigned int n){
n = n - 1;
unsigned int t3 = n / (unsigned int)3,
t5 = n / (unsigned int)5,
t15 = n / (unsigned int)15;
unsigned int res_3 = 3 * (t3 * (t3 + 1)) *0.5;
unsigned int res_5 = 5 * (t5 * (t5 + 1)) *0.5;
unsigned int res_15 = 15 * (t15 * (t15 + 1)) *0.5;
return res_3 + res_5 - res_15;
}
但是这并没有为此测试用例提供正确的答案。它确实为 n = 1000 提供了正确答案。我不确定为什么它对我的测试用例失败了,有什么想法吗?
【问题讨论】:
-
与其他答案相差多远?
-
(t3 * (t3 + 1))将溢出 MAX_INT。 -
@avakar,你的意思是UINT_MAX
-
@FredrickGauss,当然。