p*2
2 楼
不是说binary search吗?
f*t
3 楼
牛顿迭代法,任取一a,然后不断a = (a + n/a)/2循环,直到某个精度。
double sqrt(int n)
{
if(n <= 0)
return 0;
double ans = (double)n / 2.0;
int round = 0;
const double epsilon = 0.00001;
while(true) {
round++;
ans = (ans + (double)n / ans) / 2;
double diff = ans * ans - (double)n;
if(diff < 0)
diff = -diff;
if(diff < epsilon)
break;
}
printf("Calculated %d rounds\n", round);
return ans;
}
double sqrt(int n)
{
if(n <= 0)
return 0;
double ans = (double)n / 2.0;
int round = 0;
const double epsilon = 0.00001;
while(true) {
round++;
ans = (ans + (double)n / ans) / 2;
double diff = ans * ans - (double)n;
if(diff < 0)
diff = -diff;
if(diff < epsilon)
break;
}
printf("Calculated %d rounds\n", round);
return ans;
}
a*h
6 楼
格式像算除法,
2011 是“被除数”, 两位一分,
20 > 4×4 <5*5, 所以第一位是 4,
余数加11 是 411, 第一位的结果 4 × 20 加上 a, 计算 8a*a < 411, a=4 是第
二位,
类推。
2011 是“被除数”, 两位一分,
20 > 4×4 <5*5, 所以第一位是 4,
余数加11 是 411, 第一位的结果 4 × 20 加上 a, 计算 8a*a < 411, a=4 是第
二位,
类推。
d*y
10 楼
using binary search
public double sqrt(int n)
{
if(n<=0)
return 0;
int low = 0;
int high = n + 1;
while((high - low) > 1)
{
middle = (high + low)/2;
if(middle*middle<=n)
low = middle;
else
high = middle;
}
}
public double sqrt(int n)
{
if(n<=0)
return 0;
int low = 0;
int high = n + 1;
while((high - low) > 1)
{
middle = (high + low)/2;
if(middle*middle<=n)
low = middle;
else
high = middle;
}
}
d*y
11 楼
using binary search
public double sqrt(int n)
{
if(n<=0)
return 0;
int low = 0;
int high = n + 1;
while((high - low) > 1)
{
middle = (high + low)/2;
if(middle*middle<=n)
low = middle;
else
high = middle;
}
return low;
}
public double sqrt(int n)
{
if(n<=0)
return 0;
int low = 0;
int high = n + 1;
while((high - low) > 1)
{
middle = (high + low)/2;
if(middle*middle<=n)
low = middle;
else
high = middle;
}
return low;
}
d*y
13 楼
I didn't test my code. just to practice.Any comments will be appreciated
I think you may know the concept of this type of question.
e.g. if n = 100 you choose 0 - 101 boundary. 50 * 50 > n so the boundary
turns into 0 - 50 keep doing this.
BTW : what is Gover's security clearance and DoD secret security clearance?
I think you may know the concept of this type of question.
e.g. if n = 100 you choose 0 - 101 boundary. 50 * 50 > n so the boundary
turns into 0 - 50 keep doing this.
BTW : what is Gover's security clearance and DoD secret security clearance?
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