[Algorithm] - binary search

int type of binary search

The basic idea : half basic idea is to find a time to find the narrow half Look

int LowerBound(int a[], int target, int left, int right)
{
    while(right >= left)
    {
        int mid = left + (right - left) / 2;//防止left + right导致上溢
        if(a[mid] > target) right = mid - 1;
        else if(a[mid] < target) left = mid + 1;
        else if(a[mid] == target) return mid;
    }
    return -1;
}

In order to prevent the occurrence of overflow left + right, can be exchanged for left + (right - left) / 2

Trying to find examples where the array element position

#include <stdio.h>

int LowerBound(int a[], int target, int left, int right)
{
    while(right >= left)
    {
        int mid = left + (right - left) / 2;//防止left + right导致上溢
        if(a[mid] > target) right = mid - 1;
        else if(a[mid] < target) left = mid + 1;
        else if(a[mid] == target) return mid;
    }
    return -1;
}
int main(void)
{
    int a[105] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
    int ans;
    ans = LowerBound(a, 11, 0, 9);
    if(ans >= 0) printf("%d", ans);
    else printf("NO");
    return 0;
}

a double binary search

Int type and similar, the only difference is the type double
while - (right left> 0.0000001)
function tz example of big brother

Topic background

tz chiefs from distant Himalayas dug up a NN-order functions, in order to study the magic of the function, tzdalao the functions delegated to the programming club.

Input Format

The first line, a positive integer N and two real numbers ll, rr, showing the closed interval range.

Second row, N + 1N + 1 real number, from left to right represents the coefficients of the functions.

Output Format

Output value of x, rounded to five decimal places.

Sample input

3 -0.9981 0.5
1 -3 -3 1

Sample Output

-0.41421

Explanation

=. 3 above sample n-, X ^{. 3} - 3x ² - 3x +. 1

Problem-solving ideas: talk coefficients derivation function into an array, using the dichotomy to find a derivative is zero point
AC Code

include <stdio.h>
#include <math.h>

double a[100] = {0}, n, ans1, mid;
double y(double x)//判断一阶导数的大小
{
    double j = n-1, ans = 0;
    for(int i = 0 ; i < n && j >= 0; i++, j--)
    {
        ans += a[i]*pow(x, j);
    }
    return ans;//返回一阶导数的值
}
int main(void)
{
    double temp;
    double l, r;
    scanf("%lf%lf%lf", &n, &l, &r);
    temp = n;
    for(int i = 0; i <= n; i++)
    {
        scanf("%lf", &a[i]);
        a[i] = a[i]*temp;//求一阶导数的系数
        temp -= 1;
    }
    while(r - l > 0.0000001)。。二分查找
    {
        mid = (r+l)*1.0/2;
        if(y(mid) > 0) l = mid;
        else  r = mid;
    }
    printf("%.5lf", mid);
    return 0;
}