例题 1
代码:
#include <iostream>
using namespace std;
const int N = 100010;
int q[N];
int n,m;
int main()
{
cin >> n >> m;
for(int i = 0; i < n; i ++) scanf("%d", &q[i]);
while(m --)
{
int x;
scanf("%d", &x);
int l = 0, r = n - 1; // 确立边界
while(l < r) // 二分模板
{
int mid = r + l >> 1;
if(q[mid] >= x) r = mid;
else l = mid + 1;
}
if(q[l] != x) cout << "-1 -1" << endl; // 判断是否存在x,若不存在则输出“-1 -1”
else //若存在
{
cout << l << " "; //先输出左边界,再用二分求右边界。
int l = 0, r = n - 1; //二分模板
while(l < r)
{
int mid = r + l + 1 >> 1;
if(q[mid] <= x) l = mid;
else r = mid - 1;
}
cout << r << endl;
}
}
}
习题:
代码:
#include <iostream>
using namespace std;
int main()
{
double n;
cin >> n;
double l = -10000,r = 10000; //二分思想,确立边界
// 二分模板
while(r - l > 1e-8)
{
double mid = (r + l) / 2;
if(mid * mid * mid < n) l = mid;
else r = mid;
}
printf("%.6lf",r);
return 0;
}