一、二分答案：

二、题目类型：

三、模板：yxc  

 1.求符合条件的最小值：

//求符合条件中的最小值
bool check(int x){}// 检查x是否满足条件
// 区间[l, r]被划分成[l, mid]和[mid + 1, r]时使用
int sreach(int l,int r)
{
    while(l < r)
      {
        int mid = (l + r) >> 1;
        if(check(mid)) r = mid;
        else l = mid + 1;
      }
    return l;
}

2.求符合条件的最大值：

//求符合条件中的最大值
bool check(int x){}// 检查x是否满足条件
// 区间[l, r]被划分成[l, mid - 1]和[mid, r]时使用：
int sezrch(int l, int r) {
    while (l < r) {
        int mid = l + r >> 1;
        if (check(mid))l = mid;
        else r = mid - 1;
    }
    return 1;
}

3.浮点数二分算法模板：

bool check(double x) {} // 检查x是否满足某种性质
double search(double l, double r)
{
    const double eps = 1e-6;   // eps 表示精度，取决于题目对精度的要求
    while (r - l > eps)
    {
        double mid = (l + r) / 2;
        if (check(mid)) r = mid;
        else l = mid;
    }
    return l;
}

四、例题：

[NOIP2001 提高组] 一元三次方程求解 - 洛谷

#include<iostream>
#include<cstring>
using namespace std;
double a, b, c, d;
int s = 0;
const double eps = 1e-6;   // eps 表示精度，取决于题目对精度的要求
double f(double x) { return a * x * x * x + b * x * x + c * x + d; } 
double search(double l, double r)
{
    while (r - l > eps)
    {
        double mid = (l + r) / 2.0;
        if (f(l)*f(mid)<0) r = mid;
        else l = mid;
    }
    return l;
}
int main() {
    cin >> a >> b >> c >> d;
    for (double i = -100; i <= 100; i++) {
        double z1 = f(i);
        double z2 = f(i + 1);
        if (z1 == 0) 
            printf("%.2f ", i);
        else if (z1 * z2 < 0) 
            printf("%.2f ",search(i, i + 1));
    }
	return 0;
}

数列分段 Section II - 洛谷

#include<iostream>
#include<algorithm>
using namespace std;
int m, n, z1 = 0, z2 = 0;
int a[100005];
bool check(int x) {
	int p = 0, num = 0;
	for (int i = 0; i < n; i++) {
		if (p + a[i] <= x)p += a[i];
		else p = a[i], num++;
	}
	return num < m;
}
int search(int l, int r) {
	while (l < r) {
		int mid = l + r >> 1;
		if (check(mid))r = mid;
		else l = mid + 1;
	}
	return l;
}
int main() {
	cin >> n >> m;
	for (int i = 0; i < n; i++) {
		cin >> a[i];
		z1 = max(z1,a[i]);
		z2 += a[i];
	}
	cout << search(z1, z2);
	return 0;
}

 书的复制 - 洛谷

#include<iostream>
#include<algorithm>
using namespace std;
int m, n, z1 = 0, z2 = 0;
int a[550];
struct r {
	int s, e;
}b[550];
bool check(int x) {
	int p = 0, num = 0;
	for (int i = 1; i <= n; i++) {
		if (p + a[i] <= x)p += a[i];
		else {
			p = a[i];
			num++;
		}
	}
	return num < m;
}
int search(int l, int r) {
	while (l < r) {
		int mid = l + r >> 1;
		if (check(mid))r = mid;
		else l = mid + 1;
	}
	return l;
}
int main() {
	cin >> n >> m;
	for (int i = 1; i <= n; i++) {
		cin >> a[i];
		z1 = max(z1, a[i]);
		z2 += a[i];
	}
	int f = search(z1, z2);
	int p = 0, j = m;
	b[1].s = 1;
	b[m].e = n;
	for (int i = n; i >= 1; i--) {
		if (p + a[i] > f) {
			p = a[i];
			b[j].s = i+1;
			b[--j].e = i;
		}
		else p += a[i];
	}
	for (int i = 1; i <= m; i++) {
		cout << b[i].s << " " << b[i].e << endl;
	}
	return 0;
}