尺取+枚举,推出公式以后就是一个枚举加尺取 但是这题的尺取不是对一个值尺取,而是在一个区间内,所以固定左边界,尺取右边界即可 #include<bits/stdc++.h> #define maxn 100005 #define ll long long using namespace std; int t,n,m,k,x,y,z,l,tot; ll cnt,sum[maxn]; int a[maxn]; ll solve(ll L,ll R){//求区间[l,R)内i+j的和 ll ans=
Given the radius and x-y positions of the center of a circle, write a function randPoint which generates a uniform random point in the circle. Note: input and output values are in floating-point. radius and x-y position of the center of the circle is
1.在高德地图上绘制的多边形:经纬度逗号分隔格式:上面是用来方便存坐标的对象:下面是方法测试:直接复制代码即可运行 public class Point { private Double x; private Double y; public Point (Double x , Double y) { this.x = x; this.y = y; } public Double getX() { return x; } public void setX(Double x) { this.x =