题目背景
题目大意(by:曹彦臣)：
平面上有(0,0)到(n,m)的(n+1)*(m+1)个点。问有多少点对所连的线段不过其他点，且长度在[l,h]范围内。
征求翻译。如果你能提供翻译或者题意简述，请直接发讨论，感谢你的贡献。
题目描述
Bessie is returning from a long trip abroad, and Farmer John wants to erect a nice 'Welcome Home' banner in her pasture for her arrival. The banner will hang between two poles on a wire whose length is in the range L1..L2 (1 <= L1 <= L2; L1 <= L2 <= 1,500).
The pasture's size is W x H (1 <= W <= 1,000; 1 <= H <= 1,000), and Farmer John has installed a post at every point with integer
coordinates. Of these (W + 1) * (H + 1) points, Farmer John must pick just two that will hold either end of the wire from which he will hang the banner.
FJ wants no interference with his banner as it hangs and requires that no post be directly under the tight wire he stretches between the two chosen posts.
Farmer John needs your help to figure out how many possible ways he can hang the banner. He knows the number is large and that a 32-bit integer might not be sufficient to compute the answer.
Consider the example pasture below, with W = 2 and H = 1:
* * * * * * The banner size is in the range 2..3. This pasture contains (2+1) * (1+1) = 6 points and has (6 take 2) = (6*5)/(2*1) = 15 different potential pairs of points between which the banner-holding wire might stretch:
(0,0)-(0,1) (0,0)-(2,1) (0,1)-(2,1) (1,1)-(2,0)
(0,0)-(1,0) (0,1)-(1,0) (1,0)-(1,1) (1,1)-(2,1)
(0,0)-(1,1) (0,1)-(1,1) (1,0)-(2,0) (2,0)-(2,1)
(0,0)-(2,0) (0,1)-(2,0) (1,0)-(2,1) Of these pairs, only four have a length in the range 2..3: Len Len
(0,0)-(2,0) 2.00 (0,1)-(2,0) 2.24
(0,0)-(2,1) 2.24 (0,1)-(2,1) 2.00
Of these four, the pairs (0,0)-(2,0) and (0,1)-(2,1) both have a post directly on the line between the endpoints, and thus are
unsuitable.
So, just two pairs of points out of 15 are acceptable candidates for hanging the banner wire.
输入输出格式
输入格式：
* Line 1: Four space-separated integers: W, H, L1, and L2
输出格式：
* Line 1: A single integer denoting the number of possible banners
输入输出样例
2 1 2 3
2
思路：枚举三角形的长和宽
来自大神
这样的题目一定要规划好枚举的方向，我们尝试先枚举两维。 因为这个图的左右是对称的，我们可以只考虑所有斜率>0的线段：根据这条，我们可以分出当线段与水平/竖直方向平行的情况来。显然，只有全部长度为1的线段满足要求，此时如果1在[L,R][L,R]内，则需要加上这些情况。 我们还可以算出一个格点图内与当前枚举的线段相同的线段个数：将这条线段视作与横竖直线构成了一个Rt△Rt△，那么问题可以转化为求在格点图中与该Rt△Rt△相同的个数，这个只需要O(1)O(1)即可完成（对于直角边分别为x,y的Rt△Rt△，在纵方向上有(n-x+1)种可能性，在横方向上有(m-y+1)种可能性，根据乘法原理得到个数），所以重点是在枚举不同的Rt△Rt△。 通过分析两个条件，我们可以知道一个符合要求的Rt△Rt△（设直角边分别为x,y）必须满足： L2≤x2+y2≤R2，gcd(x,y)=1
#include<cmath> #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> using namespace std; int n,m,l,r; long long ans; int gcd(int x,int y){ return x==0?y:gcd(y%x,x); } int main(){ scanf("%d%d%d%d",&n,&m,&l,&r); if(l==1){ ans+=1ll*(n+1)*m+1ll*(m+1)*n; } for(int i=1;i<=n;i++) for(int j=1;j<=m;j++){ double Len=sqrt(i*i+j*j); if(Len<l||Len>r||gcd(i,j)!=1) continue; ans+=2LL*(n-i+1)*(m-j+1); } cout<<ans; }