传送门
做出一个好几个星期屯下来的题目的感觉就是一个字：
爽！
上图的黄点部分就是我们需要求的点
两边的部分很好算
求圆的地方有一个优化，由于圆心是整数点，我们可以把圆分为下面几个部分,阴影部分最难算，最后乘就好了
代码如下所示
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2005;
const int INF = 0x3f3f3f3f;
typedef long long ll;
typedef long double Double;
const Double tiny = 1e-20;ll Ceil(Double x) {ll tt = ceil(x);if(abs(tt - x) < tiny) tt ++;return tt;
}ll Floor(Double x) {ll tt = floor(x);if(abs(tt - x) < tiny) tt --;return tt;
}int main() {
#ifdef LOCALfreopen("in.txt", "r", stdin);
#endifint r, a, b, n;while(~scanf("%d %d %d %d", &r, &a, &b, &n)) {Double leftEdge = Double(a*1.0)/b;Double rightEdge = 2*n - Double(a*1.0)/b;Double radiusTo2 = n - Double(a*1.0)/b;if(leftEdge > rightEdge) swap(leftEdge, rightEdge); ll sum = 0;Double leftDouble = r + 2*leftEdge;int leftInt = ceil(leftDouble);Double rightDouble = r - 2*rightEdge;int rightInt = ceil(rightDouble);sum += 1ll * (rightInt + leftInt) * r;if(rightInt % 2) sum += r & 1;if(leftInt % 2) sum += r & 1;// printf("%d %d %lld\n", leftInt, rightInt, sum);Double cirRadius = (rightEdge - leftEdge) / sqrt(2);// printf("%.3f\n", cirRadius);ll tmpSum = 0;for(int i = Floor(cirRadius), edge = ceil(radiusTo2); i >= edge; --i) {tmpSum += Floor(sqrt( (rightEdge - leftEdge)*(rightEdge - leftEdge) / 2 - 1ll*i*i));// tmpSum += Floor(sqrt( cirRadius * cirRadius - 1ll*i*i));}sum += tmpSum * 8;// printf("%lld\n", sum);sum += 1ll * Floor(cirRadius) * 4;// printf("%lld\n", sum);// printf("%.9f\n", (rightEdge - leftEdge)/2.0);sum += 1ll* Floor(radiusTo2) * Floor(radiusTo2) * 4;sum -= 1ll * Floor(n - leftEdge) * 2;printf("%lld\n", sum);}return 0;
} 
