第一题太坑了。所以只做出来一道题TAT
Problem A Blocked Points
题意：求直径为n的圆边界上有多少个点。
思路：根据对称性只需求出四分之一圆再乘四就可以了。枚举x轴坐标。题目本身一点都不难有木有！！但是我被坑了有木有！！！第五组测试数据0有木有！！！我一直到还剩二十分钟才发现的。so sad
代码如下：
1 #include <iostream> 2 #include <cstdio> 3 #include <cstdlib> 4 #include <cmath> 5 #include <cstring> 6 #include <algorithm> 7 #include <queue> 8 #include <stack> 9 #include <vector> 10 #include <set> 11 #include <map> 12 #define MP(a, b) make_pair(a, b) 13 #define PB(a) push_back(a) 14 15 using namespace std; 16 17 typedef long long ll; 18 typedef pair<int ,int> pii; 19 typedef pair<unsigned int, unsigned int> puu; 20 typedef pair<int ,double> pid; 21 typedef pair<ll, int> pli; 22 typedef pair<int, ll> pil; 23 24 const int INF = 0x3f3f3f3f; 25 const double eps = 1e-6; 26 27 int main() 28 { 29 // freopen("in.txt", "r", stdin); 30 31 ll n, sum; 32 while(cin >> n){ 33 if(!n) { 34 cout << 1 << endl; 35 continue; 36 } 37 ll loc = n; 38 sum = 1; 39 for(int i=1; i<n; i++){ 40 ll x = n*n - (ll)i*i; 41 double y = sqrt(x); 42 sum += (loc - (ll)y); 43 if((loc - (ll)y)==0) sum++; 44 loc = (ll)y; 45 } 46 sum += (loc-1); 47 cout << 4*sum << endl; 48 } 49 return 0; 50 }