小易有一个圆心在坐标原点的圆,小易知道圆的半径的平方。小易认为在圆上的点而且横纵坐标都是整数的点是优雅的,小易现在想寻找一个算法计算出优雅的点的个数,请你来帮帮他。
例如:半径的平方如果为25
优雅的点就有:(+/-3, +/-4), (+/-4, +/-3), (0, +/-5) (+/-5, 0),一共12个点。
输入描述:
输入为一个整数,即为圆半径的平方,范围在32位int范围内。
输出描述:
输出为一个整数,即为优雅的点的个数
输入例子:
25
输出例子:
12
重要case:
input output
25 12
65 16
85 16
425 24
204864885 32 //方法1错了
16 4 // 方法2错了,在没有质数的情况下,怎么处理
//1st submit
#include <iostream>
#include <vector>
#include <cmath>
using namespace std;
int totalElegantPoints1(int r2){
int total = 0;
int r = sqrt(r2);
int mid = sqrt(r2 / 2.0);
if (r*r == r2) total += 4;
if (2*mid*mid == r2) total += 4;
for (int i = mid + 1; i < ceil(sqrt(r2)); ++i){
int j = sqrt(r2 - i*i);
if (i*i + j*j == r2) {
total += 8;
cout << i << ' ' << j << endl;
}
}
return total;
}
bool isPrime(int n, int f1){
for (int i = f1; i <= sqrt(n); i++)
if (n%i == 0) return false;
return true;
}
void getOneFactor(int &n, int &f1, int &f1_num){
int f;
for (int i = f1; i <= sqrt(n); i += 2){
// without 2, only consider odd factors
if (n % i == 0){
f1 = i;
do{
f1_num++;
n /= f1;
} while (n % f1 == 0);
break;
}
}
}
void getFactors(int n, vector<int>& v1, vector<int>& v2, int factor1){
if (isPrime(n, factor1)){
if (n % 4 == 1) v1.push_back(1);
if (n % 4 == 3) v2.push_back(1);
return;
}
int f1 = factor1, f1_num = 0;
getOneFactor(n, f1, f1_num);
if (f1 % 4 == 1) v1.push_back(f1_num);
if (f1 % 4 == 3) v2.push_back(f1_num);
if (1 == n) return;//first version
getFactors(n, v1, v2, f1);
}
int totalElegantPoints2(int n){
vector<int> vf41;// f41: factors (f mod 4 == 1)
vector<int> vf43;// f43: factors (f mod 4 == 3)
int pair_num; // number of int pairs
while (n % 2 == 0) n /= 2; // preprocess: drop 2
getFactors(n, vf41, vf43, 3);// update f41 and f43
// traverse f43, if any element is odd, return 0
for (vector<int>::iterator iter = vf43.begin(); iter != vf43.end(); iter++)
if (*iter % 2 == 1) return 0;
// compute int pair number
pair_num = 4;
for (vector<int>::iterator iter = vf41.begin(); iter != vf41.end(); iter++)
pair_num *= *iter + 1;
return pair_num;
}
int main() {
int r2;
while (cin >> r2)
cout << totalElegantPoints2(r2) << endl;
return 0;
}
// failure case: 204864885
failure case:
16 4
分析:s==0
// 2nd submit
#include <iostream>
#include <vector>
#include <cmath>
using namespace std;
bool isPrime(int n, int f1){
for (int i = f1; i <= sqrt(n); i++)
if (n%i == 0) return false;
return true;
}
void getOneFactor(int &n, int &f1, int &f1_num){
int f;
for (int i = f1; i <= sqrt(n); i += 2){// without 2, only consider odd factors
if (n % i == 0){
f1 = i;
do{
f1_num++;
n /= f1;
} while (n % f1 == 0);
break;
}
}
}
void getFactors(int n, vector<int>& v1, vector<int>& v2, int factor1){
if (1 == n) return;
if (isPrime(n, factor1)){
if (n % 4 == 1) v1.push_back(1);
if (n % 4 == 3) v2.push_back(1);
return;
}
int f1 = factor1, f1_num = 0; //
getOneFactor(n, f1, f1_num);
if (f1 % 4 == 1) v1.push_back(f1_num);
if (f1 % 4 == 3) v2.push_back(f1_num);
getFactors(n, v1, v2, f1);
}
int totalElegantPoints2(int n){
vector<int> vf41;// f41: factors (f mod 4 == 1)
vector<int> vf43;// f43: factors (f mod 4 == 3)
int pair_num; // number of int pairs
while (n % 2 == 0) n /= 2; // preprocess: drop 2
getFactors(n, vf41, vf43, 3);// update f41 and f43
// traverse f43, if any element is odd, return 0
for (vector<int>::iterator iter = vf43.begin(); iter != vf43.end(); iter++)
if (*iter % 2 == 1) return 0;
// compute int pair number
pair_num = 4;
for (vector<int>::iterator iter = vf41.begin(); iter != vf41.end(); iter++)
pair_num *= *iter + 1;
return pair_num;
}
int main() {
int r2;
cin >> r2;
cout << totalElegantPoints2(r2) << endl;
return 0;
}
totalElegantPoints2
提交时间:2018-05-20 语言:C++ 运行时间: 5 ms 占用内存:480K 状态:答案正确
表现并没有更好,为什么呢?
不可靠的sqrt
问题: ,判断r是不是整数。
思路一: if(r == int(r))
思路二: if( )
失败:$n=85137521, r=9227.00000(float类型)$