hdu 3835(R(N)) 一道简单方程折射出的编程思维
R(N)
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1458 Accepted Submission(s): 762
Problem Description
We know that some positive integer x can be expressed as x=A^2+B^2(A,B are integers). Take x=10 for example, 10=(-3)^2+1^2. We define R(N) (N is positive) to be the total number of variable presentation of N. So R(1)=4, which consists of 1=1^2+0^2, 1=(-1)^2+0^2, 1=0^2+1^2, 1=0^2+(-1)^2.Given N, you are to calculate R(N).
Input
No more than 100 test cases. Each case contains only one integer N(N<=10^9).
Output
For each N, print R(N) in one line.
Sample Input
2 6 10 25 65
Sample Output
4 0 8 12 16
Hint
For the fourth test case, (A,B) can be (0,5), (0,-5), (5,0), (-5,0), (3,4), (3,-4), (-3,4), (-3,-4), (4,3) , (4,-3), (-4,3), (-4,-3) The hint in the problem is very importtant,make sure read is carefully!
Problem analysis:
Give you one integer N,then you need to calculate the occurrence number of the equation "i*i+j*j=N";
本题虽小,但代表了一类问题的解法。也能展现一个人的面对问题时的思维:
本题有多种遍历方法:其中i可以从0到sqrt(1.0*N)遍历;也可以从0到sqrt(1.0*N/2)遍历;
第二种遍历的方法更高效,故采用第二种遍历方法;
View Code
1 #include<iostream> 2 #include<cmath> 3 using namespace std; 4 int main() 5 { 6 int N; 7 while(cin>>N) 8 { 9 int t=(int)sqrt(N/2.0); 10 int i,j; 11 int count=0; 12 for(i=0;i<=t;i++) 13 { 14 j=(int)sqrt(1.0*N-i*i); 15 if(i*i+j*j==N) 16 { 17 if(i==0||j==0||i==j) 18 count+=4; 19 else 20 count+=8; 21 } 22 23 } 24 cout<<count<<endl; 25 } 26 return 0; 27 }