R(N)
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2761 Accepted Submission(s):
1420
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).
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)Source
Recommend
xubiao
暴力题,注意不要超时就行了,注意sqrt的数一定得是int类型的。
include <iostream> #include <math.h> #include <stdio.h> using namespace std; int main() { int n; while(scanf("%d",&n)!=EOF) { if(n==0) { cout<<1<<endl; continue; } int c=sqrt(n/2); int ans=0; int i,j; for(i=0;i<=c;i++) { j=sqrt(n-i*i); if(i*i+j*j==n) { if(i==0||j==0||i==j) ans=ans+4; else ans=ans+8; } } printf("%d ",ans); } return 0; }