R(N)
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1815 Accepted Submission(s): 929
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
水题，数学题。
题意是求一个正整数N分解成 N=A^2+B^2 的形式有多少种。A、B可以从0开始。
思路：遍历即可，注意两层循环会超时，尽量用一层循环。用cin、cout不会超时。
代码：
1 #include <iostream>
2 #include <stdio.h>
3 #include <cmath>
4 using namespace std;
5
6 int main()
7 {
8 int n;
9 while(scanf("%d",&n)!=EOF){
10 int sum = 0;
11 for(int i=0;i<=sqrt(n);i++){
12 int j = n-i*i;
13 double t = sqrt(j);
14 if(t-int(t)==0){ //开方成功
15 if(i!=0 && j!=0)
16 sum+=4;
17 else if(i==0 || j==0)
18 sum+=2;
19 }
20 }
21 printf("%d
",sum);
22 }
23 return 0;
24 }
Freecode : www.cnblogs.com/yym2013