問題：請使用C/C++寫一個程式實現將一個整數拆分成兩個整數的平方和，把所有的可能的組合都要計算出來。

答：假定輸入的整數為n，則掃描1-(n的平方根)之間的整數，令row=1，column＝(int)(sqrt((double)given)+0.5)，使得row*row+column*column=n的數輸出即可。

程式碼如下所示：

//
//  main.cpp
//  MyProjectForCPP
//
//  Created by labuser on 11/2/11.
//  Copyright 2011 __MyCompanyName__. All rights reserved.
//

#include <iostream>
#include <math.h>

using namespace std;


int main (int argc, const char * argv[])
{
    int given;
    int row,column;
    int count;
    char line[100];

    printf("\nRepressenting a Given Number as the the Sum of Two Squares");
    printf("\n==========================================================\n");
    printf("\nAn Integer Please ---> ");
    gets(line);

    given = atoi(line);
    printf("\nCount     X       Y");
    printf("\n------  -----  ------");

    row =1;
    column=(int)(sqrt((double)given)+0.5);
    while (row<=given && column>0) {
        if(row*row+column*column==given){
            ++count;
            printf("\n%5d%7d%7d",count,row,column);
            ++row;
            --column;
        }
        else if(row*row+column*column>given)
        {
            --column;
        }else{
            ++row;
        }
    }

    if(count==0){
        printf("\n\nSorry, NO ANSWER found.");
    }else{
        printf("\n\nThere are %d possible answers.\n",count);
    }

    return 0;
}


 

執行結果：
Repressenting a Given Number as the the Sum of Two Squares
==================================================
An Integer Please ---> 200

Count X Y
------ ----- ------
1 2 14
2 10 10
3 14 2

There are 3 possible answers.