将一个整数拆分成两个整数的平方和算法
问题：请使用C/C++写一个程序实现将一个整数拆分成两个整数的平方和，把所有的可能的组合都要计算出来。答：假定输入的整数为n，则扫描1-(n的平方根)之间的整数，令row=1，column＝(int)(sqrt((double)given)+0.5)，使得row*row+column*column=n的数输出即可。
代码如下所示：
////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; 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;}
运行结果：
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.
页:
[1]