[欧拉习题] ProjEuler[39]:找勾股数

upc1984

原帖http://forum.simwe.com/thread-911102-1-1.htmlMathematica 解法用matlab实现 If p is the perimeter of a right angle triangle with integral length sides, {a,b,c}, there are exactly three solutions for p = 120. {20,48,52}, {24,45,51}, {30,40,50} For which value of p ≤ 1000, is the number of solutions maximised? 120可以写为三组勾股数（整数）之和，分别是{20,48,52}, {24,45,51}, {30,40,50}。要求在不超过1000的数字中求出可以写为最多组勾股数之和的数。 clear;clc;close all; t=1; for i=1:500     for j=i:1000         for k=j:1000             if i^2+j^2==k^2&&i+j+k<=1000                 a(t,:)=[i,j,k];                 t=t+1;             end         end     end end [m,n]=mode(sum(a,2)) [code]Elapsed time is 36.998713 seconds. 复制代码 [/code]以上是我的方法，效率不高，还有其他的好的办法么，请多指导！

qibbxxt

我把我的做法再写一遍

1. clear;clc;close all
   
2. tic;
   
3. [a,b]=meshgrid(1:500);
   
4. c=a.^2+b.^2;
   
5. d=sqrt(c);
   
6. f=round(d);
   
7. e=f==d;
   
8. k=find(e);
   
9. p=a(e)+b(e)+f(e);
   
10. q=accumarray(p,1);
    
11. m=find(q==max(q))
    
12. ij=find(p==m);
    
13. re=[a(k(ij)) b(k(ij)) f(k(ij))];
    
14. re=unique(sort(re,2),'rows');
    
15. fprintf('%d^2+%d^2=%d^2\n',re');
    
16. toc

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1. 40^2+399^2=401^2
   
2. 56^2+390^2=394^2
   
3. 105^2+360^2=375^2
   
4. 120^2+350^2=370^2
   
5. 140^2+336^2=364^2
   
6. 168^2+315^2=357^2
   
7. 210^2+280^2=350^2
   
8. 240^2+252^2=348^2
   
9. Elapsed time is 0.030733 seconds.

复制代码

如果借鉴ls的mode函数
代码改进为：

1. tic;
   
2. [a,b]=meshgrid(1:500);
   
3. c=a.^2+b.^2;
   
4. d=sqrt(c);
   
5. f=round(d);
   
6. e=f==d;
   
7. k=find(e);
   
8. p=a(e)+b(e)+f(e);
   
9. m=mode(p);
   
10. ij=find(p==m);
    
11. re=[a(k(ij)) b(k(ij)) f(k(ij))];
    
12. re=unique(sort(re,2),'rows');
    
13. fprintf('%d^2+%d^2=%d^2\n',re');
    
14. toc

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1. 40^2+399^2=401^2
   
2. 56^2+390^2=394^2
   
3. 105^2+360^2=375^2
   
4. 120^2+350^2=370^2
   
5. 140^2+336^2=364^2
   
6. 168^2+315^2=357^2
   
7. 210^2+280^2=350^2
   
8. 240^2+252^2=348^2
   
9. Elapsed time is 0.023293 seconds.

复制代码