题目9：找出唯一的满足a + b + c = 1000的毕达哥拉斯三元组{a , b , c},欧拉计划（Project Euler）,数据结构与算法,鱼C论坛

永恒的蓝色梦想 发表于 2019-8-3 16:19:29

本帖最后由永恒的蓝色梦想于 2020-8-2 19:29 编辑

for b in range(1,999):
for a in range(1, b):
   c = 100 - a - b

   if a * a + b * b == c * c:
         print(a * b * c)
         exit()

1666194196 发表于 2019-10-10 08:44:04

#include <stdio.h>
#include<math.h>
void main(){
//特殊毕达哥拉斯三元组
//毕达哥拉斯三元组是三个自然数a < b < c组成的集合，并满足
//a2 + b2 = c2
//例如，3^2 + 4^2 = 9 + 16 = 25 = 5^2。
//有且只有一个毕达哥拉斯三元组满足 a + b + c = 1000。求这个三元组的乘积abc。

int a,b,c;
for(a=0;a<333;a++){ //a+b+c=1000 且 a<b<c 所有 a<333否则 333 + 334 +335 >1000了
for(b=a+1;b<500;b++){//如果 b>=500 c>bb+c就大于 1000了，所以 b<500
      c = sqrt(b*b+a*a);//c=根号下 a*a+b*b
   if(c*c==b*b+a*a&&a+b+c==1000){//满足两个条件则输出....
   printf("a=%d,b=%d,c=%d,a*b*c=%d",a,b,c,a*b*c);
   }
}

}

}

foxiangzun 发表于 2019-11-5 20:21:33

1、这三个数实际存在，且不为 0
2、这三个数的平方数符合勾股定理
3、这三个数之和刚好 1000

刚开始考虑到用穷举法，但是如果三个数都穷举，那么算法复杂度就是 O(N3)，代价太大，消耗的资源和时间都太长，所以，即使用穷举，也最好能将循环控制到最小，那么用两层循环即可，代码如下：

for i in range(1, 999) :
   for j in range(1, 999) :
            if (i * i + j * j == pow(1000 - i - j, 2)) and (i < j < (1000 - i - j)) :
                     print(i, j, 1000 - i - j)
                     print(i * j * (1000 - i - j))
                     break

来自异界的飞猪 发表于 2019-11-27 20:33:59

翅膀团发表于 2015-7-16 22:24
#include

int a=0,b=1,c=2;

函数中的那些判断条件实干吗的没看懂，方便给讲一讲吗{:10_256:}

guoquanli 发表于 2019-12-31 18:00:20

#include <stdio.h>
#include <stdlib.h>
#include "string.h"
int main(void)
{
   for(int a = 1 ; a <= 1000; a++){
         for(int b = a + 1; b <= 1000; b++){
            for(int c = b+1; c <= 1000; c++){
               if(a*a + b*b == c*c && a+b+c ==1000){
                     printf("a= %d|b = %d|c= %d|a*b*c = %d\n",a,b,c,a*b*c);
               }
            }
         }
   }
   return 0;
}
=========================
a = 200|b = 375|c =425|a*b*c = 31875000

代号-K 发表于 2020-3-12 18:57:07

本帖最后由代号-K 于 2020-3-12 19:01 编辑

answer:200, 375, 425
首先把方程式化简一下；
void getPythagorean(int *a, int *b)
{
int i, j;
for(i = 1; i < 334; i++)
{
   for(j = 500 - i; j <500; j++)
   {
         if( i*j % 1000 == 0)
         {
            if(500 + i*j/1000 - i - j == 0)
            {
               *a = i;
               *b = j;
               return;
            }
         }
   }
}
}

int main(int argc, char *argv[])
{
ElementTyperet;
int a;
int b;
getPythagorean(&a, &b);
printf("answer:%d, %d, %d\n",a, b, 1000 - (a + b));
return 0;
}

leon0149 发表于 2020-5-8 23:50:01

本帖最后由 leon0149 于 2020-5-8 23:52 编辑

#include <stdio.h>
#include <time.h>

int main(void)
{
clock_t start, finish;
double duration;
start = clock();
for (int i = 1; i < 300; ++i) {
   for (int j = i + 1;j < 1000; ++j) {
         int k = 1000 - i - j;
         if (k < j)
            goto here;
         else{
            if (i * i + j * j == k * k) {
               printf("%d * %d * %d = %d\n", i, j, k, i * j * k);
               finish = clock();
               duration = (double )(finish - start) / CLOCKS_PER_SEC;
               printf("%.3f s", duration);
               return 0;
            }
         }
   }
   here:
   1;
}

return 0;
}
输出结果:
200 * 375 * 425 = 31875000
0.000 s

q327363363 发表于 2020-6-1 11:06:34

for a in range(1,998):
for b in range(1,998):
   for c in range(1,998):
         if a**2+b**2==c**2 and a+b+c<1000:
            print(a,b,c)

数据小随从 发表于 2020-6-11 20:14:56

本帖最后由永恒的蓝色梦想于 2020-6-30 18:02 编辑

#include<stdio.h>
#include<stdlib.h>

void main()
{
int i, j;
long result;
for (i = 1; i < 1000; i++)
{
for (j = i; (j < 1000 && j < (1000 - i - j)); j++)
{
if (i * i + j * j == (1000 - i - j)*(1000 - i - j))
{
printf("%d^2+%d^2=%d^2\n", i, j, 1000 - i - j);
printf("这三个数的乘积=%ld\n", i*j*(1000 - i - j));
break;
}
}

}
system("pause");
}

输出结果为：200^2+375^2=425^2
这三个数的乘积=31875000

xuan1788 发表于 2020-7-23 17:16:33

import time
from time import time
def main():
start = time()
print({a*b*(1000-a-b): for a in range(1, 1000//3) for b in range(500) if a**2 + b**2 == (1000-a-b)**2})
print(time() - start)

if __name__ == "__main__":
main()

{31875000: }
0.2583038806915283

583164028 发表于 2020-8-10 09:49:58

幻世伽蓝发表于 2016-7-11 19:37

严重鄙视这种三重for循环的搞法

yhhpf 发表于 2020-8-24 15:14:51

for a in range(1,1000):
   for b in range(a,1000):
            if a*a+b*b == (1000-a-b)**2:
                     print (a*b*(1000-a-b))
                     break

4444567 发表于 2020-8-31 23:44:43

def bd(a,b,c):
m =
i = min(m)
j = max(m)
m.remove(i)
m.remove(j)
k = m

if (i*i + k*k == j*j):
   return True

for i in range(0,1000):
for j in range(i+1,1000):
   k = 1000 - i -j
   if bd(i,j,k) and k>j:
         print(i,j,k)
         print(i*j*k)

有马_冬巳 发表于 2020-10-4 11:27:48

'''一个毕达哥拉斯三元组是一个包含三个自然数的集合，a < b < c，满足条件：a2 + b2 = c2
例如：32 + 42 = 9 + 16 = 25 = 52。
已知存在并且只存在一个毕达哥拉斯三元组满足条件 a + b + c = 1000。
找出该三元组中 abc 的乘积。'''

def pythagorean():
for a in range(1,500):
   for b in range(1,500):
         if a**2 + b**2 == (1000-a-b)**2:
            print("该三元组是: %d,%d,%d" %(a,b,1000-a-b))
            break
   if a**2 + b**2 == (1000-a-b)**2:
         break

start_pythagorean = time.time()
pythagorean()
time_pythagorean = time.time() - start_pythagorean
print("%f秒" %time_pythagorean)

该三元组是: 200,375,425
0.062495秒

gonorth 发表于 2020-11-5 15:41:16

def spt():
for a in range(1000):
   for b in range(a , 1000):
         for c in range(b, 1000):
            if (a + b + c) == 1000 and ((a** 2 + b ** 2) == c ** 2) and a < b < c:
               print (a , b ,c , a*b*c)

a1351468657 发表于 2021-3-6 16:47:13

#include <stdio.h>

main()
{
int a, b, c;

for (a = 1; a < 1000; a++)
{
for (b = a; b < 1000; b++)
{
c = 1000 - a - b;
if ((a * a + b * b) == c * c)
{
printf("a = %d, b = %d, c = %d", a, b, c);
break;
}
if (c <= b)
{
goto Label;
}
}
Label:continue;
}
}

a = 200, b = 375, c = 425
a * b * c = 31875000

Kuri5u 发表于 2021-5-6 23:33:40

for a in range(100,800):
for b in range(100,800):
   c=1000-a-b
   if c**2==a**2+b**2 and a<b<c:
         print(f"{a} {b} {c} {a*b*c}")

200 375 425 31875000

卢本伟牛逼 发表于 2021-8-10 22:12:09

#include <stdio.h>

// 1、a^2 + b^2 = C^2
// 2、c = 1000 - a - b
// 极坐标化
// rcosx + rsinx = r^2
// r = 1000 / (sinx + cosx + 1) -> c(414 ~ 500)
// (a < b < c)

static count = 0;

void method1(void)
{
int a,b,c;

for(c = 414; c <= 500; c++)
{//a + b =1000 - c
for(a = 1; a < b; a++)
{
count += 1;
b = 1000 - c - a;
if (a*a + b*b == c*c){
printf("method1 result:\n");
printf("a = %d, b = %d, c = %d\n", a,b,c);
printf("count = %d\n", count);
return;
}
}
}
printf("Can't' find~\n'");
}

void method2(void)
{
int a,b,c;
for(c = 1;c<1000;c++){
for(a = 1;a<b;a++){
count += 1;
b = 1000 - c - a;
if (a*a + b*b == c*c){
printf("method1 result:\n");
printf("a = %d, b = %d, c = %d\n", a,b,c);
printf("count = %d\n", count);
return;
}
}
}
printf("Can't' find~\n'");
}

int main(void)
{
method1();
method2();
return 0;
}
//method1 - count = 3393, a = 200, b = 375, c = 425;
//method2 - count = 167004, a = 200, b = 375, c = 425;
{:10_256:}

ft215378 发表于 2021-10-10 21:28:57

for x in range(500):
   for y in range(x,500):
            z = 1000 - x - y
            if x**2+y**2 == z**2 and z>y:
                     print(x,y,z)

番杰发表于 2021-10-27 15:30:05

#include<stdio.h>

int mian(void)
{
int a,b,c;
int flag = 0;

for(int i = 1;i<333;i++)
{
for(int j = i+1;j<500;j++)
(
c = 1000-i-j;
if(c < j)
break;
if(i*i+j*j == c*c)
{
flag = 1;
a = i;
b = j;
break;
}
)

if(flag == 1)
break;
}

printf("(%d,%d,%d)",a,b,c);

return 0；
}

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