C语言怎么实现如下编程，谢谢： 1,2,3,4,5,6,7,8,9共9个数组成3个分数，每个数不重复使用，且分母均为两位数分子均为一位数，两个分数相加等于第三个 - C语言论坛

li362490567

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第 2 楼

得分:4

不会，同求

2015-08-23 17:06

wangweiwei22

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第 4 楼

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回复 3楼 hjx1120

循环次数会不会太多，9重循环

2015-08-23 23:20

rjsp

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第 6 楼

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1/26 + 5/78 = 4/39
1/32 + 5/96 = 7/84
1/32 + 7/96 = 5/48
1/78 + 4/39 = 6/52
1/96 + 7/48 = 5/32
2/68 + 9/34 = 5/17
2/68 + 9/51 = 7/34
4/39 + 1/78 = 6/52
4/56 + 7/98 = 3/21
5/26 + 9/78 = 4/13
5/78 + 1/26 = 4/39
5/96 + 1/32 = 7/84
6/34 + 8/51 = 9/27
7/48 + 1/96 = 5/32
7/96 + 1/32 = 5/48
7/98 + 4/56 = 3/21
8/51 + 6/34 = 9/27
9/34 + 2/68 = 5/17
9/51 + 2/68 = 7/34
9/78 + 5/26 = 4/13

如果用C++的话，因为有 std::next_permutation，所以很简单

#include <cstdio>
#include <algorithm>

int main( void )
{
    unsigned n[] = { 1,2,3,4,5,6,7,8,9 };

    do {
        unsigned a = n[0];
        unsigned b = n[1]*10+n[2];
        unsigned c = n[3];
        unsigned d = n[4]*10+n[5];
        unsigned e = n[6];
        unsigned f = n[7]*10+n[8];
        if( a*d*f+b*c*f == b*d*e )
            printf( "%d/%d + %d/%d = %d/%d\n", a,b,c,d,e,f );
    }
    while( std::next_permutation(n,n+sizeof(n)/sizeof(n[0])) );

    return 0;
}

如果是C的话，就得自己写了

程序代码：

#include <stdio.h>
#include <string.h>

int main( void )
{
    for( unsigned n=0; n!=9*8*7*6*5*4*3*2*1; ++n )
    {
        unsigned t, m[] = { 1,2,3,4,5,6,7,8,9 };

        t = n/(8*7*6*5*4*3*2*1)%9;
        unsigned a  = m[t];
        memcpy( m+t, m+t+1, (8-t)*sizeof(unsigned) );

        t = n/(7*6*5*4*3*2*1)%8;
        unsigned b  = m[t];
        memcpy( m+t, m+t+1, (7-t)*sizeof(unsigned) );

        t = n/(6*5*4*3*2*1)%7;
        unsigned c  = m[t];
        memcpy( m+t, m+t+1, (6-t)*sizeof(unsigned) );

        t = n/(5*4*3*2*1)%6;
        unsigned d  = m[t];
        memcpy( m+t, m+t+1, (5-t)*sizeof(unsigned) );

        t = n/(4*3*2*1)%5;
        unsigned e  = m[t];
        memcpy( m+t, m+t+1, (4-t)*sizeof(unsigned) );

        t = n/(3*2*1)%4;
        unsigned f  = m[t];
        memcpy( m+t, m+t+1, (3-t)*sizeof(unsigned) );

        t = n/(2*1)%3;
        unsigned g  = m[t];
        memcpy( m+t, m+t+1, (2-t)*sizeof(unsigned) );

        t = n/(1)%2;
        unsigned h  = m[t];
        memcpy( m+t, m+t+1, (1-t)*sizeof(unsigned) );

        unsigned i  = m[0];

        if( a*(e*10+f)*(h*10+i) + d*(b*10+c)*(h*10+i) == g*(b*10+c)*(e*10+f) )
            printf( "%d/%d%d + %d/%d%d = %d/%d%d\n", a,b,c,d,e,f,g,h,i );
    }

    return 0;
}

2015-08-24 09:38

moc

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第 8 楼

得分:4

遍历出来的，遍历次数 9！。

2015-08-24 16:17