Chapter 1 - the problem lies in the Example 8 wedding

Problem Description: Three couples the wedding, the groom three A, B, C, three brides X, Y, Z. Some people want to know who and whom to marry, so he asked them: A said he would marry and X; X said her fiance is C; C and Z said he would marry. They are lying. Programming who seek and whom to marry.

ALGORITHM: Construction logical expression (a = 1 && c = 1 && c = 3 && a = b && b = c && a = c!!!!!!)

Note that character output

Own algorithms have a little problem, not output characters

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

int main()
{
    int a,b,c;
    for(a=1;a<=3;a++)
        for(b=1;b<=3;b++)
            for(c=1;c<=3;c++)
          {
                if(a!=1 && c!=1 && c!=3 && a!=b && b!=c && a!=c)
                {
                    printf("A与%d结婚\n",a);
                    printf("B与%d结婚\n",b);
                    printf("C与%d结婚\n",c);
                }
          }
    return 0;
}

The output is:

standard answer can output characters:

#include<stdio.h>
#include <stdlib.h>
int main()
{
    int a,b,c;                  //a,b,c为三个男人
    for(a=1;a<=3;a++)           //给a,即第一个男人尝试匹配给每一个女人
        for(b=1;b<=3;b++)       //给a,即第一个男人尝试匹配给每一个女人
            for(c=1;c<=3;c++)   //给a,即第一个男人尝试匹配给每一个女人
                if(a!=1 && c!=3 && c!=1 && a!=b && b!=c && a!=c)    //根据他们说的话进行判断
                {
                    //符合条件，即正确的情况下，输出正确答案
                    printf("%c将嫁给A\n",'X'+a-1);
                    printf("%c将嫁给B\n",'X'+b-1);
                    printf("%c将嫁给C\n",'X'+c-1);
                    printf("\n");
                    printf("%d将嫁给A\n",a);
                    printf("%d将嫁给B\n",b);
                    printf("%d将嫁给C\n",c);
                }
                //程序计算结束。退出程序
                system("pause");
                return 0;
}

The results are: