a的取值范围是0-9，
约束条件是：
0<=a<=t1[i]+2;
0<=a<=t1[i]-2;
0<=fabs(a-b)<=t2[i]+2;
0<=fabs(a-b)<=t2[i]-2;
0<=fabs(fabs(a-b)-fabs(b-c))<=t1[i]+2;
0<=fabs(fabs(a-b)-fabs(b-c))<=t1[i]-2;
如何计算出符合约束条件的a的值？

#include<stdio.h>        
#include<math.h>        
void main()        
{        
    int     
        b=1,
        c=6;
    int a,i,j,k;

    int t1[4]={2,3,4,5     },
        t2[4]={ 1,3,4,8    },
        t3[2]={6,7     };
    for(a=0;a<10;a++)
    {
        for(i=0;i<4  ;i++)
        0<=a<=t1[i]+2;
        0<=a<=t1[i]-2;
   
    for(j=0;j<4  ;j++)
        0<=fabs(a-b)<=t2[i]+2;
        0<=fabs(a-b)<=t2[i]-2;

    for(k=0;k<2  ;k++)
        0<=fabs(fabs(a-b)-fabs(b-c))<=t3[i]+2;
        0<=fabs(fabs(a-b)-fabs(b-c))<=t3[i]-2;
    printf("%d  ",a);
    }

}        
        

[此贴子已经被作者于2017-4-11 17:49编辑过]

修改后，还是有错误
#include<stdio.h>        
#include<math.h>        
void main()        
{        
        
    double a;
    int i,j,k;
   
    int b=1,c=6;

    int t1[4]={2,3,4,5     },
        t2[4]={ 1,3,4,8    },
        t3[2]={6,7     };

    for(a=0;a<6;a++)
    {
        for(i=0;i<4;i++)
        {
            for(j=0;j<4;j++)
            {
                for(k=0;k<2;k++)
                {

        if(a=fabs(t1[i]+2)
            &&a=fabs(t1[i]-2)
            &&fabs(a-b)=fabs(t2[j]+2)
            &&fabs(a-b)=fabs(t2[j]-2)
            &&fabs(fabs(a-b)-fabs(b-c))=fabs(t3[k]+2)
            &&fabs(fabs(a-b)-fabs(b-c))=fabs(t3[k]-2))
    printf("%lf  ",a);
                }
            }
        }
    }


}        
        

程序修改后，基本可以运行计算了；
#include<stdio.h>        
#include<math.h>        
void main()        
{        
        
    double a[10]={0,1,2,3,4,5,6,7,8,9};
    int f,i,j,k;
   
    double b=1,c=6;

    double t1[4]={2,3,4,5     },
           t2[4]={ 1,3,4,8    },
           t3[2]={6,7     };

    for(f=0;f<6;f++)
    {
        for(i=0;i<4;i++)
        {
            for(j=0;j<4;j++)
            {
                for(k=0;k<2;k++)
                {

        if(fabs(a[f])==fabs(t1[i]+2)&&fabs(a[f]-b)==fabs(t2[j]+2)&&fabs(fabs(a[f]-b)-fabs(b-c))==fabs(t3[k]+2))
    printf("%lf  ",a[f]);
        else
            printf("  ");   
        
        if(fabs(a[f])==fabs(t1[i]-2)&&fabs(a[f]-b)==fabs(t2[j]-2)&&fabs(fabs(a[f]-b)-fabs(b-c))==fabs(t3[k]-2))
    printf("%lf  ",a[f]);
        else
            printf(" ");



                }
            }
        }
    }


}