将两个无序的一维数组排序，然后将其合并到一个一维数组中，仍保持数组有序。

Question

程序调试失败；program received signal SIGFPE

#include <stdio.h> 
int sort(int *p, int x)
{
    int m, n, i;
    for (i=0; i<x-1; i++){
        for (m=0; m<x-i-1; m++){
            if (*p+m>*p+m+1){
                n=*(p+m);
                *(p+m+1)=*(p+m);
                *(p+m)=n;
            }       
        }
    }
    return *p;
}
void print(int num[], int x)
{
    int cnt;
    for(cnt=0; cnt<x; cnt++)
        printf ("%d", num[cnt]);
}
int main()
{
    int i, x, y, z, m, n;
    printf ("请输入两组数组的长度：");
    scanf ("%d %d", &x, &y);
    z=x+y;
    int num1[x],num2[y],num3[z];
    printf ("请输入第一组数组中的数：");
    for (i=0; i<x; i++){
    scanf ("%d", &num1[i]); 
    num3[i]=num1[i];
    }
    printf ("请输入第二组数组中的数："); 
    for (i=0; i<x; i++){
    scanf ("%d", &num2[i]);
    num3[x+i]=num2[i];
    }
    sort (num1[x], x);
    sort (num2[y], y);
    sort (num3[z], z);
    print (num1[x], x);
    print (num2[y], y);
    print (num3[z], z);
    return 0;
}

Answer 1

#include <stdio.h>
#define MAX 100
void sort(int *p, int x)
{
    int m, n, i;
    for (i=0; i<x-1; i++){
        for (m=0; m<x-i-1; m++){
            if (*(p+m)>*(p+m+1)){
                n=*(p+m + 1);
                *(p+m+1)=*(p+m);
                *(p+m)=n;
            }       
        }
    }
}
void print(int num[], int x)
{
    int cnt;
    for(cnt=0; cnt<x; cnt++)
        printf ("%d ", num[cnt]);
    printf("\n");
}
int main()
{
    int i, x, y, z, m, n;
    printf ("请输入两组数组的长度：");
    scanf ("%d %d", &x, &y);
    z=x+y;
    int num1[MAX],num2[MAX],num3[MAX+MAX];
    printf ("请输入第一组数组中的数：");
    for (i=0; i<x; i++){
        scanf ("%d", &num1[i]); 
        num3[i]=num1[i];
    }
    printf ("请输入第二组数组中的数："); 
    for (i=0; i<y; i++){
        scanf ("%d", &num2[i]);
        num3[x+i]=num2[i];
    }
    sort (num1, x);
    sort (num2, y);
    sort (num3, z);
    print (num1, x);
    print (num2, y);
    print (num3, z);
    return 0;
}

请输入两组数组的长度：3 4
请输入第一组数组中的数：1 5 9
请输入第二组数组中的数：0 9 1 2
1 5 9
0 1 2 9
0 1 1 2 5 9 9
Press any key to continue . . .

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