c - 如何在 AP 中查找缺失的术语

Question

其实我是C语言的新手。谁能告诉我我哪里出错了每次我像这样输入

1 5 7 9 11 或 1 3 7 9 11

它给出两个输出
这是代码

main()
{
    int i,ino,flag=0;
    int ser[2500],dif[2499],fault[3];
    printf("Enter number of elements\n");
    scanf("%d",&ino);
    printf("Enter elements:\n");
    for(i=0;i<ino;i++)
    {
        scanf("%d",&ser[i]);
    }

    for(i=0;i<ino-1;i++)
    {
        dif[i]=ser[i+1]-ser[i];
    }

    for(i=0;i<ino-1;i++)
    {
        if(dif[i]==dif[i+1])
        flag++;
        else
        {
            if(i==0)
            {
                if(dif[i]==dif[i+2])
                printf("Missing numbera is: %d\n",ser[i+1]+dif[i]);
                else
                printf("Missing numberb is: %d\n",ser[i]+dif[i+1]);
            }
            else
            {
                if(dif[i]==dif[i-1])
                printf("Missing numberc is: %d\n",ser[i+1]+dif[i]);
                else if(dif[i+1]==dif[i-1])
                printf("Missing numberd is: %d\n",ser[i]+dif[i+1]);
            }
        }
    }
    if(flag==(ino-2))
    printf("no term missing");

}

Answer 1

#include <stdio.h>
#include <stdbool.h>

bool has_missing_term(int arr[], int size, int *x){
    int i, d = (arr[size-1] - arr[0])/(size-1);
    int n;
    bool ok = true;
    for(n=arr[0], i=1; i < size ; ++i){
        n += d;
        if(n != arr[i]){
            ok = false;
            break;
        }
    }
    if(ok)
        return false;
    d = (arr[size-1] - arr[0])/size;
    for(n=arr[0], i=1; i < size ; ++i){
        n += d;
        if(n != arr[i]){
            *x = n;
            return true;
        }
    }
    return true;//What happened?
}

int main(){
    int i, ino, x;
    int ser[2500];
    printf("Enter number of elements\n");
    scanf("%d",&ino);
    printf("Enter elements:\n");
    for(i=0;i<ino;i++)
        scanf("%d",&ser[i]);
    if(has_missing_term(ser, ino, &x))
        printf("Missing numbera is: %d\n", x);
    else
        printf("no term missing");

}