题解 | #兔子的序列#

#include <stdio.h>
#include<math.h>
void bu(int arr[], int sz) {
    int i = 0;
    for (i = 0; i < sz - 1; i++) {
        int j = 0, flag = 0;
        for (j = 0; j < sz - 1 - i; j++) {
            if (arr[j] < arr[j + 1]) {
                int tmp = arr[j];
                arr[j] = arr[j + 1];
                arr[j + 1] = tmp;
                flag = 1;
            }
        }
        if (flag == 0)
            break;
    }
}
int main() {
    int arr[1000], n;
    scanf("%d", &n);
    int i = 0;
    for (i = 0; i < n; i++) {
        scanf("%d", &arr[i]);
    }
    bu(arr, n);
    for (i = 0; i < n; i++) {
        int tmp = sqrt(arr[i]);
        if (sqrt(arr[i]) - tmp != 0) {
            printf("%d", arr[i]);
            break;
        }
    }
    return 0;
}