C练习题

斐波拉契数列

    int a[10] = {1,1};
    for (int i = 2; i < 10; i++) {
        a[i] = a[i-1]+a[i-2];
    }
    // [1,1,2,3,5,8,13,21,34,55]
    for (int i = 0; i < 10; i++) {
        printf("%d\n",a[i]);
    }

冒泡排序法

    int a[10] = {5,4,2,6,7,1,8,3,9,10};
    int count = sizeof(a) / sizeof(int);
    for (int i = 0; i < count; i++) {
        for (int j = 0; j < count-i; j++) {
            if (a[j] > a[j+1]) {
                int temp = a[j];
                a[j] = a[j+1];
                a[j+1] = temp;
            }
        }
    }
    for (int i = 0; i < 10; i++) {
        printf("%d\n",a[i]);
    }

选择排序法

    int a[10] = {5,4,2,6,7,1,8,3,9,10};
    int count = sizeof(a) / sizeof(int);
    for (int i = 0; i < count; i++) {
        for (int j = i; j < count; j++) {
            if (a[j] > a[i]) {
                int temp = a[j];
                a[j] = a[i];
                a[i] = temp;
            }
        }
    }
    for (int i = 0; i < 10; i++) {
        printf("%d\n",a[i]);
    }

二分法查找

    int a[10] = {1,2,3,4,5,6,7,8,9,10};
    int start = 0,end = 10,mid;
    int dest = 8;
    while (start <= end) {
        mid = (start + end ) /2;
        if (dest == a[mid]) {
            printf("%d %d\n",mid,a[mid]);
            break;
        }else if (dest > a[mid]) {
            start = mid+1;
        } else if (dest < a[mid]){
            end = mid-1;
        }
    }

递归

int fab(int num);
int main(){

    int mm = fab(6);
    printf("%d\n",mm);
}

int fab(int num)
{
    if (num == 1) {
        return 1;
    } else {
        return num * fab(num - 1);
    }
}

汉诺塔