1. 要求顯示以下圖形
   987654321
   *87654321
   **7654321
   ***654321
   ****54321
   *****4321
   ******321
   *******21
   ********1

int main() {
    for (int i = 0; i < 9; i++) {
        for (int j = 0; j < i; j++) {
            printf("*");
        }
        for (int j = 9 - i; j > 0; j--) {
            printf("%d", j);
        }
        printf("\n");
    }
}

2. 打印斐波那契數(shù)列的前三十個(gè)數(shù)中所有的質(zhì)數(shù)卢鹦，并統(tǒng)計(jì)個(gè)數(shù)。

bool is_prime(int n) {
    if (n <= 1) {
        return false;
    }
    int sqr = sqrt(1.0 * n);
    for (int i = 2; i <= sqr; i++) {
        if (n % i == 0) {
            return false;
        }
    }
    return true;
}
int fib(int n) {
    if (n == 1 || n == 2) {
        return 1;
    } else {
        return fib(n - 1) + fib(n - 2);
    }
}
int main() {
    int ans[30], count = 0;
    for (int i = 0; i < 30; i++) {
        ans[i] = fib(i+1);
    }
    printf("質(zhì)數(shù)如下：\n");
    for (int i = 0; i < 30; i++) {
        if (is_prime(ans[i])) {
            printf("%d\n ", ans[i]);
            count++;
        }
    }
    printf("Total number is %d.\n", count);
    return 0;
}

3. 袋子里有2紅3綠5黃球渔肩，隨機(jī)從中摸出8個(gè)贬墩，打印顯示所有組合榴嗅。

int main() {
    for (int i = 0; i <= 2; i++) {
        for (int j = 0; j <= 3; j++) {
            int k = 8 - i - j;
            if (k <= 5) {
                printf("%d個(gè)紅球, %d個(gè)綠球, %d個(gè)黃球。\n", i, j, k);
            } else {
                continue;
            }
        }
    }
    return 0;
}

4. 輸入n（1-10之間數(shù)字）震糖，將數(shù)字分解顯示录肯，如6可以顯示為6，5+1吊说，4+2论咏，4+1+1.....

int stack[10];
void divide(int n, int m, int index) {
    if (n == 0) {
        //當(dāng)符合條件時(shí)，輸出 
        for (int i = 0; i < index - 1; i++) {
            printf("%d + ", stack[i]);
        }
        printf("%d\n", stack[index - 1]);
    }
    
    for (int j = n; j > 0; j--) {
        if (j <= m) {
            stack[index] = j;
            divide(n - j, j, index + 1);
        }
    }
}
int main() {
    int n;
    while (scanf("%d", &n) != EOF) {
        divide(n, n, 0);
    }
    return 0;
}

5. 將十進(jìn)制ip轉(zhuǎn)換為二進(jìn)制顯示颁井。如輸入128.1.2.128厅贪，轉(zhuǎn)換為10000000.00000001.00000010.10000000

void change(int n) {
    int ans[8] = {0}, flag = 0;
    while (n > 0) {
        ans[flag] = n % 2;
        n = n / 2;
        flag++;
    }
    for (int i = 7; i >= 0; i--) {
        printf("%d", ans[i]);
    }
}
int main() {
    char str[100];
    int j, sum = 0;
    while(gets(str)) {
        int len = strlen(str);
        for (int i = 0; i < len; i++) {
            if (str[i] >= '0' && str[i] <= '9') {
                j = str[i] - '0';
                sum = sum * 10 + j;
            } else {
                change(sum);
                printf(".");
                sum = 0;
            }
            if (i == len - 1) {
                change(sum);
            }
        }
        printf("\n");

    }
    return 0;
}