C.5

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//1、随机产生20个[10 , 50]的正整数存到数组中，并求数组中的所有元素最大值、最小值、平均值以及各元素之和,及第二大值。

int a[20] = {0};

for (int i = 0; i < 20; i ++) {

a[i] = arc4random() % 41 + 10;

printf("%d ", a[i]);

}

int max = 0, min = a[0], secondMax = 0, sum = 0;

float avg = 0.0;

for (int i = 0; i < 20; i ++) {

max = max > a[i] ? max : a[i];

}

for (int i = 0; i < 20; i ++) {

if (a[i] < max && a[i] > secondMax) {

secondMax = a[i];

}

min = min < a[i] ? min : a[i];

sum += a[i];

}

avg = sum / 20.0;

printf("\nmax = %d min = %d avg = %f sum = %d secondMax = %d", max, min, avg, sum, secondMax);

// 2.编程在一个已知的字符串中查找最长单词，假定字符串中只含字母和空格，用空格来分隔单词。

char a[] = "asdgf gs ysgfyg aygrfjksdf jfg sygfwygrf";

int i = 0, maxlength = 0, length = 0, lengthInde = 0;

while (a[i] != '\0') {

if (a[i] != ' ') {

length ++;

} else if(length > maxlength){

maxlength = length;

lengthInde = i - maxlength;

length = 0;

}

if(length > maxlength && a[i] == ' '){

maxlength = length;

lengthInde = i - maxlength;

}

i ++;

}

printf("%d", maxlength);

printf("\n%d", lengthInde);

printf("\n");

i = 0;

// while (a[i] !='\0') {

// if (i >= lengthInde && i < maxlength + lengthInde) {

// printf("%c", a[i]);

// }

// i ++;

for (i = lengthInde; i < lengthInde + maxlength; i ++) {

printf("%c", a[i]);

}

//3、耶稣有13个门徒，其中有一个就是出卖耶稣的叛徒，请用排除法找出这位叛徒：13人围坐一圈，从第一个开始报号：1，2，3，1，2，3...。凡是报到“3”就退出圈子，最后留在圈子内的人就是出卖耶稣的叛徒。请找出它原来的序号。

int a[13] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13};

int num = 13, i = 0, count = 0;

while (num > 1) {

if (a[i] > 0) {

count ++;

if (count == 3) {

a[i] = 0;

count = 0;

num --;

}

i ++;

if (i > 12) {

i = 0;

}

for (int i = 0; i < 13; i ++) {

if (a[i] != 0) {

printf("%d", a[i]);

}

//4.将两个排好序的数组，合并到另外一个数组中，并且合并之后的数组也是有序的。

int a[10] = {0}, b[12] = {0}, c[22] = {0};

for (int i = 0; i < 10; i ++) {

a[i] = arc4random() % 21 + 10;

printf("%d ", a[i]);

}

printf("\n");

for (int i = 0; i < 12; i ++) {

b[i] = arc4random() % 21 + 10;

printf("%d ", b[i]);

}

printf("\n");

for (int i = 0; i < 9; i ++) {

for (int j = 0; j < 9 - 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 ", a[i]);

}

printf("\n");

for (int i = 0; i < 11; i ++) {

for (int j = 0; j < 11 - i; j ++) {

if (b[j] > b[j + 1]) {

int temp = b[j];

b[j] = b[j + 1];

b[j + 1] = temp;

}

for (int i = 0; i < 12; i ++) {

printf("%d ", b[i]);

}

printf("\n");

int k = 0, i = 0, j = 0;

while (i < 10 && j < 12) {

if (a[i] > b[j]) {

c[k] = b[j];

k ++;

j ++;

} else {

c[k] = a[i];

k ++;

i ++;

}

while (i < 10) {

c[k] = a[i];

k ++;

i ++;

}

while (j < 12) {

c[k] = b[j];

k ++;

j ++;

}

for (int m = 0; m < 22; m ++) {

printf("%d ", c[m]);

}

//5.有一分数序列：2/1，3/2，5/3，8/5，13/8，21/13...求出这个数列的前20项之和。

float a = 2, b = 1,fra = 0, sum = 0;

for (int i = 0; i < 20; i ++) {

fra = a / b;

int temp = a;

a = a + b;

b = temp;

sum += fra;

}

printf("%f", sum);

//6,给一个不多于5位的正整数，要求：一、求它是几位数，二、逆序打印出各位数字。

printf("请输入一个不多余5位正整数:");

int a = 0, i = 0;

scanf("%d", &a);

int b = a;

while (b) {

i ++;

b = b / 10;

}

printf("%d", i);

printf("\n");

for (int j = 0; j < i; j ++) {

printf("%d", a % 10);

a = a / 10;

}

//7.,一个5位数，判断它是不是回文数。即12321是回文数，个位与万位相同，十位与千位相同。

printf("输入一个5位数:");

int a = 0;

scanf("%d", &a);

if (a % 10 == a / 10000 && ((a / 10) % 10) == ((a / 1000) % 10))

{

printf("a是回文数");

} else {

printf("a不是回文数");

}

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