实验2

task1.c

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#include <stdio.h>
#include <stdlib.h>
#include <time.h>

#define N 5

int main()
{
	int number;
	int i;

	srand(time(0));

	for (i = 0; i < N; ++i) {
		number = rand() % 65 + 1;
		printf("2023831%04d\n", number);
	}
	return 0;
}

Line 15的作用是 生成一个1~65之间的随机数
这个程序的功能是在202383310001~202310065之间随机生成5个号码

task2.c

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#include <stdio.h>

int main() {
	char c;
	while ((c = getchar()) != EOF) {
		getchar();
		if (c == 'r') {
			printf("stop!\n");
		}
		else if (c == 'g') {
			printf("go go go\n");
		}
		else if (c == 'y') {
			printf("wait a minute\n");
		}
		else {
			printf("something must be wrong...\n");
		}
	}
	return 0;
}

task3.c

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#include <stdio.h>
#include <stdlib.h>
#include <time.h>

int main() {
	int lucky_day;
	int guess;
	int cnt;

	srand(time(0));
	lucky_day = rand() % 31 + 1;
	printf("猜猜2024年5月哪一天会是你的lucky_day\n");
	printf("开始喽,你有三次机会,猜吧(1~31):");

	for (cnt = 0; cnt < 3; ++cnt) {
		scanf_s("%d", &guess);

		if (guess == lucky_day) {
			printf("哇,猜中了：-)\n");
			return 0;
		}
		else if (guess < lucky_day) {
			printf("你猜的日期早了,你的lucky_day还没到呢\n");
		}
		else {
			printf("你猜的日期晚了,你的lucky_day在前面哦\n");
		}
		if (cnt < 2)
		{
			printf("\n再猜(1~31):");
		}

	}
	if (cnt == 3) {
		printf("\n\n次数用完了,偷偷告诉你,5月你的lucky_day是%d号\n", lucky_day);
	}
	return 0;
}

task4.c

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#include <stdio.h>
#include <stdlib.h>

int main() {
	int n, a, j, i,k;
	double s;

	while (scanf_s("%d%d", &n, &a) != EOF) {
		s = 0.0;
		for (i = 1; i <= n; i++) {
			k = 0;
			for (j = 0; j < i; j++) {
				k = k * 10 + a;
		}
			s += 1.0 * i / k;
		}
		printf("n=%d,a=%d,s=%1f\n\n", n, a, s);
	}
	return 0;
}

task5.c

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#include <stdio.h>

int main() {
	int i, j;
	for (i = 1; i <= 9; i++) {
		for (j = 1; j <= i; j++) {
			printf("%d*%d=%d\t", i, j, i * j);
		}
		printf("\n");
	}
	return 0;
}

task6.c

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#include <stdio.h>
#include <stdlib.h>

int main() {
	int n, i, m, j;
	printf("input n:");
	scanf_s("%d", &n);

	for (i = 0; i < n; i++)
	{
		for (j = 0; j < i; j++)
		{
			printf("\t");
		}
		for (m=2*(n-i)-1;m>0;m--)
		{
			printf(" O \t");
		}
		printf("\n");

			for (j = 0; j < i; j++)
			{
				printf("\t");
		}
		for (m=2*(n-i)-1;m>0;m--) {
			printf("<H>\t");
		}
		printf("\n");

		for (j = 0; j < i; j++)
		{
			printf("\t");
		}
		for (m = 2 * (n - i) - 1; m > 0; m--)
		{
			printf("I I\t");
		}
		printf("\n");
	}
	return 0;
}