实验2

task1;

#include <stdio.h>
#include <stdlib.h>
#include <time.h>
#define N 5
int main()
{
int number;
int i;
srand( time(0) ); // 以当前系统时间作为随机种子
for(i = 0; i < N; ++i) {
number = rand()%65 + 1;
printf("20238331%04d\n", number);
}
system("pause");
return 0;

}

 

问题1:生成1~65之间的随机数

问题2:生成5个学号

task2

#include <stdio.h>

int main() {
char input;

printf("请输入交通信号灯的颜色\n");

while ((input = getchar()) != EOF) {
switch (input) {
case 'r':

printf("stop!\n");
break;
case 'g':

printf("go go go\n");
break;
case 'y':
printf("wait a minute\n");
break;
default:
printf("something must be wrong...\n");
}

printf("请输入下一个交通信号灯的颜色(r/g/y)或按CTRL+Z结束:\n");
getchar();

}
return 0;
}

 

task3

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

int main()
{
srand((unsigned int)time(NULL));
int luckyday=rand()%31+1;
int guess;
int try=0;
int m=0;
printf("开始喽,你有三次机会,猜吧(1~31)\n");
while(try<3 && !m)
{
scanf("%d",&guess);
try++;
if (guess<luckyday)
{

printf("你猜的日期早了,你的lucky day还没到\n");
printf("再猜(1~31)\n");
}
if(guess>luckyday){

printf("你猜的日期晚了,你的lucky day在前面哦\n");
printf("再猜(1~31)\n");
}
if(guess==luckyday) {

printf("哇,猜中了:-\n");
m=1;}
}
if(!m)
printf("次数用完了,偷偷告诉你,5月你的lucky day是%d号\n",luckyday);
system("pause");
return 0;
}

 

task4

#include <stdio.h>

#include <math.h>

int main()

{

int a, n;

 

while (scanf("%d%d", &n, &a) != EOF)

{

 

double s = 0;

double down=0;

int i = 1;

do {

 

double up = i;

down = down * 10 + a;

s += up / down;

i++;

} while (i<=n);

 

printf("n=%d a=%d s=%lf", n, a, s);

 

}

return 0;

}

 

task5

#include <stdio.h>

 

int main() {

    int i, j;

    for (i = 1; i <= 9; i++) {

        for (j = 1; j <= i; j++) {

            printf("%d*%d=%-2d ", j, i, i * j);

        }

        printf("\n"); 

    }

    return 0;

}

 

task6

#include <stdio.h>

int main()

{

int n;

while(scanf("%d",&n)!=EOF)

{

int i=1;

int j=1;

for(i=n;i>=1;i--)

{

for(j=1;j<=(n-i);j++)

{

  printf("\t");

}

for(j=1;j<=(2*i-1);j++){

printf(" O\t");

}

printf("\n");

for(j=1;j<=(n-i);j++){

printf("\t");

}

 

for(j=1;j<=(2*i-1);j++){

printf("<H>\t");

}

printf("\n");

for(j=1;j<=(n-i);j++){

printf("\t");

}

for(j=1;j<=(2*i-1);j++){

printf("I I\t");

}

printf("\n");

 

}

 

}

return 0;

 } 

 

 

 
posted @ 2024-04-09 19:51  HanXT  阅读(26)  评论(0)    收藏  举报