实验任务1
源代码
#include<stdio.h> #include<stdlib.h> #include<time.h> #define N 5 #define R1 586 #define R2 701 int main() { int number; int i; srand(time(0)); for(i=0; i<N; ++i) { number=rand()%(R2-R1+1)+R1; printf("20228330%04d\n",number); } return 0; }
运行结果截图
问题1:生成586到701之间到随机数
问题2:随机生成5个20228330586到202283300701之间到学号
实验任务2
源代码
#include<stdio.h> int main() { double x, y; char c1, c2, c3; int a1, a2 ,a3; scanf("%d%d%d", &a1, &a2, &a3);//修改 printf("a1 = %d, a2 = %d, a3 = %d\n", a1,a2,a3); getchar();//补充 scanf("%c%c%c", &c1, &c2, &c3); printf("c1 = %c, c2 = %c, c3 = %c\n",c1,c2,c3); scanf("%lf,%lf", &x, &y);//修改 printf("x = %lf, y = %lf\n",x,y);//修改 return 0; }
运行截图
实验任务3
task3_2.c源代码
#include<stdio.h>
#include<math.h>
int main()
{
double x, ans;
while(scanf("%lf",&x)!=EOF)
{
ans = pow(x,365);
printf("%.2f的365次方:%.2f\n",x,ans);
printf("\n");
}
return 0;
}
运行截图
task3_3.c源代码
#include<stdio.h> #include<math.h> int main() { double c, f; while(scanf("%lf",&c)!=EOF) { f = 9/5*c+32; printf("摄氏度c = %.2f时,华氏度f = %.2f\n",c,f); printf("\n"); } return 0; }
运行截图
实验任务4
源代码
#include<stdio.h> int main() { char light; while(scanf("%c",&light)!=EOF){ getchar(); switch(light){ 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");break; } } return 0; }
运行截图
实验任务5
源代码
#include<stdio.h> #include<stdlib.h> #include<time.h> #define N 3 #define R1 1 #define R2 30 int main() { int n1,n2,i,flag; srand(time(0)); flag=0; n1 = rand()%(R2-R1+1)+R1; printf("猜猜2023年4月哪一天会是你的lucky day\n"); printf("开始喽,你有三次机会,猜吧(1-30):"); for(i=1;i<=N;i++) { scanf("%d",&n2); if(n2==n1){ printf("哇,猜中了:-\n"); break; } else if(n2>n1) printf("你猜的日期晚了,你的lucky day已经过啦\n"); else printf("你猜的日期早了,你的lucky day还没到呢\n"); printf("再猜(1-30):"); flag++; } if(flag==3) printf("次数用完啦,偷偷告诉你:4月,你的lucky day是%d号\n",n1); return 0; }
运行截图
实验任务6
源代码
#include<stdio.h> int main() { int line,column; for(line=1;line<=9;line++){ for(column=1;column<=line;column++){ printf("%d*%d=%d\t",column,line,line*column); } printf("\n"); } return 0; }
运行截图
实验任务7
规律:当输入n时,第i行需要打印2*(n-i)+1个小人,前面需要i-1个空白(\t)
源代码
#include<stdio.h> int main() { int n,i,j; printf("input n:"); scanf("%d",&n); for(i=1;i<=n;i++){ for(j=1;j<i;j++) printf("\t"); for(j=1;j<=2*(n-i)+1;j++) printf(" o \t"); printf("\n"); for(j=1;j<i;j++) printf("\t"); for(j=1;j<=2*(n-i)+1;j++) printf("<H>\t"); printf("\n"); for(j=1;j<i;j++) printf("\t"); for(j=1;j<=2*(n-i)+1;j++) printf("I I\t"); printf("\n"); } return 0; }
运行截图
标签:截图,include,源代码,int,实验,printf,main From: https://www.cnblogs.com/yuetong/p/17223586.html