实验二

任务一

#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;
}

18行在计算输出number的值

任务二

#include<stdio.h>
int main()
{
double x, y;
char c1, c2, c3;
int a1, a2, a3;
scanf("%d %d %d", &a1, &a2, &a3);
printf("a1 = %d, a2 = %d, a3 = %d\n", a1,a2,a3);
scanf("%c %c %c", &c1, &c2, &c3);
printf("c1 = %c, c2 = %c, c3 = %c\n", c1, c2, c3);
scanf("%lf %lf", &x, &y);
printf("x = %lf, y = %lf\n",x, y);
return 0;
}

任务三

#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;
}