实验1
`#include int main() { double a, b, c; scanf("%lf%lf%lf", &a, &b, &c); if(a+b>c&a-b<c) printf("能构成三角形\n"); else printf("不能构成三角形\n"); return 0; }`
#include <stdio.h> int main() { double a, b, c; scanf("%lf%lf%lf", &a, &b, &c); if(a+b>c&a-b<c) printf("能构成三角形\n"); else printf("不能构成三角形\n"); return 0; }



#include<stdio.h>
int main(){
printf(" o \n<H>\nI I\n");
return 0;
}

#include<stdio.h> int main(){ for(int i=0;i<2;i++){ printf(" O\n"); printf("<H>\n"); printf("I I\n"); } return 0; }

#include <stdio.h> int main() { printf(" O O\n"); printf("<H> <H>\n"); printf("I I I I\n"); return 0; }

#include <stdio.h> int main() { char ans1, ans2; printf("每次课前认真预习、课后及时复习了没? (输入y或Y表示有,输入n或N表示没有) : "); ans1 = getchar(); getchar(); printf("\n动手敲代码实践了没? (输入y或Y表示敲了,输入n或N表示木有敲) : "); ans2 = getchar(); if ((ans1=='y' || ans1=='Y') &&(ans2=='y' || ans2=='Y'));{ printf("\n罗马不是一天建成的, 继续保持哦:)\n"); }else{ printf("\n罗马不是一天毁灭的, 我们来建设吧\n");} return 0; }

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

#include <stdio.h> int main() { int year; long long seconds=1000000000; int seconds_per_year=365*24*60*60; year=seconds/seconds_per_year; printf("10亿秒约等于%d年\n", year); return 0; }

#include <stdio.h> #include <math.h> int main() { double x, ans; scanf("%lf", &x); ans = pow(x, 365); printf("%.2f的365次方: %.2f\n", x, ans); return 0; }



#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; }

#include<stdio.h> int main(){ double C,F; while (scanf("%lf",&C)!=EOF){ F=9.0/5*C+32; printf("摄氏度=%.2f,华氏度%。2f\n",C,F); } return 0; }

#include <stdio.h> #include <math.h> int main() { double a, b, c, s, area; while (scanf("%lf%lf%lf", &a, &b, &c) != EOF) { s = (a + b + c) / 2.0; area = sqrt(s * (s - a) * (s - b) * (s - c)); printf("a = %.0f, b = %.0f, c = %.0f, area = %.3f\n", a, b, c, area); } return 0; }

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