C语言基础编程练习

1.屏幕上输入：This is a C program#include int main()printf(“This is a C program.n“);return 0;2、求两个整数之和#include void main()int a,b,sum;a=123;b=456;sum=a+b;printf(“%dn“,sum);3、求两个整数中的较大者#include void main()int max(int x,int y);int a,b,c;scanf(“%d,%d“,c=max(a,b);printf(“max=%d“,c);int max(int x,int y)int z;if(xy)z=x;elsez=y;return(z);4、编写一个程序，输入 a,b,c 三个值，输出其中最大者#include void main()int a,b,c,max;printf(“input a,b,c:n“);scanf(“%d,%d,%d“,max=a;if(maxvoid main()int t=1,i=2;while(ivoid main()int sign=1;double deno=2.0,sum=1.0,term;while(denovoid main()float f,c;f=64.0;c=(5.0/9.0)*(f-32);printf(“c=%fn“,c);8、1000 元存一年：（1）活期，年利率为 r1；（2）一年期顶起，年利率 r2；（3）存两次半年定期，年利率为 r3.分别计算 3 种方法本息和#include void main()float p0=1000,r1=0.0036,r2=0.0225,r3=0.0193,p1,p2,p3;p1=p0*(1+r1);p2=p0*(1+r2);p3=p0*(1+r3/2)*(1+r3/2);printf(“p1=%fnp2=%fnp3=%fn“,p1,p2,p3);9、给定一个大写字母 A，输出其小写字母#include void main()char c1,c2;c1=A;c2=c1+32;printf(“%cn“,c2);printf(“%dn“,c2);10、给出三角形三边长，求面积#include #include int main()double a,b,c,s,area;a=3.67;b=5.43;c=6.21;s=(a+b+c)/2;area=sqrt(s*(s-a)*(s-b)*(s-c);printf(“a=%ftb=%ftc=%fn“,a,b,c);printf(“area=%fn“,area);11、输入一个 double 类型的数，使该数保留小数点后两位，对第三位小数进行四舍五入后处理，然后输出此数，以便验证处理是否正确。#includevoid main()double x;printf(“Enter x:“);scanf(“%lf“,printf(“(1)x=%fn“,x);x=x*100+0.5;x=(int)x;x=x/100;printf(“(2)x=%fn“,x);12、编程从键盘输入两个整数给变量 x 和 y，然后输出 x 和 y 的值，最后交换 x 和 y 的值并输出。#includevoid main()int x,y,t;printf(“Enter x,y:n“);scanf(“%d %d“,printf(“x=%d y=%dn“,x,y);t=x;x=y;y=t;printf(“x=%d y=%dn“,x,y);13、从键盘输入两个两位的正整数给变量 x 和 y，并将 x 和 y 合并形成一个整数放在变量z 中。合并的方式是：将数 x 的十位和各位依次放在 z 的千位和十位，将 y 的十位和个位放在 z 的个位和百位上。#includevoid main()int x,y,z;printf(“Input x,y:“);scanf(“%d,%d“,x=x%100;y=y%100;z=(x/10)*1000+(x%10)*10+y/10+(y%10)*100;printf(“x=%d,y=%d,z=%dn“,x,y,z);14、输入 3 个整数，分别放入在变量 a，b，c 中，程序把输入的数据重新按由从小到大的顺序放在变量 a，b，c 中，最后输出 a，b，c 中的值#includevoid main()int a,b,c,t;printf(“input a,b,c:“);scanf(“%d%d%d“,printf(“a=%d,b=%d,c=%dn“,a,b,c);if(ab)t=a;a=b;b=t;if(ac)t=a;a=c;c=t;if(bc)t=b;b=c;c=t;printf(“a=%d,b=%d,c=%dn“,a,b,c);15、根据输入的学生成绩，给出相应的等级，90 分以上的等级为 A，60 以下为 E，其余每10 分一个等级。#includevoid main()int g;printf(“Enter g:“);scanf(“%d“,printf(“g=%d:“,g);if(g=90) printf(“An“);else if(g=80) printf(“Bn“);else if(g=70) printf(“Cn“);else if(g=60) printf(“Dn“);else printf(“En“);#includevoid main()int g;printf(“Enter g:“);scanf(“%d“,printf(“g=%d:“,g);switch(g/10)case 9: printf(“An“);break;case 8: printf(“Bn“);break;case 7: printf(“Cn“);break;case 6: printf(“Dn“);break;default: printf(“En“);16、平面点 M(x,y),若 M 落在圆心在坐标原点单位圆上，key=1；圆外，key=2；圆内 key=0#include#includevoid main()float x,y,r;int key;printf(“input x,y:“);scanf(“%f,%f“,r=sqrt(x*x+y*y);if(r1) key=2;else if(r=1) key=1;else if(r0#includevoid main()int x,y;printf(“Enter x:“);scanf(“%d“,if(xvoid main()int x,y;printf(“Enter x:“);scanf(“%d“, y=-1;if(x!=0) if(x0)y=1;elsey=0;printf(“x=%d,y=%dn“,x,y);18、判定某年是否闰年。#includevoid main()int year,leap;printf(“input the year:“);scanf(“%d“,if(year%4=0else leap=0;if(leap=1)printf(“%d is a leap yearn“,year);elseprintf(“%d is not a leap yearn“,year);#includevoid main()int year,leap;printf(“input the year:“);scanf(“%d“,if(year%4=0)if(year%100!=0)leap=1;elseif(year%400=0)leap=1;elseleap=0;elseleap=0;if(leap=1)printf(“%d is a leap yearn“,year);elseprintf(“%d is not a leap yearn“,year); #includevoid main()int year,leap;printf(“input the year:“);scanf(“%d“,if(year%400=0) leap=1;elseif(year%4=0)if(year%100=0)leap=0;else leap=1;else leap=1;if(leap=1)printf(“%d is a leap yearn“,year);elseprintf(“%d is not a leap yearn“,year);#includevoid main()int year,leap;printf(“input the year:“);scanf(“%d“,if(year%4!=0) leap=0;else if(year%100=0) leap=0;else if(year%400=0) leap=1;else leap=1;if(leap=1)printf(“%d is a leap yearn“,year);elseprintf(“%d is not a leap yearn“,year);19、求一元二次方程 ax2+bx+c=0 的解。分析：a=0，非二次方程b*b-4*a*c=0 ,两个相等实根0,两个不等实根#includevoid main()float a,b,c,disc,x1,x2,realpart,imagpart;printf(“input a,b,c:“);scanf(“%f,%f,%f“,printf(“The equation“);if(fabs(a)1E-6)x1=(-b+sqrt(disc)/(2*a);x2=(-b+sqrt(disc)/(2*a);printf(“has distinct real roots:%8.4f and %8.4fn“,x1,x2);elserealpart=-b/(2*a);imagpart=sqrt(-disc)/(2*a);printf(“has complex roots:n“);printf(“%8.4f+%8.4fin“,realpart,imagpart);printf(“%8.4f-%8.4fin“,realpart,imagpart);20、编写程序，从键盘输入一个以秒为单位的时间数，将其换算成几小时几分几秒，然后输出。#includevoid main()int x,h,m,s;scanf(“%d“,h=x/3600;m=x%3600/60;s=x%60;printf(“%ds=%d:%d:%d!n“,x,h,m,s);21、任意输入一个字符，若是大写字母将其转换成小写字母#includevoid main()char ch;ch=getchar();ch=(ch=Aprintf(“Enter y/m/d:“);scanf(“%d/%d/%d“,printf(“%d/%d/%d“,y,m,d);switch(m-1)case 11: dd=dd+30;case 10: dd=dd+31;case