C语言基础编程练习

1.屏幕上输入：ThisisaCprogram#includestdio.hintmain(){printf(ThisisaCprogram.\n);return0;}2、求两个整数之和#includestdio.hvoidmain(){inta,b,sum;a=123;b=456;sum=a+b;printf(%d\n,sum);}3、求两个整数中的较大者#includestdio.hvoidmain(){intmax(intx,inty);inta,b,c;scanf(%d,%d,&a,&b);c=max(a,b);printf(max=%d,c);}intmax(intx,inty){intz;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#includestdio.hvoidmain(){inta,b,c,max;printf(inputa,b,c:\n);scanf(%d,%d,%d,&a,&b,&c);max=a;if(maxb)max=b;if(maxc)max=c;printf(最大数是%d\n,max);}5、求5！#includestdio.hvoidmain(){intt=1,i=2;while(i=5){t=t*i;i=i+1;}printf(%d\n,t);}6、求多项式1-1/2+1/3-1/4+````+1/99-1/100的值#includestdio.hvoidmain(){intsign=1;doubledeno=2.0,sum=1.0,term;while(deno=100){sign=-sign;term=sign/deno;sum=sum+term;deno=deno+1;}printf(%f\n,sum);}7、将华氏温度64°F转换为摄氏度#includestdio.hvoidmain(){floatf,c;f=64.0;c=(5.0/9.0)*(f-32);printf(c=%f\n,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#includestdio.hvoidmain(){floatp0=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=%f\np2=%f\np3=%f\n,p1,p2,p3);}9、给定一个大写字母A，输出其小写字母#includestdio.hvoidmain(){charc1,c2;c1='A';c2=c1+32;printf(%c\n,c2);printf(%d\n,c2);}10、给出三角形三边长，求面积#includestdio.h#includemath.hintmain(){doublea,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=%f\tb=%f\tc=%f\n,a,b,c);printf(area=%f\n,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。#includestdio.hvoidmain(){doublex;printf(Enterx:);scanf(%lf,&x);printf((1)x=%f\n,x);x=x*100+0.5;x=(int)x;x=x/100;printf((2)x=%f\n,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。#includestdio.hvoidmain(){intx,y,t;printf(Enterx,y:\n);scanf(%d%d,&x,&y);printf(x=%dy=%d\n,x,y);t=x;x=y;y=t;printf(x=%dy=%d\n,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。#includestdio.hvoidmain(){intx,y,z;printf(Inputx,y:);scanf(%d,%d,&x,&y);x=x%100;y=y%100;z=(x/10)*1000+(x%10)*10+y/10+(y%10)*100;printf(x=%d,y=%d,z=%d\n,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#includestdio.hvoidmain(){inta,b,c,t;printf(inputa,b,c:);scanf(%d%d%d,&a,&b,&c);printf(a=%d,b=%d,c=%d\n,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=%d\n,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。#includestdio.hvoidmain(){intg;printf(Enterg:);scanf(%d,&g);printf(g=%d:,g);if(g=90)printf(A\n);elseif(g=80)printf(B\n);elseif(g=70)printf(C\n);elseif(g=60)printf(D\n);elseprintf(E\n);}#includestdio.hvoidmain(){intg;printf(Enterg:);scanf(%d,&g);printf(g=%d:,g);switch(g/10){case9:printf(A\n);break;case8:printf(B\n);break;case7:printf(C\n);break;case6:printf(D\n);break;default:printf(E\n);}}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#includestdio.h#includemath.hvoidmain(){floatx,y,r;intkey;printf(inputx,y:);scanf(%f,%f,&x,&y);r=sqrt(x*x+y*y);if(r1)key=2;elseif(r==1)key=1;elseif(r1)key=0;printf((%.1f,%.1f):%d\n,x,y,key);}-1，x017、编写一个程序实现符号函数y=｛0,x=0的功能1，x0#includestdio.hvoidmain(){intx,y;printf(Enterx:);scanf(%d,&x);if(x0)y=-1;elseif(x==0)y=0;elsey=1;printf(x=%d，y=%d\n,x,y);}#includestdio.hvoidmain(){intx,y;printf(Enterx:);scanf(%d,&x);y=-1;if(x!=0){if(x0)y=1;}elsey=0;printf(x=%d,y=%d\n,x,y);}18、判定某年是否闰年。#includestdio.hvoidmain(){intyear,leap;printf(inputtheyear:);scanf(%d,&year);if((year%4==0&&year%100!=0)||(year%400==0))leap=1;elseleap=0;if(leap==1)printf(%disaleapyear\n,year);elseprintf(%disnotaleapyear\n,year);}#includestdio.hvoidmain(){intyear,leap;printf(inputtheyear:);scanf(%d,&year);if(year%4==0)if(year%100!=0)leap=1;elseif(year%400==0)leap=1;elseleap=0;elseleap=0;if(leap==1)printf(%disaleapyear\n,year);elseprintf(%disnotaleapyear\n,year);}#includestdio.hvoidmain(){intyear,leap;printf(inputtheyear:);scanf(%d,&year);if(year%400==0)leap=1;elseif(year%4==0)if(year%100==0)leap=0;elseleap=1;elseleap=1;if(leap==1)printf(%disaleapyear\n,year);elseprintf(%disnotaleapyear\n,year);}#includestdio.hvoidmain(){intyear,leap;printf(inputtheyear:);scanf(%d,&year);if(year%4!=0)leap=0;elseif(year%100==0)leap=0;elseif(year%400==0)leap=1;elseleap=1;if(leap==1)printf(%disaleapyear\n,year);elseprintf(%disnotaleapyear\n,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0,两个共轭复根#includestdio.h#includemath.hvoidmain(){floata,b,c,disc,x1,x2,realpart,imagpart;printf(inputa,b,c:);scanf(%f,%f,%f,&a,&b,&c);printf(Theequation);if(fabs(a)=1E-6){printf(isnotquadratic\n);}elsedisc=b*b-4*a*c;if(fabs(disc)=1E-6)printf(hastwoequalroot:%8.4f,-b/(2*a));elseif(disc1E-6){x1=(-b+sqrt(disc))/(2*a);x2=(-b+sqrt(disc))/(2*a);printf(hasdistinctrealroots:%8.4fand%8.4f\n,x1,x2);}else{realpart=-b/(2*a);imagpart=sqrt(-disc)/(2*a);printf(hascomplexroots:\n);printf(%8.4f+%8.4fi\n,realpart,imagpart);printf(%8.4f-%8.4fi\n,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。#includestdio.hvoidmain(){intx,h,m,s;scanf(%d,&x);h=x/3600;m=x%3600/60;