res-转载-2012春学期语言上机练习参考答案

第1周(1-4题略 显示"ProgramminginCis 显示图案（复习printf()的字符串输出 第2 20011求华氏温度100°F对应的摄氏温度 20012求华氏温度150°F对应的摄氏温度 求摄氏温度26°C对应的华氏温 计算三门课程的平均成 20015当n为152时，分别求出n的个位数字(digit1)、十位数字(digit2)和百位数字(digit3)的值 20016计算x的平方（改错题 20021计算分段函数的值（调试示例 20023计算存款利 20024计算分段函 200262num1num2数 第3 求1+2+3+......+100（调试示例 求 求 求1－1/4＋1/7－1/10＋……的前n项之 求x的n次 生成3的乘方 求 计算物体自由下落的距 求 阶梯电 第4 20025计算分段函数的值（改错题 20027计算旅途时间 20028数字加 求1+1/3+1/5+......的前n项 输出华氏-摄氏温度转换表（改错题 使用函数求n!/(m!*(n- 求1－2/3＋3/5－4/7＋5/9- 求 求 第5 求一元二次方程的 求分段函数的 显示五级记分制成绩所对应的百分制成绩区间（使 显示水果的价格（使用 求三角形的面积和周 计算个人所得 统计学生成 分段计算水费（使用嵌套的if-else语句 出租车计 30062输出21世纪所有闰 第6 求最小公倍数和最大公约数（调试示例 求 求整数的位 求 换硬 找出各位数字的立方和等于它本身的 找完数（改错题 验证歌德巴赫猜想(选作 从高位开始逐位输出一个整数的各位数字(选作 求平均成绩并统计不及格人 第7 使用函数判断数的符 使用函数求奇数 使用函数统计素数并求 使用函数统计一个整数中数字的个 使用函数找水仙花 使用函数求1!+2!+…+m!（改错题 使用函数求余弦函数的近似 使用函数找最大 使用函数输出指定范围内的Fibonacci 使用函数找出指定范围内的完 第8 求x＋x*x/2!＋x*x*x/3!＋x*x*x*x/4!＋……的 使用函数计算两点间的距 50051数字金字 使用函数求 使用函数求PI（调试示例源程序 整数的十进制、八进制和十六进制表现形 分类统计字 使用函数输出整数的逆序 统计单 简单计算 1周(1-4题略 显示"ProgramminginCis编写程序，在屏幕上显示一个短句"ProgramminginCisintmain(void){printf("ProgramminginCisfun!");return0;} 显示图案（printf()的字符串输出*******intmain(void){printf("***"***\n" * }220011求华氏温度100°F对应的摄氏温度。#include#include<stdio.h>intmain(void){intcelsius, printf("fahr=%d,celsius=%d\n",fahr,}}20012150°F计算公式：C5*F/9-5*32/9，式中：C表示摄氏温度，F表示华氏温度。intmain(void){intcelsius,celsius=5*fahr/9-printf("fahr=%d,celsius=%d\n",fahr,celsius);return0;} 求摄氏温度26°C对应的华氏温度摄氏温度，f表示华氏温度。celsius=26,fahr=#include#include<stdio.h>intmain(void){intcelsius, celsius=26;fahr=fahr=9*celsius/5+printf("celsius=%d,fahr=%d\n",celsius,} 计算三门课程的平均成绩math=87,eng=72,comp=93,average=intmain(void){intmath,eng,comp,average;math=87;eng=72;comp=93; average=printf("math=%d,eng=%d,comp=%d,average=%d\n",math,eng,comp,}intmain(void){printf("整数%d的个位数字是%d,十位数字是%d,百位数字是%d\n",n,digit2,return}x的值为3xy，并分别以"yx*x"和"x*xy"xy的9=3*3*3=#include#include<stdio.h>intmain(void){printf("%d=%d*%d\n",y,x,x);printf("%d*%d=%d\n",x,x,y);return0;}20021计算分段函数的值（error02_3）输入一个正整数repeat(0<repeat<10)，做repeat次下列运算：输入x，计算并输出下列分段函数f(x)的值(1位小数)。x0时，yf(x)1/xx0时，yf(x0。 f(10.00)=f(0.00)=#include#include<stdio.h>intmain(void){intrepeat,ri;doublex,scanf("%d",&repeat);for(ri=1;ri<=repeat;ri++){scanf("%lf",if(x!=0)y=1/x; printf("f(%.2f)=%.1f\n",x,}}20023输入存款金额moneyyearrate，根据下列公式计算存款到期时的利息interest(税前)，输出时保留2位小数。100030.025(money=1000,year=3,rate=interest=#include#include<stdio.h>#include<math.h>intmain(void){intmoney,year;doubleinterest,rate;interest=money*pow((1+rate),year)-money;printf("interest=%.2f\n",interest);}20024xf(x)的值(2位小数)sqrt函数求平方根，调用pow函数求幂。x0时，f(xx^0.5x0时，f(xx+1)^22x1/x。-0f(10.00)=f(-0.50)=-f(0.00)=#include#include<stdio.h>#include<math.h>intmain(void){intrepeat,doubledoublex,scanf("%d",&repeat);for(ri=1;ri<=repeat;ri++){if(x>=0)y=elsey=pow((x+1),2)+2*x+1.0/x;printf("f(%.2f)=%.2f\n",x,y);}}200262num1num2printf("%d%d%d\n",num1,num2,5 5+3=85-3=25*3=5/3=15%3=2#include#include<stdio.h>intmain(void){int printf("%d+%d=%d\n",num1,num2,num1+num2);printf("%d-%d=%d\n",num1,num2,num1-num2);printf("%d*%d=%d\n",num1,num2,num1*num2);printf("%d/%d=%d\n",num1,num2,num1/num2);printf("%d%%%d=%d\n",num1,num2,num1%num2);return}3 求1+2+3100（调试示例123100sum=intmain(void){inti,sum;} 求 0+1+2+. (计算 (计算 sum=sum=sum=intmain(void){inti,m,sum;for(ri=1;ri<=ri++){scanf("%d",&m);sum=0; }} 求 2mn(m<=n)，1/m1/(m+1)1/(m+2) 5101sumsumsumintmain(void){inti,m,n;doublesum;for(ri=1;ri<=ri++){&m,&n);sum=0;for(i=m;i<=n;i++)}} 求1－1/4＋1/7－1/10＋……的前n项之n1－1/4＋1/7－1/10n留3位小数。 3sum=sum=intmain(void){intflag,i,n,t;intrepeat,doubleitem,for(ri=1;ri<=ri++){scanf("%d",&n); }}intmain(void){intflag,i,n,t;intrepeat,ri;doubleitem,for(ri=1;ri<=ri++){scanf("%d",&n);{}}} xn不允许调用pow函数求幂。222intmain(void){inti,intrepeat,ri;for(ri=1;ri<=ri++){scanf("%lf%d",&x,&n);mypow=1; printf("%.2f\n",}} 3计算3的乘方。printf("pow(3,%d)%.0f\n",i, =1=3=9=#include#include<stdio.h>#include<math.h>intmain(void){inti,doublescanf("%d",&n);for(i=0;i<=n;i++)}} sum=#include#include<stdio.h>#include<math.h>intmain(void){int } 计算物体自由下落的距离intmain(void){doubleheight=0.5*10*3* } 求 编写程序，输入一个正整数n，求 的前n项之和，输出 sum=sum=intmain(void){inti,doublesum;for(ri=1;ri<=ri++){scanf("%d",}}} 阶梯电价留2位小数。 cost=cost=intmain(void){intrepeat,ri;doublecost,for(ri=1;ri<=ri++){scanf("%lf",&e); }}420025计算分段函数的值（error02_4）xf(x)1位小数。当x10时，y=f(x)=x，当x10时，y=f(x)=1/x。 f(10.0)=f(234.0)=#include#include<stdio.h>intmain(void){intrepeat,ri;doublex,for(ri=1;ri<=repeat; printf("f(%.1f)=%.1f\n",x,}}20027计算旅途时间。输入2个整数time1time2，表示火车的出发时间和到达时间，计算并输出旅途时间。Thetrainjourneytimeis6hours59#include#include<stdio.h>intmain(void){inttime1,time2,hour,minutes,t1,t2,result;scanf("%d%d",&time1,&time2);t1=(time1/100)*60+time1%100; /*minutes*/t2=(time2/100)*60+time2%100;result=t2-hour=result/60;minuteshour=result/60;minutes=result%60;printf("Thetrainjourneytimeis%dhours%dminutes.\n",hour,}#include<stdio.h>intmain(void)#include<stdio.h>intmain(void){inttime1,time2,hour,intth1,th2,tm1,tm2; //printf(“\n”)也能通过？scanf("%d%d",&time1,&time2);}}printf("Thetrainjourneytimeis%dhours%dminutes.\n",hour,minutes);return0;}20028#include#include<stdio.h>intmain(void){intnumber,digit1,digit2,digit3,digit4, /*取个位＋910取余，然后变成百位(个位与百位交换)/*取百位+910取余，变成个位，百位和个位上的数字互换*//*取千位+910printf("Theencryptednumberis%d\n",newnum);return0;}#include<stdio.h>intmain(void){#include<stdio.h>intmain(void){intdigit1,digit2,digit3,digit4,newnum,number; /*个位*/ /*十位*/ 百位*/ /*千位*/printf("Theencryptednumberis%d\n",newnum);return0;} 求1+1/3+1/5 的前n项n，11/31/5n出时保留6位小数。 (计算 sum=sum=#include#includeintint{inti,intrepeat,ri;doublesum;for(ri=1;ri<=ri++){scanf("%d",for(i=1;i<=n;i++){}}} 输出华氏-摄氏温度转换表（改错题取值范围是[lower,upper]，每次增加2F。计算公式：c5*(f32)9，其中：c，f输出请使用语句printf("%3.0f%6.1f\n",fahr,celsius); 32 40 fahr fahrintmain(void){intrepeat,ri;doublecelsius,for(ri=1;ri<=&upper);printf("fahr{celsius=5*(fahr-32)/9; }} nm!*n-2m和n(m<=n)，计算n!/(m!*(n-m)!)。是double。2 (m=2,5 (m=5,result=result=intmain(void){intm,doubles;doublefact(intn);for(ri=1;ri<=ri++){scanf("%d%d",&m,printf("result=%.0f\n",s);}}{returnresult;} 求 m和n(m<=n)，求sum=312255===intmain(void){inti,m,n;doublesum;for(ri=1;ri<=repeat;ri++){scanf("%d%d",&m,&n); }} 求1－2/3＋3/5－4/7＋5/9-出时保留3位小数。 sum=sum=sum=intmain(void){intflag,i,intrepeat,ri;doubleitem,sum;for(ri=1;ri<=ri++){scanf("%d",&n);for(i=1;i<=n;{item=flag*i*1.0/denominator;} }} #include#include<stdio.h>int{{intintrepeat,ri;doublesum;for(ri=1;ri<=ri++){scanf("%d",&n); }} n，e0!＋1!＋2!＋……＋n!，要求定义和调用函数fact(n)计算n!，函数类型是double。 124sum=sum=sum=intmain(void){intintrepeat,ri;doublesum;for(ri=1;ri<=ri++){scanf("%d",i);sum=sum+fact(i);} }}{intj;}5 求一元二次方程的根printf("x=%0.2f\n",-c/b);printf("x1=%0.2f\n",(-printf("x2=%0.2f\n",(-b-printf("x1=%0.2f+%0.2fi\n",-b/(2*a),sqrt(-printf("x2=%0.2f-%0.2fi\n",-b/(2*a),sqrt(-500000102413ab0，c0，方程不成立x=-2.00x1=-x2=-x1=-1.00+1.41ix2=-1.00-#include#include<stdio.h>#include<math.h>intmain(void){intrepeat,ri;doublea,b,c,d;for(ri=1;ri<=ri++){scanf("%lf%lf%lf",&a,&b,&c);d=b*b-4*a*c;if(c==0)printf("参数都为零，方程无意义}elseprintf("x=%0.2f\n",-}printf("x1=%0.2f\n",(-}}}}}参考#include#include<stdio.h>#include<math.h>intmain(void){intrepeat,ri;doubleintrepeat,ri;doublea,b,c,d;for(ri=1;ri<=ri++){scanf("%lf%lf%lf",&a,&b,&c);d=b*b-4*a*c; printf("参数都为零，方程无意义!\n"); printf("x=%0.2f\n",-c/b);printf("x1=%0.2f\n",(-}}}} 求分段函数的值repeat(0<repeat<10)，repeat输入一个整数x，计算并输出下列分段函数sign(x)的值。- x<y=sign(x)= x= x> - (x=-sign(10)=1 (x=10时y=1)sign(0)=0 (x=0时y=0)sign(-98)=-1 (x=-98时y=-1)intmain(void){intrepeat,intintx,for(ri=1;ri<=ri++){scanf("%d",&x);if(x==0)y=0;elseif(x>0)y=1;elsey=-1;}} 显示五级记分制成绩所对应的百分制成绩区间（使 repeat(0<repeat<10repeat和E(0-59),如果输入不正确的成绩，显示"Invalidinput"。printf("90-printf("80-printf("70-printf("60-printf("Invalid 字符j)90-80-70-60-0-Invalidinputintmain(void){charintrepeat,for(ri=1;ri<=repeat;ri++){for(ri=1;ri<=repeat;ri++){ch=getchar();}}return} 显示水果的价格（使用55输入其他编号，显示价格为0。 price=[0]intmain(void){intchoice,i;for(i=1;i<=5;i++){printf("[1]printf("[0]Exit\n");if(choice==0)case1:casecasecase4:}price=3.00;}}}30007repeat(0<repeat<10repeat输入三角形的3a,b,c，如果能构成一个三角形，输出面积area和周长perimeter(2Thesesidesdonotcorrespondtoavalidtriangle"。在一个三角形中，任意两边之和大于第三边。三角形面积计算公式：area(s(s-a)(s-b)(s-c))^0.5，其中s=(a+b+c)/2printf("area=%.2f,perimeter=%.2f\n",area,printf("Thesesidesdonotcorrespondtoavalid25514area=7.15,perimeter=Thesesidesdonotcorrespondtoavalid#include#include<stdio.h>#include<math.h>intmain(void){inta,b,c;for(ri=1;ri<=ri++){scanf("%d%d%d",&a,&b,if((a+b>c)&&(a+c>b)&&area=sqrt(s*(s-a)*(s-b)*(s-printf("area=%.2f,perimeter=%.2f\n",area,}elseprintf("Thesesidesdonotcorrespondtoavalid}} 计算个人所得税repeat(0<repeat<10repeat计算公式：tax=rate*(salary-850)当salary≤850时，rate=当850＜salary≤1350时，rate=当salary≤2850，rate当salary5850，rate当＜salary时，rate= =====intmain(){intri,doublerate,salary,scanf("%d",&repeat);{scanf("%lf",&salary);if(salary<=850) }elseif{rate=0.05;}elseif{rate=0.10;}elseif{rate=0.15;}{rate=0.20;}printf("tax=%0.2f\n",}}} 统计学生成绩成绩等级分为五级，分别为A(90-100)、B(80-89)、C(70-79)、D(60-69)和 77549273NumberofA(90-100):1NumberofB(80-89):0NumberofC(70-79):2NumberofD(60-69):1NumberofE(0-59):1intmain(void){intmark,n,intfor(i=1;i<=n;i++){ if(mark>=90&&mark<=100)elseif(mark>=80&&mark<=89)nb++;elseif(mark>=70&&mark<=79)nc++;elseif(mark>=60&&mark<=69)nd++;elsene++;}printf("NumberofB(80-89):%d\n",nb);printf("NumberofC(70-79):%d\n",nc);printf("NumberofD(60-69):%d\n",nd);printf("NumberofE(0-59):%d\n",ne);return0;} 分段计算水费（使用嵌套的if-else语句y(x(数)。要求用嵌套的if-else语句。 y=f(x)= 2.5x- - (x=- f(-0.50)=f(9.50)=f(21.30)=intmain(void){doublex,y;scanf("%d",&repeat);for(ri=1;ri<=repeat;ri++){ elsey=2.5*x-10.5;printf("f(%.2f)=%.2f\n",x,y);}} 出租车计费repeat(0<repeat<10repeat1.62 6 11.8230 cost=cost=cost=intmain(void){intrepeat,intminutes,seconds;doublecost,mile;scanf("%d",&repeat);for(ri=1;ri<=repeat;ri++){scanf("%lf%d%d",&mile,&minutes, elseif(mile<=10) printf("cost=%.0f\n",}3006221400intmain(void){intfor(year=2000;year<=2099;year++){if((year%4==0&&}6 求最小公倍数和最大公约数（调试示例repeat(0<repeat<10)，repeat输入两个正整数m和n，输出它们的最小公倍数和最大公约数。 3 24 24 21istheleastcommonmultipleof3and7,1isthegreatestcommondivisorof3and7.24istheleastcommonmultipleof24and4,4isthegreatestcommondivisorof24and4.72istheleastcommonmultipleof24and18,6main(void){intgcd,lcm,m,n,c;intrepeat,ri;scanf("%d",&repeat);for(ri=1;ri<=repeat;ri++){for(lcm=1;} lcmprintf("%distheleastcommonmultipleof%dand%d,%disthegreatestcommondivisorof%dand%d.\n",lcm,m,n,gcd,m,n);}return}intmain(void){intgcd,lcm,m,intintrepeat,scanf("%d",for(ri=1;ri<=repeat;{scanf("%d",scanf("%d",if(m<=n)c=m; elsec=n;for(gcd=c;;gcd--)}lcm=printf("%distheleastcommonmultipleof%dand%d,%disgreatestcommondivisorof%dand%d.\n",lcm,m,n,gcd,m,}return} eps(保留6位小数)。请使用while语句实现循环。2E- (eps=2E- sum=sum=#include#include<stdio.h>#include<math.h>intmain(void){intdenominator,flag;intrepeat,ri;doubleeps,item,for(ri=1;ri<=repeat;}}return} 求整数的位数repeat(0<repeat<10repeat - (in=-- (in=- count= count= count= count= intmain(void){intcount,in;for(ri=1;ri<=ri++){scanf("%d",if(in<0)in=-in;}while}return} repeat(0<repeat<10repeat输入一个正整数n，输出2/1＋3/2＋5/3＋8/5n项之和，保留2位小数。(该序列从第2项起，每一项的分子是前一项分子与分母的和，分母是前一 sum sum sum= intmain(void){inti,intrepeat,for(ri=1;ri<=ri++){scanf("%d",sum+=tempdenominator*}}} 换硬币repeat(0<repeat<10repeat将一笔零钱（大8分1精确到分）换521分的硬币。要求：硬币面值按5分、2分、1分顺序，各类硬币数量依次从大到小的顺序，输出使用语句：printf("fen5:%d,fen2:%d,fen1:%d,total:%d\n",fen5,fen2,fen1,fen5+fen2+fen1); count=2 count= intmain(void){intcount,fen1,fen2,fen5,money;intrepeat,ri;for(ri=1;ri<=ri++){scanf("%d",&money);printf("fen5:%d,fen2:%d,fen1:%d,total:%d\n",fen5,fen2,fen1,fen5+fen2+fen1);}return} 找出各位数字的立方和等于它本身的数repeat(0<repeat<10repeatmn(1<=m,n<=1000)，mn输出使用语句：printf("%d\n", 100 (m=100,1 (m=1, intmain(void){inti,digit,m,n,a,b,c,sum;intrepeat,ri;for(ri=1;ri<=repeat;ri++){scanf("%d%d",&m,&n);}}return}intmain(void){inti,digit,m,n,number,sum;intrepeat,ri;for(ri=1;ri<=repeat;ri++){scanf("%d%d",&m,&n);}}}} 找完数（改错题repeat(0<repeat<10repeatmn(1<=m,n<=1000)，mn其中1、2、3为因子，6为因子和。printf("%d=1",printf("+%d",factor);1 (m=1,400 (m=400,1=6=1+2+ 28=1+2+4+7+496=1+2+4+8+16+31+62+124+intmain(void){intfactor,m,n,number,sum;intrepeat,ri;for(ri=1;ri<=repeat;ri++){scanf("%d%d",&m,&n);for{forif(sum==number){printf("%d=forif(number%factor==0)printf("+%d",factor);}}}} 验证歌德巴赫猜想(选作mn(6<=m,n<=100)，mn数之和，打印时一行打印5组。输出使用语句：printf("%d=%d+%d",number,i,number-89100(m=90,90=7+8392=3+8994=5+8996=7+89#include"math.h"intmain(void){intcount,i,j,k,m,n,scanf("%d%d",&m,&n);if(m%2!=0)m=m+1;if(m>=6){ }}intprime(intm){inti,ifPrime=0;if(m==1)return0;}return}int{intcount,i,m,n,if(m%2!=0)m=m+1;if(m>=6){ prime(i)&&prime(number-i){printf("%d=%d+%d",number,i,number-i);count++;}}}}} 从高位开始逐位输出一个整数的各位数字(选作输入一个整数in，从高位开始逐位分割并输出它的各位数字。输出使用语句：printf("%-2d",digit); - (in=- 2345608intmain(void){intdigit,in,power,temp,k;intrepeat,ri;for(ri=1;ri<=repeat;if(in<0)in=-in;k={k=k*} /*temp{temp--;k=digit=power/power=power%k; /*power*/}}return} 求平均成绩并统计不及格人数678873548287437056100-Gradeaverageis72.80,thenumberoffailis1Gradeaverageis25.38,thenumberoffailis#include#includeintint{intfail,num;intrepeat,ri;doublegrade,for(ri=1;ri<=repeat;ri++){scanf("%lf",&grade);if(num>0)printf("Gradeaverageis%.2f,thenumberoffailis%d\n",total/num,fail);}return}7 使用函数判断数的符号repeat(0<repeat<10repeat1x，x0，sign(x)1x0，sign(x)0；否则，sign(x)=-1，最后输出sign(x)的值。类型是int。 - (x=- sign(10)=1 (x=10时sign(x)的值为1)sign(-5)=- sign(0)= ①intsign(intx);int{intx,intrepeat,for(ri=1;ri<=ri++){}}intsign(int{intt=0; elseif(x==0) return} 使用函数求奇数0，函数形参n的类型是int，函数类型是int。 12961721019-Thesumoftheoddnumbersis30.Thesumoftheoddnumbersisinteven(intn);int{intn,sum;intn,sum;for(ri=1;ri<=ri++){sum=0;while(n>0){if(even(n)==0)}printf("Thesumoftheoddnumbersis%d.\n",}}inteven(int{intif(n%2==0) returnt;}50005repeat(0<repeat<10repeat2mn（1<=m,n<=500），mn数以及这些素数的和。素数就是只能被1和自身整除的正整数，1不是素数，2要求定义并调用函数prime(m)m是否为素数，当m素数时返1，否则0，函数形参m的类型是int，函数类型是int。 1 (m=1,Count=4,sum= #include#include"stdio.h"#include"math.h"intmain(void){intcount,i,m,n,intintrepeat,intprime(intfor(ri=1;ri<=ri++){scanf("%d%d",&m,&n);sum=0;if /*1,for(i=m;i<=n;i++){if(}}printf("Count=%d,sum=%d\n",count,}}intprime(int{int if(m==1)returnresult=for(t=2;t<=m/2;t++)if(m%t==0)returnresult;} 使用函数统计一个整数中数字的个数repeat(0<repeat<10repeat数字digit的个数。digitnumberdigitint，函数类型是int。例如，countdigit(10090,0)的返回值是3。22(number=21252,-9(number=-1111,Number21252ofdigit2: (21252中有3intmain(void){intintmain(void){intcount,digit,in;intrepeat,ri;for(ri=1;ri<=repeat;ri++){scanf("%d%d",&in,&digit);printf("Number%dofdigit%d:%d\n",in,digit,}}{intdoif(t==digit)c++;returnc;}/*//while0050007使用函数找水仙花数repeat(0<repeat<10repeat2mn(1<=m,n<=1000)，mnis(numbernumber自身，若相等则返回1，否则返回0，函数形参number的类型是int，函数类型是int。输出使用语句：printf("%d\n", 100400(m=100,1 (m=1,result:(100400 result:(1100 intmain(void){inti,m,n;intis(intfor(ri=1;ri<=ri++){scanf("%d%d",&m,&n);for{}}}intis(int{intsum,res,t,j;while{}if(sum!=number) 1!2!+…m!（n!，nint，double。 1!+2!+...+5!=#include#include<stdio.h>doublefact(intn);intmain(void){inti,m;printf("1!+2!+...+%d!=%f\n",m,sum);}{int double (n==0)return1.0;return} 使用函数求余弦函数的近似值repeat(0<repeat<10repeat于e。cos(x)=为double，函数类型是double。20(e=0.001,-(e=0.0001,x=-sum=sum=-#include#includedoublefuncos(doublee,doublex);intmain(void){intrepeat,ri;doublee,sum,x;for(ri=1;ri<=ri++){scanf("%le%le",&e,&x);sum=funcos(e,x);printf("sum=%f\n",sum);}}doublefuncos(doublee,double{intflag,n; /*tn*/}return}求funcos(e,x)方法doublefuncos(doublee,double{intflag=1,n=0;doubledouble /*item*/}returnreturn} 使用函数找最大a、ba、bint，int。 5 (a=5,-1-10(a=-1,b=-1 (a=1,max(5,8)=max(-1,-10)=-max(1,1)=intmain(void){inta,b,maximum;intrepeat,ri;intmax(inta,intfor(ri=1;ri<=repeat;ri++){scanf("%d%d",& }}intmax(inta,int{int /*aif(c<b) /*b*/returnc; Fibonaccimn(1<=m,n<=10000)，mnFibonacci序列(第一项起)：11 5 1321nint，long。例如，fib(7)13。输出使用语句：printf("%ld",f); 1 (m=1,20 (m=20,1000 (m=1000,11235 intmain(void){inti,m,n;longf;for(ri=1;ri<=repeat;ri++){&m,&n);f=1;i=1;while(f<=n){if(f>=m)printf("%d",f);}}}longfib(int{intlonga=1,b=1,if(n==1||n==2) for(i=3;i<=n;i++){}}50063repeat(0<repeat<10repeat输入两个正整数mn(1<=m,n<=1000)，输出mn要求定义并调用函数factorsum(number)，它的功能是返number的因子和，函数形参number的类型是整型，函数类型是整型。例如，factorsum(12)的返回值是16(1+2+3+4+6)。输出使用语句：printf("%d 20 (m=201 (m=1,28 16 intmain(void){inti,m,n;for(ri=1;ri<=ri++){scanf("%d%d",&m,&n);printf("%d",i);}}}{{ else{ t<=number-1也对}}return}8 求x＋x*x/2!＋x*x*x/3!＋x*x*x*x/4!＋……的输入1个实数x，计算并输出下最后一项的绝对值小于0.00001(保留2位小数)。要求定义和调用函数fact(n)计算n的阶乘，可以调用pow函数sx＋x*x/2!＋x*x*x/3!＋x*x*x*x/4!＋……301sss#include<stdio.h>#include<stdio.h>intmain(void){intintrepeat,ri;doubleitem,s,x;for(rifor(ri=1;ri<=ri++){scanf("%lf",&x);}printf("s=%.2f\n",}}{intdoubleproduct;product=1;returnproduct;} 使用函数计算两点间的距离输入一个正整数repeat(0<repeat<10)，做repeat次下列运算：(x1,y1)(x2,y2)，求这两点之间的距离(2位小数)dist(x1,y1,x2,y2)x1、y1、x2y2的类型都是double，函数类型是double。 10 (x1=10,200 (x2=200,输#include<stdio.h>#include<stdio.h>doubledist(doublex1,doubley1,doublex2,doubley2);intmain(void){intrepeat,doubledoubledistance,x1,y1,x2,for(ri=1;ri<=ri++){scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);distance=dist(x1,y1,x2,y2);printf("Distance=%.2f\n",distance);}}doubledist(doublex1,doubley1,doublex2,double{doublea,b,c;returnc;} 数字金字塔数字金字塔，函数形参n的类型是int，函数类型是void。printf("printf("%d", (n=5时的数字金字塔2334445555 (n=2时的数字金字塔2intmain(void){intintintrepeat,ri;intnRow,nCol;for(ri=1;ri<=ri++){scanf("%d",for(nRow=1;nRow<= {for(nCol=1;nCol<=n-nRow;nCol++) for(nCol=1;nCol<=nRow*2-1;nCol++)printf("%d",nRow);}}} 使用函数求程序填空，不要改变与输入输出有关的语句。an,a＋aa＋aaa＋aa…a(na)之和。要求定义并调用函fn(an)，它的功能是返回aa…a（na）函数形an的类型输 2 (a=2,8 (a=8,输 intmain(void){inta,i,longsn;longfn(inta,intfor(ri=1;ri<=repeat;scanf("%ld%d",scanf("%ld%d",&a,&n);printf("%ld\n",}}/*思路：如2222=((2+2*10)*10)+2) *10) +2,是一个累加和*/longfn(inta,intn){}return} 使用函数求PI（调试示例源程序repeat(0<repeat<10)，repeat输入精度eps，根据下式求PI的值，直到最后一项小于eps。fact(n)n!，nint，函数类型是函数类型是double。 1E- (eps=1E-1E- (eps=1E-PI=PI=doubledenominator(intn);intmain(void){intintintrepeat,doubleeps,sum,item;for(ri=1;r