res2009春学期语言上机练习参汇总

第2周 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 20013求摄氏温度26°C对应的华氏温度 20015n152n的个位数字(digit1)、十位数字(digit2) 200262个整数num1num2第3周 求 求 求1+1/3+1/5+......的前n项 求x的n次 生成3的乘方 求 计算物体自由下落的距 计算分段函 阶梯电 求 求 第4周 显示图案（复习printf()的字符串输出 生成阶乘 使用函数求n!/(m!*(n- 求平均 求 求 求最小 求 第5周 求一元二次方程的 求分段函数的 分类统计字 显示水果的价格（使用 求三角形的面积和周 计算个人所得 判断闰 统计学生平均成绩与及格人 第6周 求 求整数的位 换硬 找出各位数字的立方和等于它本身的 找完数（改错题 判断素 逆序输出整 输出那契序 第7周 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周 求奇数 使用函数计算两点间的距 使用函数求 分类统计字 验证猜 使用函数输出整数的逆序 统计单 简单计算 2周20011100°F#include#include<stdio.h>intmain(void){intcelsius, printf("fahr=%d,celsius=%d\n",fahr,return}20012求华氏温度150°F计算：C=5*F/9-5*32/9，式中：C表示摄氏温度，F表示华氏温度。fahr=150,#include#include<stdio.h>intmain(void){intcelsius,printf("fahr=%d,celsius=%d\n",fahr,celsius);return0;}2001326°C计算如下，f=9*c/5+32，式中：c表示摄氏温度，f表示华氏温度。#include#include<stdio.h>intmain(void){intcelsius, printf("celsius=%d,fahr=%d\n",celsius,fahr);return0;}20015n152n的个位数字(digit1)、十位数字(digit2)和百位数字(digit3)#include#include<stdio.h>intmain(void){intn,digit1,digit2,digit3;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}200262个整数num1num2printf("%d%d%d\n",num1,num2,5 5+3=85-3=25*3=5/3=15%3=2#include#include<stdio.h>intmain(void){intprintf("%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=return}}3周 求 123100sum=#include#include<stdio.h>intmain(void){inti,sum;printf("sum=%d\n",} 求 repeat(0<repeat<10)，repeat输入一个正整数m(0<=m<=100)，计算表达式m+(m+1)+(m+2) 100 (计算 (计算 (计算 sum=sum=sum=#include#include<stdio.h>intmain(void){inti,m,sum;intrepeat,scanf("%d",for(ri=1;ri<=repeat;scanf("%d",&m);scanf("%d",&m); printf("sum=%d\n",}} 求 repeat(0<repeat<10repeat2m和n(m<=n)，计算表达式1/m+1/(m+1)+1/n3。35101sumsumsum#include#include<stdio.h>intmain(void){inti,m,n;intrepeat,ri;doublesum;scanf("%d",for(ri=1;ri<=repeat;ri++){scanf("%d%d",&m,&n);for(i=m;i<=n;i++)printf("sum=%.3f\n",}} 求1+1/3+1/5 的前n项repeat(0<repeat<10repeatn，11/31/5n6 (计算 sum=sum=#include#include<stdio.h>intmain(void){inti,intdenominator;intrepeat,ri;doublesum;scanf("%d",for(ri=1;ri<=repeat;ri++){scanf("%d",&n);{}printf("sum=%.6f\n",}} repeat(0<repeat<10repeatn，1－1/4＋1/7－1/10＋……n3 3sum=sum=#include#include<stdio.h>intmain(void){intflag,i,n,t;intrepeat,ri;doubleitem,scanf("%d",for(ri=1;ri<=repeat;ri++){scanf("%d",&n); printf("sum=%.3f\n",}} repeat(0<repeat<10repeat2lowerupper，输出一张华氏—摄氏温度转换表，华氏温度的取值范围是[lower,upper2F。计算：c=5*(f-32)/9，其中：c表示摄氏温度，f表示华氏温度。输出请使用语句printf("%3.0f%6.1f\n",fahr,celsius); 32 40 #include#include<stdio.h>intmain(void){intrepeat,ri;doublecelsius,fahr;scanf("%d",&repeat);for(ri=1;ri<=repeat;ri++){ { }} xnrepeat(0<repeat<10)，repeat1个实数xn(n<=50)，xn2pow1.5 2 #include#include<stdio.h>intmain(void){inti,intrepeat,ri;doublemypow,x;scanf("%d",for(ri=1;ri<=repeat;ri++){scanf("%lf%d",for(ri=1;ri<=repeat;ri++){scanf("%lf%d",&x,&n); }} 生成3输入一个正整数n，33^0~3^n3输出使用语句printf("pow(3,%d)=%.0f\n",i,=1=3=9=#include#include<stdio.h>#include<math.h>intmain(void){inti,double/powerscanf("%d",&n);for(i=0;i<=n;i++){printf("pow(3,%d)=%.0f\n",i,}} 计算100^0.5＋101^0.5＋……＋1000^0.52sum=#include<stdio.h>#include#include<stdio.h>#include<math.h>intmain(void){intdoublesum; printf("sum=} 一个物体从100m的高空自由落下，编写程序，求它3s内下落的垂直距210^2。#include#include<stdio.h>intmain(void){doubleheight=0.5*10*3* printf("height=%.2f\n",} repeat(0<repeat<10repeat编写程序，输入x，f(x)2sqrt()pow()函数求幂。x>=0，f(x)x^0.5，x0f(x)x^52x1/x。 - (x=- f(10.00)=f(-0.50)=-f(0.00)=#include<stdio.h>#include#include<stdio.h>#include<math.h>intmain(void){intrepeat,ri;doublex,y;scanf("%d",for(ri=1;ri<=repeat;ri++){ if(x>=0)else printf("f(%.2f)=%.2f\n",x,}} repeat(0<repeat<10repeat500.53500.05e(千瓦时)，计算并输出该用户应支付的电费(元)，结果2 cost=cost=#include#include<stdio.h>intmain(void){intrepeat,ri;doubleintrepeat,ri;doublecost,e;scanf("%d",for(ri=1;ri<=repeat;ri++){scanf("%lf",&e); printf("cost=%.2f\n",}} 求 repeat(0<repeat<10repeat输入两个正整数m和n(m<=n)，求sum= ＋n*n＋1/n，6312255===#include#include<stdio.h>intmain(void){inti,m,n;intrepeat,ri;doublesum;for(ri=1;ri<=repeat;ri++){scanf("%d%d",&m,&n); printf("sum=%.6f\n",}} repeat(0<repeat<10repeatn，1－2/3＋3/5－4/7＋5/9-6/11+……n3 sum=sum=sum=#include#include<stdio.h>intmain(void){intflag,i,floatdenominator;intrepeat,ri;doubleitem,sum;scanf("%d",for(ri=1;ri<=repeat;ri++){scanf("%d",&n);denominator=1.0;for(i=1;i<=n;{item=flag*i*1.0/denominator;flag=-denominator=denominator} printf("sumprintf("sum=%.3f\n",}} #include<stdio.h>#include#include<stdio.h>#include<math.h>intmain(void){intintrepeat,ri;doublesum;scanf("%d",for(ri=1;ri<=repeat;ri++){scanf("%d",&n); printf("sum=%.0f\n",}}4周 显示图案（printf()的字符串输出）*******#include#include<stdio.h>intmain(void){printf("***"***\n" * }} 输入一个正整数n，1!~n!fact(n)计算n!，double。输出使用语句printf("%d!=%.0f\n",i,31!12!23!6#include#include<stdio.h>intmain(void){inti,doublemyfact;doublefact(intn);scanf("%d",&n);for(i=1;i<=n;i++)printf("%d!=%.0f\n",i,}}doublefact(int{doubleresult;intj;return} nm!*n-repeat(0<repeat<10)，repeat2m和n(m<=n)，计算n!/(m!*(n-m)!)。fact(n)nnint，函数类型是double。2 (m=2,5 (m=5,result=result=#include#include"stdio.h"intmain(void){intm,intrepeat,ri;doubles;doublefact(intn);scanf("%d",&repeat);for(ri=1;ri<=repeat;ri++){scanf("%d%d",&m,&n);printf("result=%.0f\n",s);}}doublefact(int{doubleresult;inti;return} 3212average=#include#include<stdio.h>intmain(void){inta,b,c;doubleaverage;average=(a+b+c)/3.0;printf("average=%.2f\n",average);} 求 repeat(0<repeat<10repeat编写程序，输入一个正整数n，求 的前n项之和，输出6 sum=sum=#include#include<stdio.h>intmain(void){inti,intrepeat,ri;doublesum;scanf("%d",for(ri=1;ri<=repeat;ri++){scanf("%d",&n);printf("sum=%.6f\n",sum);}} repeat(0<repeat<10repeatn，e0!＋1!＋2!＋……＋n!，要求定fact(n)n!，double。124sum=sum=sum=#include#include<stdio.h>intmain(void){intintrepeat,ri;doublesum;doublefact(intn);scanf("%d",for(ri=1;ri<=repeat;ri++){scanf("%d",&n);} printf("sum=%.0f\n",}}doublefact(int{doubleresult;intj;return} repeat(0<repeat<10repeatn,再输入n 4-2-1231004-9-11-353min=-min=-min=#include#include<stdio.h>intmain(void){inti,min,n,x;intrepeat,ri;scanf("%d",for(ri=1;ri<=repeat;ri++){scanf("%d",&n);scanf("%d",&x);scanf("%d",&x); } printf("min=%d\n",}} ：tn=a+a*10+a*100…+a*10^(n-=t(n-1)+a*10^(n-#include<stdio.h>#include#include<stdio.h>#include<math.h>intmain(void){inta,i,n,sn,tn;intri,repeat;scanf("%d",for(ri=1;ri<=repeat;ri++){scanf("%ld%d",&a,&n);for(i=0;i<n;i++){} printf("sum=%d\n",}}5周 repeat(0<repeat<10repeata,b,c，a*x*x＋b*x＋c＝02printf("ab0，c0，方程不成立\n");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(-50000010248.9 12 ab0，c0，方程不成立x=-2.00x1=-x2=-x1=-1.00+1.41ix2=-1.00-#include<stdio.h>#include#include<stdio.h>#include<math.h>intmain(void){intrepeat,ri;doublea,b,c,d;scanf("%d",for(ri=1;ri<=repeat;ri++){scanf("%lf%lf%lf",&a,&b,&c);}}}printf("x=%0.2f\n",-}}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(-}}}}} repeat(0<repeat<10repeat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)#include#include<stdio.h>intmain(void){intrepeat,ri;intx,y;scanf("%d",for(ri=1;ri<=repeat;ri++){scanf("%d",&x);if(x==0)y=0;elseif(x>0)y=1;elsey=-1;printf("sign(%d)=%d\n",x,}} 10Reoldletter=5,blank=1,digit=3,other=#include#include<stdio.h>intmain(void){charintblank,digit,i,letter,blank=digit=letter=other=0;for(i=1;i<=10;i++){c=getchar();if(c=='')blank++;letter++;digit++;other}printf("letter=%d,blank=%d,digit=%d,other=%d\n",letter,blank,digit,} 输入一个正整数repeat(0<repeat<10)，做repeat输入五级制成绩(A－E)，输出相应的百分制成绩(0－100)switch五级制成绩对应的百分制成绩区间为：A(90-100)、B(80-89)、C(70-79)D(60-69)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 (repeat=6，A、B、C、D、Ej)90-80-70-60-0-Invalidinput(输入数据不合法)#include#include<stdio.h>intmain(void){charintrepeat,scanf("%d",for(ri=1;ri<=repeat;ri++){ch=getchar();printf("Invalidinput\n");}}return} 显示水果的价格（4(apples)、梨(pears)、桔子(oranges)和葡萄(grapes)，3.00，2.50，4.1010.20551～4，10，退出查0。 (oranges price=[0]#include#include<stdio.h>intmain(void){intchoice,i;doubleprice;for(i=1;i<=5;i++){printf("[1]apples\n");printf("[2]printf("[3]printf("[4]printf("[0]Exit\n");scanf("%d",&choice);if(choice==0)switch(choice){case1:price=casecasecasecase}printf("price=%0.1f\n",}}}30007输入一个正整数repeat(0<repeat<10)，做repeat3a,b,c，areaperimeter(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：area=(s(s-a)(s-b)(s-c))^0.5，其中s=(a+b+c)/2printf("area=%.2f,perimeter=%.2f\n",area,perimeter);printf("Thesesidesdonotcorrespondtoavalidtriangle\n"); 55 14 area=7.15,perimeter=Thesesidesdonotcorrespondtoavalid#include<stdio.h>#include#include<stdio.h>#include<math.h>intmain(void){inta,b,intrepeat,scanf("%d",for(ri=1;ri<=repeat;ri++){scanf("%d%d%d",&a,&b,&c);if((a+b>c)&&(a+c>b)&&(b+c>a)printf("area=%.2f,perimeter=%.2f\n",area,}printf("Thesesidesdonotcorrespondtoavalid}} 输入一个正整数repeat(0<repeat<10)，做repeatsalary，tax(2计算：tax=rate*(salary-850)当salary≤850时，rate=当850＜salary≤1350时，rate=当1350＜ 2850时，rate=当2850＜salary≤5850时，rate=当5850＜salary时，rate= =====#include#include<stdio.h>intmain(){intri,doublerate,salary,scanf("%d",&repeat);for(ri=1;scanf("%d",&repeat);for(ri=1;ri<=repeat;ri++){scanf("%lf",&salary);if(salary<=850){}elseif{}elseif{}elseif{}{}printf("tax=%0.2f\n",}} repeat(0<repeat<10repeatyear，4被100整除，或者能被400整除。printf("%disaleapyear.\n",year);printf("%disn'taleapyear.\n",year); 2000isaleapyear.2010isn'taleapyear.2011isn'taleap#include#include<stdio.h>intmain(void){intrepeat,ri;intyear;scanf("%d",for(ri=1;ri<=repeat;ri++){printf("%disaleapyear.\n",year);printf("%disn'taleap}} n，n1 77549273average=count=#include#include<stdio.h>intmain(void){intcount,i,doubleaverage,grade,}}average=1.00*total/n;printf("average=%.1f\n",average);\printf("count=%d\n",count);} repeat(0<repeat<10repeaty(x(吨)的函数关系式如下。x(y(2if-else y=f(x)= 2.5x- - (x=- f(-0.50)=f(9.50)=f(21.30)=#include#include<stdio.h>intmain(void){intrepeat,ri;doublex,y;scanf("%d",&repeat);scanf("%d",&repeat);for(ri=1;ri<=repeat;ri++){scanf("%lf",&x); elseif(x>=0&&x<=15)y=4.0*x/3;elsey=2.5*x-10.5;printf("f(%.2f)=%.2f\n",x,y);}}6周 repeat(0<repeat<10repeat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 21istheleastcommonmultipleof3and7,1isthegreatestcommondivisorof3and7.24istheleastcommonmultipleof24and4,4isthegreatestcommondivisorof24and4.72istheleastcommonmultipleof24and18,6isthegreatestcommondivisorof24and18.#include#include<stdio.h>intmain(void){ ,lcm,m,n,c;intrepeat,ri;scanf("%d",&repeat);for(ri=1;ri<=repeat;ri++){if(m<=n)c=m; /*mm*/elsec=n;for(lcm=1;;lcm++)}=c;--)}printf("%distheleastcommonmultipleof%dand%d,%disthegreatestcommondivisorof%dand%d.\n",lcm,m,n, ,m,n);}return}参考方法2：#includestdio.h>intmain(void){ ,lcm,m,n,i;intrepeat,ri;scanf("%d",for(ri=1;ri<=repeat;ri++){scanf("%d",&m);scanf("%d",&n);if(m<=0||n<=printf("m<=0orn<={} printf("%distheleastcommonmultipleof%dand%d,%disthegreatestcommondivisorof%dand%d.\n",lcm,m,n, ,m,n);}}} repeat(0<repeat<10repeat1eps，计算并输出下式的值，精确到最后一项的绝对值小于eps(6while语句实现循环。 2E- (eps=2E- sum=sum=#include<stdio.h>#include#include<stdio.h>#include<math.h>intmain(void){intrepeat,ri;doubleeps,item,scanf("%d",for(ri=1;ri<=repeat;ri++){scanf("%le",&eps);}printf("sum=%.6f\n",}return} 输入一个正整数repeat(0<repeat<10)，做repeatin，1233do-while - (in=-- (in=- count= (12345count= (-100count= (-1count= (99#include#include<stdio.h>intmain(void){intcount,in;intrepeat,scanf("%d",for(ri=1;ri<=repeat;ri++){scanf("%d",&in);}whileprintf("count=%d\n",}return} 输入一个正整数repeat(0<repeat<10)，做repeat将一笔零钱（81）521521输出使用语句：printf("fen5:%d,fen2:%d,fen1:%d,total:%d\n",fen5,fen2,fen1,fen5+fen2+fen1); (money=10 (money=13count=2 (102count= (134#include#include"stdio.h"intmain(void){intcount,fen1,fen2,fen5,money;intrepeat,ri;scanf("%d",for(ri=1;ri<=repeat;ri++){scanf("%d",&money); printf("fen5:%d,fen2:%d,fen1:%d,total:%d\n",fen5,fen2,fen1,fen5+fen2+fen1);printf("count=%d\n",}return} 输入一个正整数repeat(0<repeat<10)，做repeatmn(1<=m,n<=1000mn1531^3＋5^3＋3^3＝153a^bab输出使用语句：printf("%d\n", 100 (m=100,1 (m=1, #include#include"stdio.h"intmain(void){inti,digit,m,n,a,b,c,sum;intrepeat,ri;for(ri=1;ri<=repeat;ri++){scanf("%d%d",&m,&n);if(sum==i)printf("%d\n",i);}}return}2：#includestdio.h"intmain(void){inti,digit,m,n,number,sum;intrepeat,ri;for(ri=1;ri<=repeat;ri++){scanf("%d%d",&m,&n);}if(sum==i)}}} 找完数（输入一个正整数repeat(0<repeat<10)，做repeatmn(1<=m,n<=1000mn6=1+2+3，1、2、3，6printf("%d1",number);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+#include#include<stdio.h>intmain(void){intfactor,m,n,number,intintrepeat,for(ri=1;ri<=repeat;ri++){scanf("%d%d",&m,&n);for(number=m;number<=n;number++){forif(number%factor==0)sum=sum+factor;if(sum==number){printf("%d=forif(number%factor==0)printf("+%d",factor);}}}} repeat(0<repeat<10repeat输入一个整数in，从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=- 2345608#include#include<stdio.h>intmain(void){intdigit,in,power,temp,k;intrepeat,ri;scanf("%d",for(rifor(ri=1;ri<=repeat;ri++){scanf("%d",&in);if(in<0)in=-in;k={k=k*} /*temp{temp--;k=k/10;digit=power/power=power% /*power*/}}return} 输入一个正整数n，再输入n112printf("%disaprime\n",m);printf("%dis'ntaprime\n",m); 1is'nta2isa9is'nta17isa#include<stdio.h>#include#include<stdio.h>#include<math.h>intmain(void){intflag,i,j,k,m,scanf("%d",&n);if(m==1)printf("%dis'ntaprime\n",m);else{if(m%j==0)printf("%dis'ntaprime\n",m);}printf("%disaprime\n",}}return} repeat(0<repeat<10repeatin，将其逆序输出。假设正数和负数逆序输出的结果一样。输出使用语句printf("%d",digit); - (in=-005432#include#include<stdio.h>intmain(void){intdigit,intintrepeat,for(ri=1;ri<=repeat;ri++){scanf("%d",&x);if(x<0)x=-x;printf("%d",digit);}}} 输出那契序输入一个正整数n(1<n<20)，输出那契(Fibonacci)序列112358…的前n项。输出使用语句：printf("%d",112358132134#include#include<stdio.h>intmain(void){inti,n,x1,x2,scanf("%d",&n);x1=1;x2=printf("%d%d",x1,x2);printf("%d",x);}return}7周 输入一个正整数repeat(0<repeat<10)，做repeat1x，x0，sign(x)1x0，sign(x)0；否则，sign(x)=-1，sign(x)的值。sign(x)实现该分段函数,函数形参xint，函数int。 - (x=- sign(10)1(x=10sign(x)1)sign(-5)1(x=-5sign(x)的值为-1)sign(0)=0(x=0sign(x)0)①#include#include<stdio.h>intsign(intx);int{intx,intrepeat,scanf("%d",for(ri=1;ri<=repeat;ri++){printf("sign(%d)=%d\n",x,}}intsign(int{int elseif(x==0) return} repeat(0<repeat<10repeateven(n)判断数的奇偶性，当n1，否则返0，函数形参nint，int。 12961721019-Thesumoftheoddnumbersis30.Thesumoftheoddnumbersis#include#include<stdio.h>inteven(intn);int{intn,sum;intri,repeat;scanf("%d",for(ri=1;ri<=repeat;ri++){whileif(even(n)==0)}printf("Thesumoftheoddnumbersis%d.\n",}}inteven(int{intif(n%2==0) returnt;} 输入一个正整数repeat(0<repeat<10)，做repeat2mn（1<=m,n<=500），mn11数，2prime(m)判断mm1，否0，函数形参mint，int。1 (m=1,Count=4,sum= (1104#include#include"stdio.h"#include"math.h"intmain(void){intcount,i,m,n,sum;intrepeat,ri;intprime(intscanf("%d",for(ri=1;ri<=repeat;ri++){scanf("%d%d",&m,&n);if /*1,可直接判定下一个数}}printf("Count=%d,sum=%d\n",count,}}intprime(int{intt,limit, /**/for(t=2;t<=m-1;t++)if(m%t== /**/}return} 输入一个正整数repeat(0<repeat<10)，做repeat1in，digit(0≤digit<10)，统计并输出整数中数字digitcountdigit(number,digit)，它的功能是统计整数numberdigitnumberdigitint，函数int。例如，countdigit(10090,03。22-9(number=-Number21252ofdigit2: (212523Number-1111ofdigit9: (-11110#include#include"stdio.h"intmain(void){intcount,digit,in;intrepeat,ri;intcountdigit(intnumber,intfor(ri=1;ri<=repeat;ri++){scanf("%d%d",&in,&digit);printf("Number%dofdigit%d:%d\n",in,digit,}}{int /**/while(number!=0){if(t==digit)c++;}return} 输入一个正整数repeat(0<repeat<10)，做repeat2mn(1<=m,n<=1000mnis(number)numberint。输出使用语句：printf("%d\n",100 (m=100,1 (m=1, (100400 (1100 #include#include"stdio.h"intmain(void){inti,m,n;intrepeat,ri;intis(intscanf("%d",for(ri=1;ri<=repeat;ri++){scanf("%d%d",&m,&n);for(i=m;i<=n;i++){printf("%d\n",i);}}}intis(int{intsum,res,t,j;while{}if(sum!=number)res=0;returnres;} 输入一个正整数repeat(0<repeat<10)，做repeat输入精度e和x，用下列求cos(x)的近似值，精确到最后一项的绝对值小于e。cos(x)=funcos(e,x)cos(x)的值，函数形参exdouble，double。 (e=0.001,0.0001- (e=0.0001,x=-sum=sum=-#include#include"stdio.h"#include"math.h"doublefuncos(doublee,doublex);intmain(void){intrepeat,ri;doublee,sum,x;scanf("%d",for(ri=1;ri<=repeat;ri++){scanf("%le%le",&e,&x);printf("sum=%f\n",sum);}}doublefuncos(doublee,double{intflag,n;doubleintflag,n;doublesum,t;}return/*tn}doublefuncos(doublee,double{intflag=1,n=0;doubleitem=1.0;doubledoubledemoninator=1.0;}return} repeat(0<repeat<10repeat输入两个整数ab，max(a,b)找出a、b中较大的数，函数形参a、bint，int。5 (a=5,-1- (a=-1,b=-1 (a=1,max(5,8)=max(-1,-10)=-max(1,1)=#include#include<stdio.h>intmain(void){inta,b, intrepeat,ri;intmax(inta,intscanf("%d",for(ri=1;ri<=repeat;ri++){scanf("%d%d",&a,&b);printf("max(%d,%d)=%d\n",a,b,}}intmax(inta,int{int /*a是大数if(c<b) /*b*/returnc;} Fibonaccirepeat(0<repeat<10repeatmn(1<=m,n<=10000mnFibonacciFibonacci序列(第一项起 21fib(n)，nFibonacci参nint，long。例如，fib(7)13。输出使用语句：printf("%ld",31(m=1,20(m=20,1000(m=1000,11238(110Fibonacci (20100Fibonacci (10006000Fibonacci#include#include"stdio.h"intmain(void){inti,m,n;intrepeat,ri;longf;longfib(intfor(ri=1;ri<=repeat;ri++){scanf("%d%d",&m,&n);f=1;i=1;while(f<=n)if(f>=m)printf("%d",f);}}}longfib(int{intlonga=1.0,b=1.0,c;if(n==1||n==2)return1;}return} 输入一个正整数repeat(0<repeat<10)，做repeatmn(1<=m,n<=1000mnfactorsum(numbernumbernumberfactorsum(12)的返16(1+2+3+4+6)。输出使用语句：printf("%d",20 (m=201 (m=1,28 (2050016 (1100#include#include"stdio.h"intmain(void){inti,m,n;intrepeat,ri;intfactorsum(intscanf("%d",for(ri=1;ri<=repeat;ri++){scanf("%d%d",&m,&n);printf("%d",i);}}}intfactorsum(int{intt,sum,j; else{}}return}8周 (读入一批正整数(以零或负数为结束标志)while语句实现循环。 1390787437056101-Thesumoftheoddnumbersis11.Thesumoftheoddnumbersis116.#include#include<stdio.h>intmain(void){intx,sum;intrepeat,ri;scanf("%d",for(ri=1;ri<=repeat;ri++){scanf("%d",&x);while(x>0)scanf("%d",&x);}printf("Thesumoftheoddnumbersis%d.\n",}} repeat(0<repeat<10repeat1x，计算并输出下式的值，直到最后一项的绝对值小于0.00001(2fact(n)npowsx＋x*x/2!＋x*x*x/3!＋x*x*x*x/4!＋…… s=s=s=#include<stdio.h>#define#include<stdio.h>#defineMY_EPS0.00001intmain(void){intintrepeat,ri;doubleitem,s,x;doublefact(intscanf("%d",for(ri=1;ri<=repeat;ri++){scanf("%lf",&x);while(item*x/fact(i)>=MY_EPS){}printf("s=%.2f\n",}}doublefact(int{intdoubleproduct;product=1;product=product*j;returnproduct;} ((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doubledouble。 10 (x1=10,200 (x2=200,#include<stdio.h>#includedoubledist(double#include<stdio.h>#includedoubledist(doublex1,doubley1,doublex2,doubley2);intmain(void){intrepeat,doubledistance,x1,y1,x2,scanf("%d",for(ri=1;ri<=repeat;ri++){scanf("%lf%lf%lf%lf",&x1,&y1,&x2,&y2);distance=dist(x1,y1,x2,y2);printf("Distance=%.2f\n",}}doubledist(doublex1,doubley1,doublex2,double{doublea,b,c;returnc;} repeat0<repeat<10)repeat次下列运算：a和n,a＋aa＋aaa＋aa…a(na) 2 (a=2,8 (a=8, #include#include<stdio.h>intmain(void){inta,i,intri,repeat;longsn;longfn(inta,intscanf("%d",for(ri=1;ri<=repeat;ri++){scanf("%ld%d",&a,&n);}}/*思路：如2222=( (2+2*10)*10)+2) *10) +2,是一个累加和*/longfn(inta,intn){intsum,j,b;{}return} repeat0<repeat<10)repeat 3111 Thedecimalis31,theoctalis37,thehexadecimalis1f. (31的十进制、八进制和31,37,1f)Thedecimalis9,theoctalis11,thehexadecimalis9. （11的十进制、八进制和9,11,9)Thedecimalis26,theoctalis32,thehexadecimalis1a.（1a的十进制、八进制26,32,