程序设计基础试卷第一学期最终答案答案

《程序设计基础试卷第一学期最终答案答案》由会员分享，可在线阅读，更多相关《程序设计基础试卷第一学期最终答案答案（23页珍藏版）》请在装配图网上搜索。

1、1.#includeint main()int x,y;printf(x=);scanf(%d,&x);if(x100|x0)y=-1;else if(x=0)y=0;else y=1;printf(y=%dn,y);return 0;2.#includeint main()int a,b,c,min;scanf(%d%d%d,&a,&b,&c);min=a;if(minb)min=b;if(minc)min=c;printf(min=%dn,min);return 0;3.#include#includeint main()double a,b,c,s,p;printf(输入三角形三边a,b

2、,c:);scanf(%lf,%lf,%lf,&a,&b,&c);if(c(a+b)&a(b+c)&b(a+c)p=(a+b+c)/2; s=sqrt(p*(p-a)*(p-b)*(p-c);printf(三角形面积:%5.2lfn,s);else printf(三边不能构成三角形n);return 0;4.#includeint main()int a,b,c,d,t;printf(a,b,c,d:);scanf(%d,%d,%d,%d,&a,&b,&c,&d);if(ab)t=a;a=b;b=t;if(ac)t=a;a=c;c=t;if(ad)t=a;a=d;d=t;if(bc)t=b;b

3、=c;c=t;if(bd)t=b;b=d;d=t;if(cd)t=c;c=d;d=t;printf(max=%dnmin=%dn,d,a);return 0;5.#includeint main()char c;printf(输入成绩等级:);scanf(%c,&c);if(c68|c65)printf(enter data errorn);elseif(c=A)printf(分数段:80-100n);else if(c=B)printf(分数段:70-79n);else if(c=C)printf(分数段:60-69n);else printf(分数段:60以下n);return 0;6.#

4、includeint main()int x,y;printf(x=);scanf(%d,&x);if(x1)y=x;printf(y=x,y=%dn,y);else if(x=10)y=3*x-11;printf(y=3*x-11,y=%dn,y);else y=2*x-1;printf(y=2*x-1,y=%dn,y);return 0; 7.#includeint main()int n;printf(输入成绩:);scanf(%d,&n);if(n100)printf(errorn);else if(n=90)printf(成绩等级: An); else if(n=80)printf(

5、成绩等级: Bn);else if(n=70)printf(成绩等级: Cn);else if(n=60)printf(成绩等级: Dn);else if(n0)printf(成绩等级: En);else printf(errorn);return 0;8.#include#includeint main()double a,b,c,disc,x1,x2,r,i;scanf(%lf,%lf,%lf,&a,&b,&c);printf(The equation );if(fabs(a)=1e-6)printf(is not a quadraticn);elsedisc=b*b-4*a*c;if(f

6、abs(disc)1e-6)x1=(-b+sqrt(disc)/(2*a);x2=(-b-sqrt(disc)/(2*a);printf(has distinct real roots:58.2fn,x1,x2);elser=-b/(2*a);i=sqrt(-disc)/(2*a);printf( has complex roots:n);printf(%8.4f+%8.4fin,r,i);printf(%8.4f-%8.4fin,r,i);return 0;9.#includeint main()int n;scanf(%d,&n);if(n%4=0&n%100!=0)|(n%400=0)p

7、rintf(%d is a leap yearn,n);elseprintf(%d is not a leap yearn,n);return 0;*10.#include#includeint main()int a,i,j=0,k;for(a=3;a=1000;a=a+2)k=sqrt(a);for(i=2;i=k+1)printf(%d ,a);j+; if(j%5=0) printf(n); printf(n);return 0;11.#includeint main()int i,j=1;for(i=5,j;i=100;i+,j+)if(i%5=0|i%7=0)&(i%5!=0|i%

8、7!=0)printf(%d ,i);if(j%5=0)printf(n);j+;return 0;12.#include#includeint main()int i,k,l;for(i=0;i=2000;i+)k=sqrt(i+100); l=sqrt(k+168);if(k*k=i+100)&(l*l=k+168)printf(%d ,i);printf(n);return 0;13.#includeint main()int f20=1,1,i,j;for(i=2;i20;i+)fi=fi-1+fi-2;for(i=0,j=1;i20;i+)printf(%d ,fi);if(j%5=

9、0) printf(n);j+;printf(n);return 0;14.#includeint main()double s=0,t=1;int n;for(n=1;n=20;n+)t=t*n;s=s+t;printf(1!+2!+.+20!=%22.15en,s);return 0;15.#includeint main()int i,n=30;double a=1,b=2,s=0,t;for(i=1;i=n;i+)s=s+a/b;t=b;b=a+b;a=b;printf(sun=%16.10fn,s);return 0;16.#includeint main()int i,sum=1,

10、s=1;for(i=1;i=10;i+)s=3*s;sum=sum+s;printf(1+3+.3,20=%dn,sum);return 0;17.#include#includeint main()double x1,x0,f,f1; x1=1.5;dox0=x1;f=(2*x0-4)*x0+3)*x0-6;f1=(6*x0-8)*x0+3;x1=x0-f/f1;while(fabs(x1-x0)=1e-5);printf(The root of equation is %5.2fn,x1);return 0;18.#includeint main()int i,a,b,c;for(i=10

11、0;i500;i+)a=i/100;b=i/10-a*10;c=i%10;if(i=a*a*a+b*b*b+c*c*c)printf(%d ,i);printf(n);return 0;19.#includeint main()int i;printf(输入一个8进制数:);scanf(%o,&i);printf(转为10进制为:%dn,i);return 0;20.#includeint main()int p,r,n,m,t;printf(请输入两个正整数:);scanf(%d,%d,&n,&m);if(nm)t=n;n=m;m=t;p=n*m;while(m!=0)r=n%m;n=m;m

12、=r;printf(最大公约数为:%dn,n);printf(最小公被数为:%dn,p/n);return 0;21.#includeint main()double sn=100,hn=50,i;for(i=2;i=10;i+)sn=sn+2*hn;hn=hn/2;printf(第10次落地共经过%lf米n,sn);printf(第10次落地反弹%lf米n,hn);return 0;22.#includeint main()int i,j;for(i=1;i=9;i+)for(j=i;j=9;j+)printf(%d*%d=%dt,i,j,i*j);printf(n);23.#include

13、int main()int i,j,k;for(i=0;i=3;i+)for(j=0;j=2-i;j+)printf( );for(k=0;k=2*i;k+)printf(*);printf(n);for(i=0;i=2;i+)for(j=0;j=i;j+)printf( );for(k=0;k=4-2*i;k+)printf(*);printf(n);return 0;24.#includeint main()int m,s,i;for(m=2;m500;m+)s=0;for(i=1;im;i+)if(m%i)=0)s=s+i;if(s=m)printf(%d,its factors are

14、 ,m);for(i=1;im;i+)if(m%i=0)printf(%d,i);printf(n);return 0;25.#includeint main()int a5,i,t;printf(enter array a:n);for(i=0;i=0;i-)printf(%d ,ai);printf(n);return 0;26.#includeint main()int a,b,c,A,B,C,n=0;for(a=1;a=4;a+) A=a*100;for(b=1;b=4;b+)if(b=a) continue;B=b*10; for(c=1;c=4;c+)if(c=b|c=a)cont

15、inue; printf(%d ,A+B+c);n+;printf(n共有%d个数n,n);printf(n);return 0;27.#includeint main()int a10,i,j,t;for(i=0;i10;i+)scanf(%d,&ai);printf(n);for(i=0;i9;i+)for(j=0;jaj+1)t=aj;aj=aj+1;aj+1=t;for(i=0;i10;i+)printf(%d ,ai);printf(n);return 0;28.#includeint main()int a10,i,t,min,j;for(i=0;i10;i+)scanf(%d,&

16、ai);printf(n);for(i=0;i9;i+)min=i;for(j=i+1;jai)min=j;t=ai;ai=ai+1;ai+1=t;for(i=9;i=0;i-)printf(%d ,ai);printf(n);return 0;29.#includeint main()int a;scanf(%d,&a);if(a100000|a=0&a=10&a=100&a=1000&a10000)printf(%d是4位数n,a);printf(%d %d %d %dn,a/1000,a/100-a/1000*10,a/10-a/100*10,a%100-10*(a/10-a/100*1

17、0);printf(%d%d%d%dn,a%100-10*(a/10-a/100*10),a/10-a/100*10,a/100-a/1000*10,a/1000);elseprintf(%d是5位数n,a);printf(%d %d %d %d %dn,a/10000,a/1000-a/10000*10,a/100-a/1000*10,a/10-10*(a/100),a%10);printf(%d%d%d%d%dn,a%10,a/10-10*(a/100),a/100-a/1000*10,a/1000-a/10000*10,a/10000);return 0;30.#include#incl

18、udeint main()int i,j,k;for(i=3;i=100;i=i+2)k=sqrt(i);for(j=2;j=k+1)printf(%d ,i);printf(n);return 0;31. #include int main() int a11=1,4,6,9,13,15,17,36,68,100;int temp1,temp2,num,end,i,j;printf(array a:n);for(i=0;iend)a10=num;elsefor(i=0;inum)temp1=ai;ai=num;for(j=i+1;j11;j+)temp2=aj;aj=temp1;temp1=

19、temp2;break;printf(Now array:n);for(i=0;i11;i+)printf(%5d,ai);printf(n);return 0; array a: 1 4 6 9 13 15 17 36 68 100insert data:54Now array: 1 4 6 9 13 15 17 36 54 68 10032.#includeint main() int a23=1,2,3,4,5,6;int b32,i,j;printf(array a:n);for(i=0;i=1;i+)for(j=0;j=2;j+)printf(%5d,aij);bji=aij;pri

20、ntf(n);printf(array b:n);for(i=0;i=2;i+)for(j=0;j=1;j+)printf(%5d,bij);printf(n);return 0;33.#includeint main() int i,j,row=0,colum=0,max;int a34=1,2,4,5,23,54,6,7,0,3,5,6;max=a00;for(i=0;i=2;i+)for(j=0;jmax)max=aij;row=i;colum=j;printf(max=%dnrow=%dncolum=%dn,max,row,colum);return 0;34.include#defi

21、ne N 4 #define M 5 /数组为4行5列int main()int i,j,k,aNM,max,maxj,flag;printf(please input matrix:n);for(i=0;iN;i+) /输入数组for(j=0;jM;j+)scanf(%d,&aij);for(i=0;iN;i+)max=ai0; /开始时假设ai0最大maxj=0; /将列号0赋给maxj保存for(j=0;jmax)max=aij; /将本行的最大数存放在max中maxj=j; /将最大数所在的列号存放在maxj中flag=1; /先假设使鞍点，以flag为1代表for(k=0;kakma

22、xj) /将最大数和其同列元素比flag=0; /如果max不是，表示不是鞍点，令flat等0continue;if(flag) /如果flag为1表示是鞍点printf(a%d%d=%dn,i,maxj,max); /输出鞍点值和所在行列号break;if(!flag) /如果flag为0并表示鞍点不存在printf(It is not exist!n);return 0;35. #include #include int main() char str320; /定义二维字符数组char string20; /定义以为字符数组，作为交换字符串时的临时字符数组int i;for(i=0;i0

23、) /若str0str1strcpy(string,str0); /把str0的字符串赋给字符数组stringelsestrcpy(string,str1); /否则把str1的字符串赋给stringif(strcmp(str2,string)0) /若str2大于stringstrcpy(string,str2); /把str2的字符串赋给字符数组stringprintf(nThe largest string is:n%sn,string);return 0; Holland China America The largest string is : Holland36.#includei

24、nt main() void fun(char str,int i,int n);int i,n;char str100;printf(请输入一串字符:);gets(str);printf(i,n:); scanf(%d,%d,&i,&n);fun(str,i,n);puts(str);printf(n);return 0;void fun(char str,int i,int n) int j;for(n;n0;n-)for(j=i;j=100;j+) /注意要引用另一个变量做循环条件，并使每次循环初始值为istrj-1=strj;请输入一串字符:zzxkzjxkjzxci,n:3,5zzk

25、jzxcPress any key to continue37.#includeint main()void concatenate(char string1,char string2,char string);char s1100,s2100,s100;printf(input string1:);scanf(%s,s1);printf(input string2:);scanf(%s,s2);concatenate(s1,s2,s);printf(nThe new string is %sn,s);return 0;void concatenate(char string1,char st

26、ring2,char string)int i,j;for(i=0;string1i!=0;i+)stringi=string1i;for(j=0;string2j!=0;j+)stringi+j=string2j;stringi+j=0;input string1:countryinput string2:sideThe new string is countrysidePress any key to continue38. #include int main() char string100;int i,num=0,word=0;char c;gets(string); /输入一个字符串

27、给字符组string for(i=0;(c=stringi)!=0;i+) /只要字符不是0就继续执行循环if(c= ) word=0; /如果是空格符，使word置0else if(word=0) /如果不his空格符且word原值为0word=1; /使word置1 num+; /sum累加1，表示增加一个单词printf(There are %d words in this line.n,num);/输出单词数return 0; 39.#includeint main()char c;int letters=0,space=0,digit=0,other=0;printf(请输入一行字符

28、:n); while(c=getchar()!=n)if(c=a&c=A&c=0&c=9)digit+;else other+;printf(字母数:%dn空格数:%dn数字数：%dn其他字符数：%dn,letters,space,digit,other);return 0;40.#includeint main()int i,j,upp,low,dig,spa,oth;char text440;upp=low=dig=spa=oth=0;for(i=0;i4;i+)printf(please input line %d:n,i+1);gets(texti);for(j=0;j=A&texti

29、j=a&textij=0&textij=9)dig+;else if(textij= )spa+;else oth+;printf(nupper case :%dn,upp);printf(low case :%dn,low);printf(digit :%dn,dig);printf(space :%dn,spa);printf(other :%dn,oth);return 0;41.#includeint main()int j,n;char ch90,tran90;printf(input cipher code:);gets(ch);printf(ncipher sode :%s,ch

30、);j=0;while(chj!=0)if(chj=A)&(chj=a&(chj=z)tranj=219-chj;elsetranj=chj;j+;n=j;printf(noriginal text:);for(j=0;jn;j+)putchar(tranj);printf(n);return 0;42.#include #include math.h main() int a,b,c,d; scanf(%d,&a); for(b=3;b=a/2;b+=2) for(c=2;csqrt(b) d=a-b; else break; for(c=2;csqrt(d) printf(%d=%d+%d

31、n,a,b,d); 43.#includeint main() int ge,shi,qian,wan,x;scanf(%d,&x);wan=x/10000;qian=x%10000/1000;shi=x%100/10;ge=x%10;if(ge=wan&shi=qian) /*个位等于万位并且十位等于千位*/printf(%d是回文数n,x);elseprintf(%d不是回文数n,x);return 0;1232112321是回文数Press any key to continue44.#includeint main() int n,i,ind,M=3; printf(请输入总人数:);

32、 scanf(%d,&n); ind=1; for(i=2;i=n;i+) ind=(ind+M)%i; if(ind=0) ind=i; printf(最后留下是原来第%d号n,ind);解析。事实上，约瑟夫问题存在着时间复杂度为O(n)的解法。要解决这个问题，要用到同余这个数学工具。下面，假设目前还剩下K个人，这K个人从1到M报数，那么，当第M个人被杀后，剩下的人将按照怎样的规则报数呢？如果我们将第M个人被杀后，第P个人报的数计做Q，那么可以得到下面的同余式：（P - M % K) % K = Q将上式变形可以得出：P = （ Q + M）% K那么可以知道，在第（N - K + 2）轮报

33、数为Q的人，在第（N - K + 1）轮的报数为 （ Q + M）% K.又易知，最后剩下的人，在最后一轮的报数必然为1，那么可以利用同余式倒推其原始编号：（1 + M）% 2 + M） % 3）+ M % 4。程序源代码： #define nmax 50 main() int i,k,m,n,numnmax,*p; printf(please input the total of numbers:); scanf(%d,&n); p=num; for(i=0;i *(p+i)=i+1; i=0; k=0; m=0; while(m if(*(p+i)!=0) k+; if(k=3) *(p+i)=0; k=0; m+; i+; if(i=n) i=0; while(*p=0) p+; printf(%d is leftn,*p); 45.#includeint main() double i=0.01;int n=0;while(i=8844)i=i*2;n+;printf(对折%d次，可超得过珠穆朗玛峰高度n,n);return 0;