語言期末考試編程題

1、1.打印出所有的“水仙花數(shù)”，并按照一行5個(gè)的格式輸出。所謂“水仙花數(shù)”是指一個(gè)三位數(shù)，其各位數(shù)字立方和等于該數(shù)本身。例如：153是一個(gè)“水仙花數(shù)”，因?yàn)?53=1*1*15*5*53*3*3。#include<stdio.h>void main() int a,b,c,s,i=0; for(s=100;s<1000;s+) c=s%10; b=s/10%10; a=s/100; if(s=a*a*a+b*b*b+c*c*c) printf("%dt",s); +i; if(i%5=0) printf("n"); 2.求1+2!+3!+

2、.+20!的和#include<stdio.h>void main() double fact=1.0,sum=0.0; int i,j; for(i=1;i<=20;i+) for(j=1;j<=i;j+) fact*=j; sum+=fact; fact=1; printf("%.0fn",sum);3.輸入一個(gè)5位數(shù)，判斷它是不是回文數(shù)。即12321是回文數(shù)，個(gè)位與萬位相同，十位與千位相同。#include<stdio.h>#include<stdlib.h>void main() long num; int a,b,c

3、,d; printf("Please input the number:"); scanf("%d",&num); if(num<=9999|num>=100000) printf("%The number is error!n"); exit(0); a=num/10000; b=num/1000%10; c=num/10%10; d=num%10; if(a=d&&b=c) printf("nYes,the %ld is palindrome!n",num); else pr

4、intf("nNo,the %ld is not palindrome!n",num);4. 求出1到1000之內(nèi)能被7或11整除、但不能同時(shí)被7和11整除的所有整數(shù)并并按照一行5個(gè)的格式輸出。#include<stdio.h>void main() int num,i=0; for(num=1;num<=1000;num+) if(num%7|num%11)=1)&&(num%7&&num%11)=0) printf("%dt",num); +i; if(i%5=0) printf("n&qu

5、ot;); 5.編程列出200以內(nèi)所有的素?cái)?shù)，按照一行10個(gè)的格式輸出，并求出所有素?cái)?shù)的平均值，#include<stdio.h>#include<math.h>void main() int i,j,k=0,sum=0,avg=0; for(i=2;i<200;i+) for(j=2;j<=sqrt(i);j+) if(i%j=0) break; if(j>sqrt(i) printf("%dt",i); +k; sum+=i; if(k%10=0) printf("n"); avg=sum/k; printf

6、("nThe average is %dn",avg);6、編寫一個(gè)函數(shù)，輸入n為偶數(shù)時(shí)，調(diào)用函數(shù)求1/2+1/4+.+1/n,當(dāng)輸入n為奇數(shù)時(shí)，調(diào)用函數(shù)1+1/3+1/5.+1/n#include<stdio.h>void main() void odd(int n); void even(int n); int n; printf("Please input the number:"); scanf("%d",&n); if(n%2) odd(n); else even(n);void odd(int n) f

7、loat sum=0.0; float j; int i; for(i=1;i<=n;i+=2) j=1.0/i; sum+=j; printf("%fn",sum);void even(int n) float sum=0.0; int i; float j; if(n=0) printf("%fn",sum); else for(i=2;i<=n;i+=2) j=1.0/i; sum+=j; printf("%fn",sum); 7、 已知abc+cba=1333，其中a,b,c均為一位數(shù)，編程求出滿足條件的a,b,c

8、所有組合。#include<stdio.h>void main() int a,b,c; for(a=1;a<=9;a+) for(b=0;b<=9;b+) for(c=1;c<=9;c+) if(100*a+10*b+c+100*c+10*b+a)=1333) printf("a=%d,b=%d,c=%dn",a,b,c);8、用戶輸入12個(gè)0100之間的整數(shù)，統(tǒng)計(jì)出小于60，60到79，80到100三個(gè)范圍的整數(shù)各有多少個(gè)？#include<stdio.h>#include<stdlib.h>void main()

9、int i,j,k,n,num; i=j=k=0; printf("Please input the 12 numbers with 0100:"); for(n=1;n<=12;n+) scanf("%d",&num); if(!(num>=0&&num<=100) exit(0); if(num<60) +i; else if(num<80) +j; else +k; printf("nThere are %d numbers less than 60,%d numbers betwee

10、n 60 and 79,%d numbers between 80 and 100.n",i,j,k);9、求這樣一個(gè)三位數(shù)并輸出該數(shù)字，該三位數(shù)等于其每位數(shù)字的階乘之和。即：abc = a! + b! + c!#include<stdio.h>void main() int a,b,c,num; int s(int); for(num=100;num<1000;num+) a=num/100; b=num/10%10; c=num%10; if(num=s(a)+s(b)+s(c) printf("The number is %dn",num)

11、; break; s(int n) int i,sum=1; for(i=1;i<=n;i+) sum*=i; return sum;10、猜數(shù)游戲：由用戶隨機(jī)輸入一個(gè)1位正整數(shù)數(shù)讓人來猜，只能猜5次，如果人猜對了，則在屏幕上顯示“You are so clever”，否則計(jì)算機(jī)給出提示，告訴人所猜的數(shù)是太大還是太小，直到人猜對為止或者5次都猜不對給出提示“Game Over”。#include<stdio.h>#include<stdlib.h>void main() int num,i,n; printf("Please input the numb

12、er with 09 for guessing:"); scanf("%d",&num); system("cls"); if(num<0|num>9) exit(0); for(i=1;i<=5;i+) printf("nPlease input the number you guess:"); scanf("%d",&n); if(num=n) printf("nYou are so clever!n"); break; else if(n<

13、num) printf("nThe number you input is little,go on please!You have %d times left!n",5-i); continue; else printf("nThe number you input is large,go on please!You have %d times left!n",5-i); continue; if(i>5) printf("Game Over!n");11、設(shè)N是一個(gè)四位數(shù)，它的9倍恰好是其反序數(shù)（例如：123的反序數(shù)是321

14、），求N的值。#include<stdio.h>void main() int n,k,a,b,c,d; for(n=1000;n<=9999;n+) a=n/1000; b=n/100%10; c=n/10%10; d=n%10; k=1000*d+100*c+10*b+a; if(k=9*n) printf("N=%dn",n); 12、100匹馬馱100擔(dān)貨，大馬一匹馱擔(dān)，中馬一匹馱擔(dān)，小馬兩匹馱擔(dān)。試編寫程序計(jì)算大、中、小馬的數(shù)目。#include<stdio.h>void main() int h1,h2,h3; for(h1=0;h

15、1<=100;h1+) for(h2=0;h2<=100;h2+) for(h3=0;h3<=100;h3+=2) if(h1+h2+h3=100&&3*h1+2*h2+1/2*h3=100) printf("The old horse has %d,middle horse has %d,young horse has %dn",h1,h2,h3); 13、一位司機(jī)酒駕撞人逃跑?，F(xiàn)場三人目擊事件，但都沒記住車號，只記下車號的一些特征。甲說：牌照的前兩位數(shù)字是相同的；乙說：牌照的后兩位數(shù)字是相同的；丙是位數(shù)學(xué)家，他說：四位的車號剛好是一個(gè)整

16、數(shù)的平方。請根據(jù)以上線索求出車號。（車號為4位數(shù)）#include<stdio.h>#include<math.h>void main() int n,a,b,c,d,i; for(n=1000;n<=9999;n+) a=n/1000; b=n/100%10; c=n/10%10; d=n%10; i=sqrt(n); if(a=b&&c=d&&i*i=n) printf("%dn",n); 14、求S=1/(1*2)+1/(2*3)+1/(3*4)+前50項(xiàng)之和并輸出結(jié)果。#include<stdio.

17、h>void main() float i; float s=0; for(i=1;i<51;i+) s+=1.0/(i*(i+1); printf("%f",s);15、編程求出所有1000到3000之間能被7、11、17同時(shí)整除的整數(shù)，并求其平均值,并輸出結(jié)果（結(jié)果保留兩位小數(shù)）。#include<stdio.h>void main() int i,k=0,sum=0; for(i=1000;i<=3000;i+) if(i%7=0&&i%11=0&&i%17=0) sum+=i; +k; printf(&q

18、uot;%dt",i); printf("n%d",sum/k);16、編程找出滿足下列條件的所有四位數(shù)的和并輸出：該數(shù)第一、三位數(shù)字之和為10，第二、四位數(shù)字之積為12。#include <stdio.h>#include <conio.h>int main(void) int i,j,x,y,z,m; for(i = 1000; i< 10000; i+) x = i/1000; y = (i/100)%10; z = (i/10)%10; m = i%10; if(x + z)=10&&(y+ m)=12) pr

19、intf("%dt",i); printf("n"); return 0;17、求并輸出所有滿足如下條件的三位正整數(shù)：它是某整數(shù)的平方，它的三位數(shù)碼有兩位是相同的。（如100是10的平方，它有兩個(gè)0，225是15的平方，它有兩個(gè)2）。#include<stdio.h>#include<math.h>void main() int n,i,a,b,c; for(n=100;n<=999;n+) a=n/100; b=n/10%10; c=n%10; i=sqrt(n); if(a=b|a=c|b=c)&&n=i*i) printf("%dt",n); 18、輸出所有大于1010的4位偶數(shù)，且該類偶數(shù)的各位數(shù)字兩兩不相同。#include<stdio.h>void main() int num,a,b,c