C语言操作题常考编程题库

1、C语言操作题常考编程题库C语言编程题覆盖到的算法:1、个人所得税计算；2、求一组数的正、负数（或奇、偶数）的个数 和平均值；3、二维数组主、次对角线之和；4、求素数（用函数实现）；5、级数有限项求和问题；6、两个一维数组相加（用函数实现）；7、求最大、最小值；8判断闰年（用函数实现）；9、求水仙花数；10、百钱百鸡；11、求最大公约数、最小公倍数。12、求回文数；13、排序；14、二维数组的转置;15、递归函数求阶乘；16、求斐波那契数列；17、求和：a+aa+aaa+aaaa+.;18、求符合条件的整数（如：是某个二位数的平 方，个位、十位、百位数各不相同等）；（包括素 数，回文，完数，等等

2、）19、字符串加密；20、统计字符串中字母和数字的个数；/*1.个人所得税计算，以实验指导P24T5为标准*/#in cludevoid mai n()int i,n;double a,b,c;n=i/400;prin tf(please in put in come: n);sca nf(%d,&i);a=(i-800)*0.05,b=20+(i-1200)*0.08,c=84+(i-2000)*0.2;switch( n)case 0:case 1:pri ntf(不需缴税n);break;case 2:pri ntf(需交纳 %lf 兀n,a);break;case 3:case 4:p

3、ri ntf(需交纳 %lf 兀n,b);break;default:pri ntf(需交纳 %lf元n,c);break;/*2.求一组数的奇数个数，偶数个数和平均值 #in clude#defi ne N 10void mai n()int aN;int i,m=0, n=0,sum=0;double average;prin tf(please in put 10 nu mbers:n);for(i=O;iN;i+)sca nf(%d, &ai);if(ai%2=0)m+;elsen+;sum+=ai;average=(double)sum/N;prin tf(wehave %d odd

4、s and %deve nsn, n,m);prin tf(average=%lfn,average);*/*2.求一组数的正数个数和负数个数与平均值*/#in clude#defi ne N 10void mai n()int aN;int i,m=0, n=0,sum=0;double average;prin tf(please in put 10 nu mbers: n);for(i=O;i0)m+;elsen+;sum+=ai;average=(double)sum/N;prin tf(wehave %dpositive n umbers and %dn agetive nu mbe

5、rsn,m, n);prin tf(the average =%lf,average);/*3.求二维数组的主次对角线之和*/#in clude #defi ne M 5#defi ne N 5void mai n()int aMN;int i,j,sum1=0,sum2=0;prin tf(please in put the matrix:n);for(i=0;iM;i+)for(j=0;jN;j+)sca nf(%d,&aij);for(i=0,j=0;iM;i+,j+)sum1+=aij;for(i=0,j=N-1;iM;i+,j-)sum2+=aij;printf(主对角线和为 dn,

6、sum1);printf(次对角线和为 %dn,sum2);/*4.调用函数判断素数*/#in clude#in cludeint isPrime(i nt n)int i;for(i=1;isqrt (n );i+)if(n %i=0)return 0;return 1;void mai n()int n,i;prin tf(please in put a n umber n);sea nf(%d,&n);i=isPrime( n);if(i)prin tf(%d is a primen, n);elseprin tf(%d is n ot a primen, n);5.级数有限项求和(以格

7、里高利公式求n为例)#in clude#in cludevoid mai n()int n=1;double p=O,t,pi;t=1/(pow(-1),( n+1)*2*( n-1);while(fabs(t)=1e-6)pi+=t;n+;pi=4*p;prin tf(%lf,pi);arrayarray6.两个一维数组相加/*# in clude #defi ne N 5void mai n()int aN,bN,cN;int i;prin tf(pleasein putthe firstaN:n);for(i=0;iN;i+)sca nf(%d, &ai);prin tf(please

8、in put the sec ondbN:n);for(i=0;iN;i+)scanf(%d,&bi);for(i=0;iN;i+)ci=ai+bi;prin tf(the sum is: n);for(i=0;iN;i+)prin tf(%dt,ci);*/6.1调用函数实现/#in clude#defi ne N 5 int f(i nt a,i nt b)int sum;sum=a+b;return sum;void mai n()int aN,bN,cN;int i;prin tf(pleasein put the firstaN:n);for(i=0;iN;i+)sca nf(%d,

9、 &ai);prin tf(pleasein putthe sec ondbN:n);arrayarrayfor(i=0;iN;i+)sea nf(%d,&bi);for(i=0;iN;i+)ei=f(ai,bi);prin tf(the sum is:n);for(i=0;iN;i+) printf(%dt,ci);/*7.求一组数的最大最小值*/*# in clude void mai n()int i,max,mi n;int a10;prin tf(please in put ten n umbers n); for(i=0;i10;i+)sca nf(%d, &ai);max=a0;

10、min=a0;for(i=1;imax)max=ai;if(aimi n)min=ai;prin tf(max is %d,min is %dn,max,mi n);*/*随机数*/#in clude#in cludevoid mai n()int i,max,mi n;int a10;for (i=0;i10;i+)ai=ra nd()%90+10;prin tf(%dt,ai);max=aO;min=a0;for(i=1;imax)max=ai;if(aimi n)min=ai;prin tf(max=%dtmi n=%dn,max,mi n);/*8.判断闰年*/#in clude in

11、t f(i nt n) if(n %4=0&n %100!=0| n%400=0) return 1;elsereturn 0;void mai n()int n;prin tf(please in put year: n);sca nf(%d,&n);if(f(n)printf(%d 是闰年 n,n);elseprintf(%d 不是闰年 n,n);/*可能会有要输出某一范围内所有闰年，思想一样，用循环调用函数即可*/*9.判断水仙花数(运用循环)#in cludevoid mai n()int n,a,b,c,m;prin tf(please in put a n umber: n);sc

12、a nf(%d,&n);a=n/100;b=n/10%10;c=n%10;m=a*a*a+b*b*b+c*c*c;if(m=n)printf(%d is the number we wantn,n);elseprin tf(%d is not the n umber n, n);*/(书P108为/*9.输出所有水仙花数(调用函数 循环做法)*/#in cludeint f(i nt n)int a,b,c,m;a=n/100;b=n/10%10;c=n%10;m=a*a*a+b*b*b+c*c*c;if(m=n)return 1;elsereturn 0;void mai n()inf Lr

13、rfor (耳 OOXOOOT+) 宀f(三=K(II巴)prin爲 d5m=. (関四自出 P15)二=H=inc-udeASs:iohvvoid main。宀inf LjKfor(niAH2i+)for(j=0;jv=33;j+)for(k=3;k=99;k+=3)if(i+j+k=lO0)&( 5*i+3*j+k/3=100)printf(“公鸡宀只，母鸡宀只，小鸡 %d只 n,i,j,k);/*11.求最大公约数和最小公倍数，函数实现*/#in cludeint f(int m,i nt n)int i=1,t; if(mn)t=m;m=n;n=twhile(i=m% n)!=0)m=

14、n;n=i;return n;int g(i nt m,i nt n) int i,j; i=f(m, n);j=m* n/i;return j; void mai n()int m, n,a,b;prin tf(please in put two n umbers:n);sea nf(%d %d,&m,&n);a=f(m, n);b=g(m, n);printf( 最大公约数为%dn,a);printf( 最小公倍数为%dn,b);12.1输出10-2000之间的回文数（循环）/*# in cludevoid mai n()int i,j,k=0,a;n);prin tf(10-2000 之

15、间的回文数有:for(i=10;iv=2000;i+)a=i;j=0;while(a0)j=j*10+a%10;a/=10;if(i=j)prin tf(%dt,i);k+;if(k%5=0)prin tf(n);prin tf(n);*/12.2输出10-2000之间的回文数(函数调用)/*# in cludeint f(i nt n)int i=0;while( n0)i=i*10+n%10;n/=10;return i;void mai n()int n,k;n);prin tf(10-2000之间的回文数有:for(n=10; n=2000; n+)if(n=f(n)prin tf(%

16、dt, n);k+;if(k%5=0)prin tf(n);prin tf(n);*/12.3判断回文数(循环)/*# in cludevoid mai n()int i,a,j=0;prin tf(please input a nu mber:n);scanf(%d,&i);a=i;while(a0)j=j*10+a%10;a/=10;if(j=i)printf(%d是回文数 n,i);elseprintf(%d 不是回文数 n,i);*/12.4判断回文数(函数调用)/#in cludeint f(i nt n)int i=0;while( n0)i=i*10+n%10;n/=10;ret

17、urn i;void mai n()int n;prin tf(please in pit a n umber:n);sea nf(%d,&n);if(n=f(n)printf(%d是回文数 n,n);elseprintf(%d不是回文数 n,n);13.排序/选择法/*# in clude#defi ne N 10void f(i nt a,i nt n)int i,j,k,t;for(i=0;i=n ;i+)k=i;for(j=i+1;j n;j+)if(ajak) k=j;if(k!=i)t=ak;ak=ai;ai=t;void mai n()int aN=8,2,34,5,12,78,

18、4,78,76,9;int i=0;int size=sizeof(a)/sizeof(i nt);for(i=O;iN;i+)prin tf(%4d,ai);prin tf(n);f(a,size);for(i=0;iN;i+)prin tf(%4d,ai);printf (n);/冒泡法/#in clude#defi ne N 5void f(i nt a,i nt n)int i,j;int tfor(i=0;i n-1;i+)for(j=0;j aj+1)t=aj;aj=aj+1; aj+1=t;*/插入法/#in clude#defi ne N 5void f(i nt a,i nt

19、 n)int i,j;int tfor(i=0;i0&aj-1t)aj=aj-1;-j；aj=t;void mai n()int aN=8,2,34,5,12;int i=0;int size=sizeof(a)/sizeof(i nt);for(i=0;iN;i+)prin tf(%4d,ai);prin tf(n);f(a,size);for(i=0;iN;i+)prin tf(%4d,ai);printf (n);/二维数组的转置/#in clude#defi ne N 3#defi ne M 3 void mai n()int aMN,bMN;int i,j;prin tf(pleas

20、e inp ut aMNn);for(i=0;iM;i+)for(j=0;jN;j+)sca nf(%d,&aij);for(i=0;iM;i+)for(j=0;jN;j+)bij=aji;prin tf(bMN=n);for(i=0;iM;i+)for(j=0;jN;j+)prin tf(%4d,bij); prin tf(n);prin tf(n);15.调用递归函数求阶乘II#in cludeint f(i nt n)if(n=1| n=0) return 1;elsereturn f(n-1)*n;void mai n()int n,m;prin tf(please in put a

21、n umber: n);sca nf(%d,&n);m=f( n);prin tf(%d!=%dn, n,m);/*16.运用递归函数求斐波拉契数列（前 30项）*/#in cludeint f(i nt n)if (n=1| n=2)return 1;elsereturn f(n-1)+f( n-2);void mai n()int i,k=0;for (i=1;iv=30;i+)prin tf(%dt,f(i);k+;if(k%5=0)prin tf(n);prin tf(n);17.求 a+aa+aaa+aaaa/*# in clude#in clude#defi ne N 5void mai n()int n ,i,j,k,a,sum1=0;prin tf(please in put a n umber: n);sca nf(%d,&n);for(i=1;i=N;i+)k=fabs(10,i-1);j=n+n *(i-1)*k;sum1+=j;prin tf(sum=%dn,sum1);*/#in clude#defi ne N 5int k(i nt a,i nt sum)sum=sum*10