C语言操作题常考编程题库.docx

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)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*/#includevoid main()int i,n;double a,b,c;n=i/400;printf(please input income:n);scanf(%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:printf(不需缴税n);break;case 2:printf(需交纳%lf元n,a);break;case 3:case 4:printf(需交纳%lf元n,b);break;default:printf(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#include#define N 10void main()int aN;int i,m=0,n=0,sum=0;double average;printf(please input 10 numbers:n);for(i=0;iN;i+)scanf(%d,&ai);if(ai%2=0)m+;elsen+;sum+=ai;average=(double)sum/N;printf(we have %d odds and %d evensn,n,m);printf(average=%lfn,average);*/*2.求一组数的正数个数和负数个数与平均值*/#include#define N 10void main()int aN;int i,m=0,n=0,sum=0;double average;printf(please input 10 numbers:n);for(i=0;i0)m+;elsen+;sum+=ai;average=(double)sum/N;printf(we have %d positive numbers and %d nagetive numbersn,m,n);printf(the average =%lf,average);/*3.求二维数组的主次对角线之和*/#include#define M 5 #define N 5void main()int aMN;int i,j,sum1=0,sum2=0;printf(please input the matrix:n);for(i=0;iM;i+)for(j=0;jN;j+)scanf(%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,sum1);printf(次对角线和为%dn,sum2);/*4.调用函数判断素数*/#include#includeint isPrime(int n)int i;for(i=1;isqrt(n);i+)if(n%i=0)return 0;return 1;void main()int n,i;printf(please input a numbern);scanf(%d,&n);i=isPrime(n);if(i)printf(%d is a primen,n);elseprintf(%d is not a primen,n);/5.级数有限项求和(以格里高利公式求为例)/#include#includevoid main()int n=1;double p=0,t,pi;t=1/(pow(-1),(n+1)*2*(n-1);while(fabs(t)=1e-6)pi+=t;n+;pi=4*p;printf(%lf,pi);/6.两个一维数组相加/*#include#define N 5void main()int aN,bN,cN;int i;printf(please input the first array aN:n);for(i=0;iN;i+)scanf(%d,&ai);printf(please input the second array bN:n);for(i=0;iN;i+)scanf(%d,&bi);for(i=0;iN;i+)ci=ai+bi;printf(the sum is:n);for(i=0;iN;i+)printf(%dt,ci);*/6.1调用函数实现/#include#define N 5int f(int a,int b)int sum;sum=a+b;return sum;void main()int aN,bN,cN;int i;printf(please input the first array aN:n);for(i=0;iN;i+)scanf(%d,&ai);printf(please input the second array bN:n);for(i=0;iN;i+)scanf(%d,&bi);for(i=0;iN;i+)ci=f(ai,bi);printf(the sum is:n);for(i=0;iN;i+)printf(%dt,ci);/*7.求一组数的最大最小值*/*#includevoid main()int i,max,min;int a10;printf(please input ten numbersn);for(i=0;i10;i+)scanf(%d,&ai);max=a0;min=a0;for(i=1;imax)max=ai;if(aimin)min=ai;printf(max is %d,min is %dn,max,min);*/*随机数*/#include#includevoid main()int i,max,min;int a10;for (i=0;i10;i+)ai=rand()%90+10;printf(%dt,ai);max=a0;min=a0;for(i=1;imax)max=ai;if(aimin)min=ai;printf(max=%dtmin=%dn,max,min);/*8.判断闰年*/#includeint f(int n)if(n%4=0&n%100!=0|n%400=0)return 1;elsereturn 0;void main()int n;printf(please input year:n);scanf(%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）#includevoid main()int n,a,b,c,m;printf(please input a number:n);scanf(%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);elseprintf(%d is not the numbern,n);*/*9.输出所有水仙花数（调用函数(书P108为循环做法)）*/#includeint f(int 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 main()int i,t;for(i=100;i1000;i+)t=f(i);if(t=1)printf(%dt,i);/10.百钱买百鸡（题目见书P115）/#includevoid main()int i,j,k;for(i=0;i=20;i+)for(j=0;j=33;j+)for(k=3;k=99;k+=3)if(i+j+k=100)&(5*i+3*j+k/3=100)printf(公鸡%d只，母鸡%d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*/#includeint f(int m,int n)int i=1,t;if(mn)t=m;m=n;n=t;while(i=m%n)!=0)m=n;n=i;return n;int g(int m,int n)int i,j;i=f(m,n);j=m*n/i;return j;void main()int m,n,a,b;printf(please input two numbers:n);scanf(%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）/*#includevoid main()int i,j,k=0,a;printf(10-2000之间的回文数有：n);for(i=10;i0)j=j*10+a%10;a/=10;if(i=j)printf(%dt,i);k+;if(k%5=0)printf(n);printf(n);*/12.2输出10-2000之间的回文数（函数调用）/*#includeint f(int n)int i=0;while(n0)i=i*10+n%10;n/=10;return i;void main()int n,k;printf(10-2000之间的回文数有：n);for(n=10;n=2000;n+)if(n=f(n)printf(%dt,n);k+;if(k%5=0)printf(n);printf(n);*/12.3判断回文数（循环）/*#includevoid main()int i,a,j=0;printf(please input a number: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）/#includeint f(int n)int i=0;while(n0)i=i*10+n%10;n/=10;return i;void main()int n;printf(please inpit a number:n);scanf(%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/*#include#define N 10void f(int a,int n)int i,j,k,t;for(i=0;i=n;i+)k=i;for(j=i+1;jn;j+)if(ajak)k=j;if(k!=i)t=ak;ak=ai;ai=t;void main()int aN=8,2,34,5,12,78,4,78,76,9;int i=0;int size=sizeof(a)/sizeof(int);for(i=0;iN;i+)printf(%4d,ai);printf(n);f(a,size);for(i=0;iN;i+)printf(%4d,ai);printf (n);/冒泡法/#include#define N 5void f(int a,int n)int i,j;int t;for(i=0;in-1;i+)for(j=0;jaj+1)t=aj;aj=aj+1;aj+1=t;*/插入法/#include#define N 5void f(int a,int n)int i,j;int t;for(i=0;i0&aj-1t)aj=aj-1;-j;aj=t;void main()int aN=8,2,34,5,12;int i=0;int size=sizeof(a)/sizeof(int);for(i=0;iN;i+)printf(%4d,ai);printf(n);f(a,size);for(i=0;iN;i+)printf(%4d,ai);printf (n);/二维数组的转置/#include#define N 3#define M 3void main()int aMN,bMN;int i,j;printf(please input aMNn);for(i=0;iM;i+)for(j=0;jN;j+)scanf(%d,&aij);for(i=0;iM;i+)for(j=0;jN;j+)bij=aji;printf(bMN=n);for(i=0;iM;i+)for(j=0;jN;j+)printf(%4d,bij);printf(n);printf(n);/15.调用递归函数求阶乘/#includeint f(int n)if(n=1|n=0)return 1;elsereturn f(n-1)*n;void main()int n,m;printf(please input a number:n);scanf(%d,&n);m=f(n);printf(%d!=%dn,n,m);/*16.运用递归函数求斐波拉契数列（前30项）*/#includeint f(int n)if (n=1|n=2)return 1;elsereturn f(n-1)+f(n-2);void main()int i,k=0;for (i=1;i=30;i+)printf(%dt,f(i);k+;if(k%5=0)printf(n);printf(n);/17.求a+aa+aaa+aaaa./*#include#include#define N 5void main()int n,i,j,k,a,sum1=0;printf(please input a number:n);scanf(%d,&n);for(i=1;i=N;i+)k=fabs(10,i-1);j=n+n*(i-1)*k;sum1+=j;printf(sum=%dn,sum1);*/#include#define N 5int k(int a,int sum)sum=sum*10+a;retu