C语言补充材料3.doc

08物理C语言补充材料3几种重要的算法与编程1. 通过键盘输入一串字符，统计数字、大写字母、小写字母、空格及其他字符出现的次数，当输入回车键时结束。#include void main( )char c; int n1,n2,n3,n4; n1=n2=n3=n4=0;while(c=getchar( )!=n)if(c=0&c=A&c=a&c=z) n3+;else if(c= = ) n4+;else n5+;printf(“ num=%d, up-case=%d, lo-case=%d, space=%d, others=%dn”, n1,n2,n3,n4,n5);2. 编写一个加法器程序，即从键盘输入N个数，求它们的和与平均值。#include void main ( )int count, n, x; float total=0, average;printf(“enter n:”);scanf(“%d”, &n);printf(“n=%dn”,n);for(count=1;count=n;count+)printf(“enter x%d:” count);scanf(“%d”,&x);total=total+x;average=total/n;printf(“total=%6.1f,average=%6.1fn”,total,average);3.计算1！2！3！ n!#include void main( )int x, n, t, s;printf(“ inter n:”);sacnf(“%d”,&n);t=1; s=0;for(x=1; x=n; x+) t=t*x; s=s+t;printf(“%dn”, s);4.求两个数的最大公约数和最小公倍数解题方法：最大公约数用大整数除小数，得到余数再用小整数除余数1，得到余数2；再用余数1整除余数2，直到余数为0最小公倍数两数相乘再除以最大公约数。main( ) int a,b,c,n1,n2; printf(“input two number:”); scanf(“%d,%d”,&n1,&n2);if(n1n2)c=n1;n1=n2;n2=c;a=n1; b=n2;while(b!=0)c=a%b; a=b; b=c;printf(“最大公约数为 %dn”,a);printf(“最小公倍数为%dn”,n1*n2/a);5. 将一个数按逆序输出。#include void main( )unsigned int number; printf(“input the number:”);scanf(“%d”. &number);do printf(“%d”, number%10); number/=10; while(number!=0);printf(“n”);6. 给出一个099999的正整数，求：(1)它是几位数；（2）分别输出每一位数字；（3）按逆序输出原数字。#include void main ( )long num;int n1, n2, n3, n4,n5;n5=num/10000;n4=num/1000%10;n3=num/100%10;n2=num/10%10;n1=num%10;if(n5!=0)printf(“place=5n”); printf(“%d,%d,%d,%d,%dn”,n5,n4,n3,n2,n1);printf(“%d,%d,%d,%d,%dn”,n1,n2,n3,n4,n5);else if (n4!=0)printf(“place=4n”); printf(“%d,%d,%d,%dn”,n4,n3,n2,n1);printf(“%d,%d,%d,%dn”,n1,n2,n3,n4,);else if(n3!=0)printf(“place=3n”); printf(“%d,%d,%dn”,n3,n2,n1);printf(“%d,%d,%dn”,n1,n2,n3);else if(n2!=0)printf(“place=2n”); printf(“%d,%dn”,n2,n1);printf(“%d,%dn”,n1,n2);else printf(“place=1n”); printf(“%dn”,n1); printf(“%dn”,n1); printf(“%d”,x7);7. 由键盘输入10个数，编写程序使其按由小到大排序输出。解法一：（选择法）#include void main( )int a, b, c, t, x10;for(a=0; a10; a+) scanf(“%d”, &xa);for(a=0; a10; a+) printf(“%4d”, xa);printf(“n”);for(a=0;a9; a+) c=a; for(b=c+1; b10; b+) if(xbxc) c=b; t=xa; xa=xc; xc=t; for( a=0;a10;a+) printf(“%4d”, xa);printf(“n”);解法二：（起泡法）#include void main( )int a, b, c, t, x10; for(a=0; a10; a+) scanf(“%d”, &xa); for(a=0; a10; a+) prinf(“%4d”, xa); printf(“n”); for(b=0; b9; b+) for(a=b+1; a=9; a+) if(xaxb) t=xa; xa=xb; xb=t; for( a=0;a10;a+) printf(“%4d”, xa);printf(“n”);8. 已知一维数组中存放互不相同的10个整数，从键盘输入一个整数，指出与该值相同的元素的下标。main( ) int a, t, x10=3, 5, 7, 9, 1, 2, 4, 6, 8 ,0; for(a=0; a10; a+) printf(“ %4d”, xa);printf(“n intput t:n”);scanf(“%d”, &t);for(a=0; a10; a+) if(t= =xa) break;printf(“ x%dn”, a);运行时显示： input输入：6结果：x79. 计算N个学生的平均成绩， 成绩如下：float average(float array , int v) int i; float a, sum=array0;for(i=1; iv; i+)sum=sum+arrayi;a=sum/v;return(a)；void main( ) float score10, aver;int i, n; printf(“input number:n”);scanf(“&d”, &n);printf(“input %d scores:n”, n);for(i=0; in;i+)scanf(“%f”, &scorei);printf(“n”);aver=average(score, n);printf(“average score is %5.2fn”, aver);10求N个数（N20）的平均值(指针的应用)，程序如下：float average( float *array, int n)int i，*p; float aver, sum=0; for(p=array; parray+n; p+)sum+=*p;aver=sum/n;return(aver);void main()float score20, ave; int a, m; printf(“input number