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1、给一个不多于4位的正整数，求出它是几位数，逆序打印出各位数字#includevoid main()int x,a,b,c,d;scanf(%d,&x);a=x/1000;b=x%1000/100;c=x%100/10;d=x%10;if(a!=0)printf(4:%d%d%d%dn,d,c,b,a);else if(b!=0)printf(3:%d%d%dn,d,c,b);else if(c!=0)printf(2:%d%dn,d,c);else if(d!=0)printf(1:%dn,d);从10至20之间的素数，自动输出（我的）#include#includevoid main()in
2、t x,k;for(x=10;x=10&x=20;x+)for(k=2;ksqrt(x)printf( %d ,x);0-9这十个数字组成多少个无重复的三位数#includevoid main()int x,a,b,c,num=0;for(x=100;x=999;x+)a=x/100;b=x/10%10;c=x%10;if(a!=b&a!=c&b!=c)num+;printf(%5d,x);printf(n number= %d,num);求300内能被17整除的最大数#includevoid main()int x;for(x=300;x=1;x-)if(x%17=0) break;prin
3、tf(x=%dn,x);斐波那契数列#includevoid main()long f1=1,f2=1,f3;int k;printf(%1dt%1dt,f1,f2);for(k=3;k=30;k+)f3=f1+f2;printf(%dt,f3);f1=f2;f2=f3;求300以内能被17整除的所有整数，用continue#includevoid main()int x;for (x=1;x=300;x+)if(x%17!=0) continue;printf(%dt,x);输出三角形星花#include void main()int k1,k2;for(k1=1;k1=4;k1+)putc
4、har(t);for(k2=1;k2=k1;k2+)putchar(b);for(k2=1;k2=k1*2-1;k2+)putchar(*);putchar(n);求s=2/1-3/2+5/3-8/5求前十项的和#includevoid main()float s=0,f1=2,f2=1,f=1,t,n;for(n=1;n=10;n+)s=s+f*f1/f2;f=f*(-1);t=f2;f2=f1;f1=f1+t;printf(s=%fn,s);逆序输出任意整数#include void main()long x,y;scanf(%d,&y);while(y!=0)x=y%10;printf(
5、%d,x);y=y/10;函数求和#includevoid main()float add(float,float);float a,b,c;scanf(%f,%f,&a,&b);c=add(a,b);printf(sum is %f,c);float add(float x,float y)float z;z=x+y;return(z);输入年月日输出该天在该年的天数#includeint leap(int year)int lp;lp=(year%4=0&year%100!=0|year%400=0)?1:0;return lp;int month_day(int year,int mon
6、th)int d;switch(month)case 1:case 3:case 5:case 7:case 8:case 10:case 12:d=31;break;case 2:d=leap(year)?29:28;break;default:d=30;return d;int days(int year,int month,int day)int i,ds=0;for(i=1;imonth;i+)ds=ds+month_day(year,i);ds=ds+day;return ds;void main()int year,month,day,t_day;scanf(%d_%d_%d,&y
7、ear,&month,&day);t_day=days(year,month,day);printf(%d_%d_%d is %dn,year,month,day,t_day);计算x的n次方#includevoid main()float power(float x,int n);float x,y;int n;scanf(%f,%d,&x,&n);y=power(x,n);printf(%f,y);float power(float x,int n)int i;float t=1;for(i=1;i=n;i+)t=t*x;return t;求和#includevoid main()floa
8、t add(float,float);float a,b,c;scanf(%f,%f,&a,&b);c=add(a,b);printf(sum is %f,c);float add(float x,float y)float z;z=x+y;return(z);交换两个数1（这个不能交换）#includevoid swap(int a,int b);void main()int x=7,y=11;printf(x=%d,ty=%dn,x,y);printf(swapped:n);swap(x,y);printf(x=%d,ty=%dn,x,y);void swap(int a,int b)in
9、t temp;temp=a;a=b;b=temp;交换两个数2（能换过来）#includevoid swap(int *p1,int *p2)int p;p=*p1;*p1=*p2;*p2=p;void main()int a,b;scanf(%d,%d,&a,&b);printf(a=%d,b=%dn,a,b);printf(swapped:n);swap(&a,&b);printf(a=%d,b=%dn,a,b);递归算法计算n！#include long fac(unsigned n)long f;if(n=0)f=1;else f=n*fac(n-1);return f;void ma
10、in()long y;int n;scanf(%d,&n);y=fac(n);printf(%d!=%ldn,n,y);例5-1 输入10个数，正序逆序输出，用数组#includevoid main()int i;int a10;printf(input 10 numbersn);for(i=0;i10;i+)scanf(%d,&ai);printf(n);for(i=0;i=0;i-)printf(%d ,ai);利用数组，求斐波拉契数列的前10项#includevoid main()int i;static int a10=1,1;for(i=2;i10;i+)ai=ai-1+ai-2;f
11、or(i=1;i10;i+)printf(%d ,ai);求若干数（不多于10个）的最大数，最小数#define N 10#includevoid main()int i,n;float aN,max,min;printf(input data numbers);scanf(%d,&n);printf(enter numbers:);for(i=0;in;i+)scanf(%f,&ai);max=min=a0;for(i=1;imax) max=ai;if(aimin) min=ai;printf(nmax=%f,min=%fn,max,min);输入10个数，求它们的和#includevoid main()int a10;int i,sum=0;for(i=0;i=9;i+)printf(enter number %d:,i+1);scanf(%d,&ai);sum+=ai;printf(the totle is :%dn,sum);冒泡排序法#includevoid
