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1、1.输入一个不超过五位的正整数,输出其逆数。例如输入12345,输出应为54321。#include"stdio.h"main(){longn;intww,qw,bw,sw,gw;printf("Pleaseinput:");scanf("%ld",&n);while(n<=0
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3、n>99999){printf("Thenumberistoolittleortoolarge.Retryplease!");scanf("%ld",&n);}if(n>=10000&&n<=99999){ww=n/10000;qw=n/1000%10;bw=n/100%10;s
4、w=n/10%10;gw=n%10;printf("Theoldnumberis:%ld",n);printf("Thenewnumberis:%d%d%d%d%d",gw,sw,bw,qw,ww);}elseif(n>=1000&&n<=9999){qw=n/1000;bw=n/100%10;sw=n/10%10;gw=n%10;printf("Theoldnumberis:%ld",n);printf("Thenewnumberis:%d%d%d%d",gw,sw,bw,qw);}elseif(n>=100&&n<=999){bw=n/100;sw
5、=n/10%10;gw=n%10;printf("Theoldnumberis:%ld",n);printf("Thenewnumberis:%d%d%d",gw,sw,bw);}elseif(n>=10&&n<=99){sw=n/10;gw=n%10;printf("Theoldnumberis:%ld",n);printf("Thenewnumberis:%d%d",gw,sw);}elseprintf("Theoldnumberequalsthenewnumber:%ld",n);}2.计算1+2+3…+n的值,n是从键盘输入的自然数。/*Not
6、e:YourchoiceisCIDE*/#include"stdio.h"main(){intn,i;longsum=0;printf("1+2+3+....+n");printf("Pleaseinputn:");scanf("%d",&n);for(i=1;i<=n;i++){sum+=i;}printf("sum=%ld",sum);}3.从终端(键盘)读入20个数据到数组中,统计其中正数的个数,并计算这些正数之和。、/*Note:YourchoiceisCIDE*/#include"stdio.h"#defineN20main(){inta[N],i,
7、sum=0,count=0;printf("Pleaseinputnumbers:");for(i=0;i0){sum+=a[i];count++;}}printf("count=%d,sum=%d",count,sum);}1.从终端(键盘)将5个整数输入到数组a中,然后将a逆序复制到数组b中,并输出b中各元素的值。/*Note:YourchoiceisCIDE*/#include"stdio.h"#defineN5main(){inta[N],b[N],i;printf("Pleaseinpu
8、tnumbers:");for(i=0;i9、rchoiceisCIDE*/#include"stdio.h"main(){inti,j,k,n;for(i=1;i<=10;i++)for(j=1;j<=25;j++)for(k=1;k<=50;k++)for(n=1;n<=100;n++)if(i*50+j*20+k*10+n*5==500)printf("num50=%d,num20=%d,num10=%d,num5=%d",i,j,k,n);}3.求n以内(不包括n)同时能被3和7整除的所有自然数之和的平方根s,n从键盘输入。例如若n为1000时,函数值应为:s=153.909064。