文档介绍：输入一个整数将各位征税反转后输出#include<iostream>usingnamespacestd; intmain() { intn,right_digit,newnum=0; cout<<"Enterthenumber:"; cin>>n; cout<<"thenumberinreverssrderis:"; do{ right_digit=n%10; cout<<right_digit; n/=10; } while(n!=0); cout<<endl; return0; }、1~10的和#include<iostream>usingnamespacestd;intmain(){ inti=1,sum=0; while(i<=10) { sum+=i; i++; } cout<<"sunm="<<sum<<endl; return0;}#include<iostream>usingnamespacestd; intmain() { inti=1,sum=0; do { sum+=i; i++; }while(i<=10); cout<<"sum="<<sum<<endl; return0; }工资#include<iostream>usingnamespacestd;intmain(){ longinti; intbouns1,bouns2,bouns4,bouns6,bouns10,bouns; scanf("%d",&i);//%ld表示这个数据的类型是longint长整形//&i表示i的地址,及输出的是i的值 bouns1=100000*; bouns2=bouns1+; bouns4=bouns2+00*; bouns6=bouns4+00*; bouns10=bouns6+400000*; if(i<=100000) bouns=i*; elseif(i<=00) bouns=bouns1+(i-100000)*; elseif(i<=400000) bouns=bouns2+(i-00)*; elseif(i<=600000) bouns=bouns4+(i-400000)*; elseif(i<=10000000) bouns=bouns6+(i-600000)*; else bouns=bouns10+(i-1000000)*; printf("bouns=%d",bouns);//输出一个数据a为整形数据。 }星期 intday; cout<<"输入数:"; cin>>day; switch(day) { case0: cout<<"sunday"<<endl; break; case1: cout<<"monday"<<endl; break; case2: cout<<"tuesday"<<endl; break; case3: cout<<"wednesday"<<endl; break; case4: cout<<"thursday"<<endl; break; case5: cout<<"friday"<<endl; break; case6: cout<<"saturday"<<endl; break; default: cout<<"dayoutofrangesunday..saturday"<<endl; break; } return0;比较XY大小#include<iostream>usingnamespacestd;intmain(){intx,y; cout<<"Enterxandy:"; cin>>x>>y; if(x!=y) if(x>y) cout<<"x>y"<<endl; if(x<y) cout<<"x<y"<<endl; else cout<<"x=y"<<endl; return0; }年可以被4或者400整除不能被100整除;#include<iostream>usingnamespacestd;intmain(){ intyear; boolisLeapYear; cout<<"Entertheyear:"; cin>>year; isLeapYear=((year%4==0&&year%100!=0||year%400==0)); if(isLeapYear) cout<<year<<"isaleapyear"<<endl; else cout<<year<<"isnotleapyear"<<endl; return0;}【程序1】题目:有1、2、3、4个数字,能组成多少个互不相同且无重复数字的三位数?都是多少?:可填在百位、十位、个位的数字都是1、2、3、4。组成所有的排列后再去掉不满足条件的排列。:#include<iostream>Intmain(