C++實(shí)驗(yàn)例題實(shí)例

問(wèn)題：給定兩個(gè)正整數(shù)p和q,求其最大公因數(shù)?#include<iostream> intmain(){ intp,q,k; std::cout<<"Inputtwonumbers:"<<std::endl; std::cin>>p>>q; if(p<q) {k=p;} else {k=q;} while(p%k!=0||q%k!=0) {k=k-1;} std::cout<<"Resultis"<<k<<"."<<std::endl; return0;}輸入一個(gè)大于等于4小于等于100的偶數(shù)X，驗(yàn)證X是否可以表示為兩個(gè)素?cái)?shù)之和。intmain(){ intPrimeList[M]; //說(shuō)明存放素?cái)?shù)表的數(shù)組 CreatPrimeList(PrimeList); //建立素?cái)?shù)表 intx;// cin>>x; intp;//p為素?cái)?shù) if(4=<x&&x<M) { //檢查偶數(shù)減去一個(gè)素?cái)?shù)后的剩余部分是否仍為素?cái)?shù) p=PrimeList[2]; while(p<x) { if(PrimeList[x-p]!=0)//PrimeList[x-p]≠0,分解成功 { cout<<"Thenum"<<x<<"="<<p<<"+"<<x-p<<endl; break; } else { p=NextPrimeNumber(p,PrimeList); } } if(p>=x) //找到了一個(gè)不能分解為兩個(gè)素?cái)?shù)和的偶數(shù) { cout<<"Greatdiscovery:Goldbachiswrong!"<<endl; } x=x+2; //檢查下一個(gè)偶數(shù) } return0;}求N階intmain(){unsignedintn,result=1;cin>>n;for(inti=1;i<=n;++i){result=result*i;}cout<<result;return0;}#include<iostream>intmain(){ constintMAXSIZE=50; inta[MAXSIZE]; intn;cin>>n; intsum,sc,count=1; a[0]=1; for(inti=2;i<=n;i++) { sc=0; for(intj=0;j<count;j++) {sum=a[j]*i+sc;sc=sum/10;a[j]=sum%10;} while(sc!=0) {a[j]=sc%10;sc=sc/10;j++;count++;} } for(i=count-1;i>=0;i--) {std::cout<<a[i];} return0;}#include<iostream>出棧usingnamespacestd;intmain(){ Containerc1;Containerc2;charch;cin>>ch; while(ch!='!') { if(ch=='#'){charx=c1.pop();c2.push(x);} elseif(ch=='&'){charx=c2.pop();c1.push(x);} elseif(ch=='@') { while(c1.m_Index!=0){c1.pop();} } else{c1.push(ch);}