西安交通大學實驗報告課程計算機程序設計實驗名稱C++集成開發(fā)環(huán)境的使用第1頁共10頁系別電氣學院 實驗日期2011年3月24日專業(yè)班級____錢學森02____組別實驗報告日期2011年3月24日姓 名____楊虬 學號2010045052報告退發(fā)（訂正、重做）同組人教師審批簽字.實驗目的.（1）掌握集成開發(fā)環(huán)境的使用方法;.（2）了解C++程序的基本特點。二.實驗內(nèi)容（一）實驗題目一：編寫字符串反轉(zhuǎn)函數(shù)mystrrev（）..要點分析略.程序源代碼#include<iostream>#include<string>#include<cstring>usingnamespacestd;voidmystrrev(charstring口)(inti,n;charp;n=strlen(string);for(i=0;i<=int(n/2-1);i++)p二string[i];string[i]=string[n-1-i];string[n-1-i]=p;main()(chars[20];inti,n;cout<<"請輸入一個字符串：";cin.get(s,20);n=strlen(s);cout<<"改變之前的字符串為："；for(i=0;i<=n-1;i++)cout<<s[i];mystrrev(s);cout<<endl<<"改變之后的字符串為："；for(i=0;i<=n-1;i++)cout<<s[i];cout<<endl;return0;).實驗結果'彳 耀手匹梢學習\C--上機.0回1—^.|請輸入一個字符串: abcdefg-Z-改變之前的字符串為二abcdefg改變之后的字符串為二gfedcbaPressanykeytocontinue.「（二）實驗題目二：編寫一組求數(shù)組中最大最小元素的函數(shù)，同時寫出主函數(shù)來檢驗之。.要點分析略.程序源代碼#include<iostream>usingnamespacestd;intimax(intarray[],intcount)(inti,max;max=array[0];for(i=0;i<=count-1;i++)if(array[i]>max)max=array[i];returnmax;)intimin(intarray[],intcount)(inti,min;min=array[0];for(i=0;i<=count-1;i++)if(array[i]<min)min=array[i];returnmin;)main()(constintnum=10;inta[num];inti,j,n;cout<<"請輸入—個"<<num<<"個數(shù)的數(shù)組："<<endl;for(i=0;i<=num-1;i++)cin>>a[i];cout<<"此數(shù)組中最大的元素為："<<imax(a,num)<<endl;cout<<"此數(shù)組中最小的元素為："<<imin(a,num)<<endl;return0;).實驗結果建存夜欄導習I=1回請輸入一個10個數(shù)的數(shù)組：-123026512-934763此數(shù)組中最大的元素為：76此數(shù)組中最小的元素為：-9Pressanykeytocontinue.-(三)實驗題目三編寫程序isprime(inta)用來判斷變量a是否為素數(shù)，用該函數(shù)找出任意給定的n個整數(shù)中的素數(shù)。要點分析略.程序源代碼#include<iostream>#include<cmath>usingnamespacestd;intisprime(inta)(inti;if(a==1)return0;for(i=2;i<=sqrt(a);i++)(if(a%i==0)return0;)return1;)main()(inti;constm=5;intn[m];cout<<"請輸入"<。<<"個正整數(shù):";for(i=0;i<=m-1;i++)cin>>n[i];cout<<”以上幾個數(shù)中的質(zhì)數(shù)有：”;for(i=0;i<=m-1;i++)if(isprime(n[i])==1)

cout<<n[i]<<"";cout<<endl;return0;)2.實驗結果■，。僧甲風欄百習a—上機u耀序,,〔o回|請輸入5個正整數(shù)：1231324523;|

以上幾個數(shù)中的質(zhì)數(shù)有.231233TOC\o"1-5"\h\zPressanvkevtocontinue /I/匚電〕 卜L 」(四)實驗題目四用弦截法求一元方程f(x)=0在區(qū)間(XO,X1)之間的一個根。.要點分析略.程序源代碼#include<iostream>#include<cmath>usingnamespacestd;doublea[10];intn;doublef(doublex)(doublef;inti;f=0;for(i=n;i>=0;i--)f+=a[i]*pow(x,i);returnf;)doublex(doublex0,doublex1)(doublet,x2;t=1E-7;do(x2=(x0+f(x0)*(x0-x1)/(f(x1)-f(x0)));if(f(x2)*f(x0)>=0)x0=x2;elsex1=x2;cout<<f(x2)<<endl;}while(fabs(f(x2))>t);returnx2;)main()(inti;doublex0,x1;cout<<”請輸入方程最高次項的次數(shù)：”；cin>>n;cout<<”請依次輸入方程各項的系數(shù)，格式為：”<<endl;for(i=n;i>=1;i--)cout<<"a["<<i<<"]X~'<<i<<"+";cout<<"a[0]=0"<<endl;for(i=n;i>=0;i--)cin>>a[i];cout<<"請輸入求根的范圍：Xo<X<X1"<<endl;cout<<"Xo=";cin>>x0;cout<<"X1=";cin>>x1;if(f(x0)*f(x1)<0)cout<<"X="<<x(x0,x1)<<endl;elsecout<<"該方程在("<<x0<<","<<x1<<")區(qū)間上無解或有多個解！"<<endl;return0;)3.實驗結果?O建序K都學習++上機褥程序建5次上機|回情輸入方程最高成球的次數(shù)：1請依次輸入方程各項的系數(shù).格式為；口a[l]X^l+a[0]=015請輸入求根的范圍:；Xo<X<XlXo=-2Xl=0該方程在《-2,0》區(qū)間上的根為：X=-0.714286Pressanvkevtocontinue .- -IIT1I'6儲字工/導習\C4；上機耳1蠲冢第5次上機加D，I口回1請輸入方程最高次項的次數(shù):3請依次輸入方程各項的系數(shù)，格式為；a[3]X"3+a[2]X"2+a[l]X"l+a[0]=017567請輸入求根的范圍：Xo<X<XlXo=-1X1=O該方程在bL0)區(qū)間上的根為:x=-o.674992PressanvkevtocontinueTOC\o"1-5"\h\z-I -I■，5儲存國州學習區(qū)-+上機1源程序爆5次上機…11=■產(chǎn)|請輸入方程最高成運的次數(shù)：4 H請依次輸入方程各項的系數(shù)，格式為； 口a[4]X"4-a閭X飛-b[2]/2+武1]工」-&[0]=0110250-36請輸入求根的范圍：Xo<X<XlXo=-10Xl=-5該方程在(TO,-5)區(qū)間上的根為：X=-6Pressanykeytocontinue^ -* 叫 I 卜(五)實驗題目五編寫計算平方根函數(shù)，并加以測試檢驗。提示：采用牛頓迭代法，計算精度10-7。函數(shù)原型：doublenmysqrl(doublex).要點分析略.程序源代碼#include<iostream>usingnamespacestd;#include<cmath>doublemysqrt(doublex)(doublex1,x2,t;x1=1;x2=1;t=1E-7;while((fabs(x1*x1-x)>t)||(fabs(x1-x2)>t))(x1=x2-(x1*x1-x)/(2*x1);x2=x1;)returnx1;)main()(doublen;cout<<"請輸入n：";cin>>n;cout<<"sqrt("<<n<<")="<<mysqrt(n)<<endl;return0;).實驗結果,. 儲存以欄\學習\J-上機,」1=11回TOC\o"1-5"\h\z|請輸入n：43 3sqrt(43)=6.55744 UPressanvkevt-ocontinue -■■L ?(六)實驗題目六編寫函數(shù)判定一個正整數(shù)是否是遞增數(shù)，即該數(shù)各位數(shù)字從左至右遞增排列，例如：1122334、123456、5599、22222222等都是遞增數(shù)。例如：32768、43987、123498都不是遞增數(shù)。顯示輸出0至9999999之間所有遞增數(shù)。.要點分析略.程序源代碼#include<iostream>#include<math.h>#include<fstream>usingnamespacestd;intp(longn)(inta[10];inti,num;num=int(log10(n))+1;a[0]=n%10;for(i=1;i<=num;i++)(a[i]=int(n/pow(10,i))%10;if(a[i-1]<a[i])return0;)return1;)main()(inti;ofstreamout("遞增數(shù)")；for(i=1;i<=9999999;i++)if(p(i)==1)out<<i<<"";out.close();return0;).實驗結果

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