大二下學(xué)期數(shù)值計(jì)算實(shí)驗(yàn)報(bào)告答案C++

#include<iostream>#include<cmath>#defineM100usingnamespacestd;intmain(){ inti,j,n,k,t; floata[M][M],x[M],temp,sum=0;cout<<"輸入方程的階數(shù):"<<endl; cin>>n; cout<<"輸入方程矩陣:"<<endl; for(i=0;i<n;i++) //輸入矩陣 for(j=0;j<n+1;j++) cin>>a[i][j];for(k=0;k<n+1;k++){ t=k; for(i=t+1;i<n;i++) //選出最大元素 if((fabs(a[t][k]))<(fabs(a[i][k]))) t=i; if(t!=k) //交換行元素 for(i=k+1;i<n+1;i++){ temp=a[k][i]; a[k][i]=a[t][i]; a[t][i]=temp; } for(i=k+1;i<n;i++){ //消元 for(j=0;j<n+1;j++) a[i][j]-=(a[i][k]/a[k][k])*a[k][j]; } } x[n-1]=a[n-1][n]/a[n-1][n-1]; //解出Xn for(i=n-2;i>=0;i--){ //解出其它解 for(j=i;j<n;j++){if(j==i)continue; sum+=a[i][j]*x[j]; } x[i]=(a[i][n]-sum)/a[i][i]; } printf("最終解得:\n"); for(i=0;i<n;i++) //輸出結(jié)果 cout<<"x["<<i<<"]="<<x[i]<<endl; return0;}#include<iostream>#include<stdlib.h>usingnamespacestd;intmain(){ intn; do{ cout<<"輸入三對(duì)角方程組的變量的數(shù)目N:"; cin>>n; }while(n<1); cout<<"請(qǐng)依次輸入三對(duì)角方程組每行的數(shù)據(jù)(0元素除外):\n"; doublea[100],b[100],c[100],f[100]; cin>>*b>>*c>>*f; for(inti=1;i<n-1;++i)cin>>*(a+i-1)>>*(b+i)>>*(c+i)>>*(f+i); cin>>*(a+i-1)>>*(b+i)>>*(f+i); for(i=0;i<n-1;++i){ *(c+i)/=(*(b+i)); *(b+i+1)-=(*(a+i))*(*(c+i)); } *b=(*f)/(*b); for(i=1;i<n;++i) *(b+i)=((*(f+i))-(*(a+i-1))*(*(b+i-1)))/(*(b+i)); *(a+n-1)=*(b+n-1); for(i=n-2;i>=0;--i) *(a+i)=*(b+i)-(*(c+i))*(*(a+i+1)); cout<<"方程的解如下:\n";for(i=1;i<n;++i)cout<<"X["<<i<<"]:"<<*(a+i-1)<<endl; return0;}#include<iostream>#include<cmath>usingnamespacestd;voidmain(){ inti,j,n,x; doublem=0,p,a[100],b[100]; cout<<"插入節(jié)點(diǎn)數(shù)："; //得到節(jié)點(diǎn)數(shù) cin>>n; cout<<"請(qǐng)輸入節(jié)點(diǎn)x值："<<endl; //得到節(jié)點(diǎn)值，即X for(i=0;i<n;i++) cin>>a[i]; cout<<"請(qǐng)輸入f(x)值："<<endl; //得到Y(jié) for(i=0;i<n;i++) cin>>b[i]; cout<<"請(qǐng)輸入x1："<<endl; //取未知X cin>>x; for(i=0;i<n;i++){ //用拉格朗日插值法求解 p=1; for(j=0;j<n;j++){ if(i!=j) p*=(x-a[j])/(a[i]-a[j]); else continue; } m+=p*b[i]; } cout<<m<<endl; //輸出結(jié)果的Y}#include<iostream>usingnamespacestd;intmain(){ intn; doublex; doublea[100],f[100]; do{ cout<<"輸入插值點(diǎn)的數(shù)目n："; cin>>n; }while(n<1); cout<<"輸入插值點(diǎn)的數(shù)據(jù)："<<endl; for(inti=0;i<n;++i) cin>>*(a+i)>>*(f+i); for(i=0;i<n;++i) for(intj=n-1;j>i;--j) *(f+j)=(*(f+j)-*(f+j-1))/(*(a+j)-*(a+j-i-1)); cout<<"輸入所求的橫坐標(biāo)："; cin>>x; doubleresult=*f; for(i=1;i<n;++i){ doubletemp=1; for(intj=0;j<i;++j) temp*=(x-*(a+j)); result+=(*(f+i))*temp; } cout<<"所求的函數(shù)值為："<<result<<endl; return1;}#include<iostream>#include<cmath>usingnamespacestd;doublefun(doublea){ if(a==0) a+=0.00000000001; returnsin(a)/a;}doubleT(doublea,doubleb,doubleE){ doubleh=(b-a)/2; doubler1=(fun(a)+fun(b))*h; intn=1; doubler0,s,e; do{r0=r1; s=0; for(inti=0;i<=(n-1);i++) s=s+fun((b-a)*i/n+a+(b-a)/(2*n)); r1=r0/2+s*h/n; e=fabs(r1-r0); n=2*n; }while(e>3*E); cout<<n<<endl; returnr1;}voidmain(){ doublea,b,e; cout<<"輸入待求積分的下屆a和上屆b及精度要求e"<<endl; cin>>a>>b>>e; cout<<"可求得積分："<<T(a,b,e)<<endl;}#include<iostream>#include<math.h>usingnamespacestd;doublefun(doublex);intmain(){ doublea=0,b=1,eps=0.0000001; doubleT[10]; intl=1; T[0]=(b-a)*(fun(a)+fun(b))/2; doubleT1,T2; intflag; do{ flag=0; intn=pow(2,l-1); doubleh=(b-a)/n; doubletemp=0; for(inti=0;i<n;++i) temp+=fun(a+(i+1.0/2)*h); T1=T[0]/2+h*temp/2; for(intm=1;m<l;++m){ T2=(pow(4,m)*T1-T[m-1])/(pow(4,m)-1); T[m-1]=T1; T1=T2; } T2=(pow(4,1)*T1-T[l-1])/(pow(4,1)-1); if(fabs(T2-T[l-1])>eps){ T[l-1]=T1; T[l]=T2; ++l; flag=1; } }while(flag); cout<<"所求積分為:"<<T2<<endl; return0;}doublefun(doublex){ if(x==0)return1; return(sin(x)/x);}#include<stdlib.h>#include<iostream.h>intmain(){ doubleh=0.25,a,b,d=5; doublex=0.0,y=2.0; cout<<"x\ty\n"; cout<<x<<"\t"<<y<<endl; while(x<d){ a=y+h*(-x*y*y); x+=h; b=y+h*(-x*a*a); y=(b+a)/2.0; cout<<x<<"\t"<<y<<endl; } return0;}#include<iostream.h>#include<stdlib.h>intmain(){ doubleh=0.25,k1=0,k2=0,k3=0,k4=0,d=5,h2=h/2; doublex=0.0,y=2.0; cout<<"x\ty"<<endl; cout<<x<<"\t"<<y<<endl; while(x<d){ k1=(-x*y*y); x+=h2; k2=-x*(y+h2*k1)*(y+h2*k1); k3=-x*(y+h2*k2)*(y+h2*k2); x+=h2; k4=-x*(y+h2*k3)*(y+h2*k3); y+=h/6*(k1+2*k2+2*k3+k4); cout<<x<<"\t"<<y<<endl; } return0;}#include<iostream.h>#include<cmath>#defineeps1e-5voidmain(){ intn=4; doublea[4][4]={{7,2,1,-2},{9,15,3,2},{-2,-2,11,5},{1,3,2,13}}; doubley[4]={4,7,-1,0}; doublex[4]={0}; doublexx[4]={0}; intM=20;//最大迭代數(shù) inti,j,k; doubledif=1.0; doublesum=0.0; doubleresidual=0.0; for(k=1;(k<=M)&&(dif>eps);k++){ dif=0.0; for(i=0;i<n;i++){ for(j=0;j<n;j++){ if(j<i)sum+=a[i][j]*x[j]; if(j>i)sum+=a[i][j]*xx[j]; } x[i]=(y[i]-sum)/a[i][i]; sum=0.0; } residual=0.0; for(intm=0;m<n;m++){ if(fabs(x[m]-xx[m])>residual) residual=fabs(x[m]-xx[m]); } dif=residual; cout<<"第"<<k<<"次迭代的結(jié)果:"; for(intp=0;p<n;p++){ cout<<x[p]<<""; xx[p]=x[p]; } cout<<endl; } cout<<endl<<"迭代計(jì)算的結(jié)果為："<<endl;