简单迭代法

在错误的地方我给你注释了

#include<stdio.h>

#include <iostream>

#include <math.h>

#define N 100

using namespace std

double g(double x)

{

return (pow((x+1),1.0/3.0))

}

void main(void)

{

double x0

cout<<"初值为x0=："

cin>>x0

int k,a=1,b=2

double x,x1,x2

while(k<N) //这里的k你没给初值

{

if(k==1)

x=g(x0)

else

{

x1=g(x0)

x2=g(x1)

x=(x0*x2-x1*x1)/(x0-2*x1+x2)

}

if((x0<1)||(x0>2))

{

cout<<"请另选初值："<<endl

exit(0)

}

else

if((fabs(x-x0)>pow(10.0,-5))&&(fabs(x-x0)<pow(10.0,-4)))

{

cout<<"迭代成功："<<endl

cout<<"迭代次数为："<<k+1<<endl

cout<<"用简单迭代法解出此方程的近似解为："<<x<<endl

break

}

else

{

k++

x0=x

}

}

if(k==N)

cout<<"迭代次数已经达到上界，不收敛。"<<endl

}

迭代法计算信道容量C++程序

#include <iostream>

#include <vector>

#include <cfloat>

#include <cmath>

using namespace std

#define FLOAT_MINUS_PRECISION 0.00001

typedef vector<float*>VEC_PFLOAT

//迭代计算信道容量,参数值为信源,信宿符号个数和信道转移概率矩阵,返回信道容量

float GetCapacity(int nSourceSymbol,int nHostSymbol,const VEC_PFLOAT&vTransMatrix)

{

//信道容量初始化为最小值

float fCapacity = FLT_MIN

//信源概率分布

float *pfSoureProb = new float[nSourceSymbol]

//初始化信源分布为均匀分布

int i

for (i = 0i <nSourceSymboli++)

{

pfSoureProb[i] = 1.0 / nSourceSymbol

}

//初始化φ函数

VEC_PFLOAT vPhi

for (i = 0i <nSourceSymboli++)

{

float *pfTemp = new float[nHostSymbol]

vPhi.push_back(pfTemp)

}

//设置精度

const float cfDelta = 0.02f

float fPrecision

//迭代计算

int j,k

float *pfSum = new float[nSourceSymbol]

do

{

for (i = 0i <nSourceSymboli++)

{

for (j = 0j <nHostSymbolj++)

{

//计算ΣPi*Pji

float fSum = 0.0f

for (k = 0k <nSourceSymbolk++)

{

fSum += pfSoureProb[i] * vTransMatrix[k][i]

}

vPhi[i][j] = pfSoureProb[i] * vTransMatrix[j][i] / fSum

}

}

float fSumDeno = 0.0f //分母求和

for (i = 0i <nSourceSymboli++)

{

float fSum = 0.0f

for (j = 0j <nHostSymbolj++)

{

fSum += vTransMatrix[j][i] * logf(vPhi[i][j])

}

pfSum[i] = expf(fSum)

fSumDeno += pfSum[i]

}

for (i = 0i <nSourceSymboli++)

{

pfSoureProb[i] = pfSum[i] / fSumDeno

}

//计算新一轮的容量

float fNewC = logf(fSumDeno)

//计算精度

fPrecision = fabs(fNewC - fCapacity) / fCapacity

fCapacity = fNewC

} while(fPrecision - cfDelta >0.0f)

//释放临时资源...

delete []pfSum

for (i = 0i <vPhi.size()i++)

{

float* pfTemp = vPhi.at(i)

delete pfTemp

}

vPhi.clear()

return fCapacity

}

int main()

{

//转移矩阵

VEC_PFLOAT vTransMatrix

int nCol,nLine

cout<<"请输入信源符号个数:"

cin>>nLine

cout<<"请输入信宿符号个数:"

cin>>nCol

cout<<"请依次输入"<<"行信道转移概率矩阵:(以空格隔开每个概率)\n"

for (int i = 0i <nLinei++)

{

float *pfTemp = new float[nCol]

Label1:

float fSum = 0.0f

cout<<"X"<<":"

for (int j = 0j <nCol - 1j++)

{

cin>>pfTemp[j]

fSum += pfTemp[j]

}

if (1.0f - fSum <0)

{

cout<<"转移概率和应该为1,请重新输入!\n"

goto Label1

}

else

{

pfTemp[j] = 1.0f - fSum

cout<<"信源符号X"<<"的转移概率分别为:"

for(int k = 0k <nColk++)

cout<<pfTemp[k]

cout<<"\n"

}

vTransMatrix.push_back(pfTemp)

}

cout<<"信道容量为:"

for (int k = 0k <vTransMatrix.size()k++)

{

float* pfTemp = vTransMatrix.at(k)

delete pfTemp

}

vTransMatrix.clear()

return 0

}

看看对你有帮助吧！！！

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