黄金分割c语言程序
Ⅰ c语言编程：用黄金分割法求 minf（x）=x2+2x+1 急！！！！
给，已经编译运行确认：
#include "math.h"
#include "stdio.h"
#define f(x) x*x+2*x+1 //一元函数，这里按照你的要求写的是：x2+2x+1
//函数功能是用黄金分割法实现求一元函数 的最优解
double hj(double *a,double *b,double e,int *n)
{ double x1,x2,s;
if(fabs(*b-*a)<=e)
s=f((*b+*a)/2);
else
{ x1=*a+0.382*(*b-*a);
x2=*a+0.618*(*b-*a);
if(f(x1)>f(x2))
*a=x1;
else
*b=x2;
*n=*n+1;
s=hj(a,b,e,n);
}
return s;
}
main()
{ double s,a,b,e;
int n=0;
scanf("%lf %lf %lf",&a,&b,&e); // 输入区间[a,b]和精度e的值
s=hj(&a,&b,e,&n); //调用hj函数，其中n代表迭代次数
printf("a=%lf,b=%lf,s=%lf,n=%d\n",a,b,s,n);
}
运行时：
输入：0.6 0.5 0.1
输出结果为：
0.6 0.5 0.1
a=0.600000,b=0.500000,s=2.402500,n=0
Ⅱ 用c语言编写黄金分割法
黄金分割点是指把一条线段分割为两部分，使其中一部分与全长之比等于另一部分与这部分之比。其比值是一个无理数，用分数表示为(√5-1)/2。黄金分割点（p）的求法，如图：①过点B作BD⊥AB，使BD=1/2AB；②连结AD，以D为圆心，CB为半径作弧，交AD于E， 则有DE=DB；③以A为圆心，AE为半径作弧，交AB于P，则有AP=AE；则点P是线段AB上的一个黄金分割点．为什么点P是线段AB上的一个黄金分割点？事实上，若设AB=2，则BD=BE=1，由作图过程可知AD=√5．则AE=AP=（√5）-1，PB/AP=AP/AB=[（√5）-1]/2。 因此点P是线段AB上的一个黄金分割点．
Ⅲ 求用c语言黄金分割数的小数后100位的程序，并显示结果
#include <stdio.h>
#include <string.h>
typedef unsigned char UCHAR;
typedef unsigned short USHORT;
#define SIZE 128 //一个大整数用个字节保存，最多表示位大整数
#define SIZE_10 2 * SIZE
typedef struct BigNum //大整数结构
{
UCHAR data[SIZE]; //空间为(SIZE * sizeof(UCHAR))，就是SIZE个字节
}BigNum;
//////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
#ifndef _BIGNUM_H_
#define _BIGNUM_H_
UCHAR atox(char ch); //将一个十六进制的字符(4位)转位数字，转换失败返回xff
BigNum Init(char* str); //初始化大整数，str为十六进制字符串
int GetByte(BigNum bignum); //判断有多少个字节不为
BigNum MovByteLetf(BigNum bignum, int n); //左移n个字节
BigNum MovByteRight(BigNum bignum, int n); //右移n个字节
int Cmp(BigNum bignum_a, BigNum bignum_b); //大整数比较大小，>返回，<返回-1，==返回
BigNum Add(BigNum bignum_a, BigNum bignum_b); //大整数加法
BigNum Sub(BigNum bignum_a, BigNum bignum_b); //大整数减法
//BigNum Mul(BigNum bignum_a, UCHAR uchar); //大整数乘法UCHAR
BigNum Mul(BigNum bignum_a, BigNum bignum_b); //大整数乘法
BigNum Div(BigNum bignum_a, BigNum bignum_b); //大整数除法
BigNum Mod(BigNum bignum_a, BigNum bignum_b); //大整数模运算
BigNum Pow(BigNum bignum, int n); //大整数乘方运算
void Print_16(BigNum bignum); //打印十六进制大整数
void Print_10(BigNum bignum); //打印十进制大整数
#endif
//////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
//////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
// BigNum.c : 定义控制台应用程序的入口点。
//
UCHAR atox(char ch) //将一个十六进制的字符(4位)转位数字，转换失败返回xff
{
UCHAR res;
if (ch >= Ɔ' && ch <= Ə')
{
res = ch - Ɔ'
}
else if (ch >= 'a' && ch <= 'f')
{
res = ch - 'a' + 10;
}
else if (ch >= 'A' && ch <= 'F')
{
res = ch - 'A' + 10;
}
else
{
printf("change error! ");
return 0xff;
}
return res;
}
int GetByte(BigNum bignum) //判断有多少个字节不为
{
int ByteOfBigNum = SIZE;
while ((bignum.data[ByteOfBigNum - 1] == 0) && (ByteOfBigNum > 0))
{
ByteOfBigNum--;
}
return ByteOfBigNum;
}
BigNum MovByteLetf(BigNum bignum, int n) //左移n个字节
{
int i;
int bignum_len = GetByte(bignum);
for (i = bignum_len - 1; i >= 0; i--)
{
if (i + n >= SIZE)
{
continue;
}
bignum.data[i + n] = bignum.data[i];
}
for (i = n - 1; i >= 0; i--)
{
bignum.data[i] = 0;
}
return bignum;
}
BigNum MovByteRight(BigNum bignum, int n) //右移n个字节
{
int i;
int bignum_len = GetByte(bignum);
for (i = 0; i < bignum_len; i++)
{
if (i + n >= SIZE)
{
bignum.data[i] = 0;
continue;
}
bignum.data[i] = bignum.data[i + n];
}
return bignum;
}
BigNum Init(char* str) //初始化大整数，str为十六进制字符串
{
BigNum bignum;
int len = (int)strlen(str);
int i;
int j = 0;
if (len > 2 * SIZE)
{
len = 2 * SIZE;
}
for (i = len - 1; i > 0; i -= 2)
{
bignum.data[j] = atox(str[i]) + atox(str[i - 1]) * 16;
j++;
}
if (i == 0)
{
bignum.data[j] = atox(str[i]);
j++;
}
while (j < SIZE)
{
bignum.data[j] = 0;
j++;
}
return bignum;
}
int Cmp(BigNum bignum_a, BigNum bignum_b) //大整数比较大小，>返回，<返回-1，==返回
{
int bignum_a_len = GetByte(bignum_a);
int bignum_b_len = GetByte(bignum_b);
int max = bignum_a_len > bignum_b_len ? bignum_a_len : bignum_b_len;
int i;
for (i = max - 1; i >= 0; i--)
{
if (bignum_a.data[i] > bignum_b.data[i])
{
return 1;
}
if (bignum_a.data[i] < bignum_b.data[i])
{
return -1;
}
}
return 0;
}
BigNum Add(BigNum bignum_a, BigNum bignum_b) //大整数加法
{
BigNum bignum_c;
USHORT temp;
UCHAR carry = 0;
int i;
for (i = 0; i < SIZE; i++)
{
temp = bignum_a.data[i] + bignum_b.data[i] + carry;
bignum_c.data[i] = temp & 0x00ff;
carry = (temp >> 8) & 0xff;
}
return bignum_c;
}
BigNum Sub(BigNum bignum_a, BigNum bignum_b) //大整数减法
{
BigNum bignum_c;
USHORT temp;
UCHAR carry = 0;
int i;
for (i = 0; i < SIZE; i++)
{
temp = bignum_a.data[i] - bignum_b.data[i] - carry;
bignum_c.data[i] = temp & 0x00ff;
carry = (temp >> 15) & 0x01;
}
return bignum_c;
}
BigNum Mul(BigNum bignum_a, BigNum bignum_b) //大整数乘法
{
BigNum bignum_c = Init("0");
USHORT temp;
UCHAR carry;
int i, j;
for (i = 0; i < SIZE; i++)
{
carry = 0;
for (j = 0; j < SIZE; j++)
{
temp = bignum_a.data[i] * bignum_b.data[j] + bignum_c.data[j + i] + carry;
bignum_c.data[j + i] = temp & 0x00ff;
carry = (temp >> 8) & 0xff;
}
}
return bignum_c;
}
BigNum Div(BigNum bignum_a, BigNum bignum_b) //大整数除法
{
BigNum bignum_c = Init("0");
BigNum B;
int bignum_a_len;
int bignum_b_len;
int bignum_c_len;
if (Cmp(bignum_b, bignum_c) == 0)
{
printf("错误！除数为 ");
return bignum_c;
}
bignum_a_len = GetByte(bignum_a);
bignum_b_len = GetByte(bignum_b);
bignum_c_len = bignum_a_len - bignum_b_len;
while (bignum_c_len >= 0)
{
B = MovByteLetf(bignum_b, bignum_c_len);
while (Cmp(bignum_a, B) != -1)
{
bignum_a = Sub(bignum_a, B);
bignum_c.data[bignum_c_len]++;
}
bignum_c_len--;
}
return bignum_c;
}
BigNum Mod(BigNum bignum_a, BigNum bignum_b) //大整数模运算
{
BigNum bignum_c = Init("0");
BigNum B;
int bignum_a_len;
int bignum_b_len;
int bignum_c_len;
if (Cmp(bignum_b, bignum_c) == 0)
{
printf("错误！除数为 ");
return bignum_c;
}
bignum_a_len = GetByte(bignum_a);
bignum_b_len = GetByte(bignum_b);
bignum_c_len = bignum_a_len - bignum_b_len;
while (bignum_c_len >= 0)
{
B = MovByteLetf(bignum_b, bignum_c_len);
while (Cmp(bignum_a, B) != -1)
{
bignum_a = Sub(bignum_a, B);
}
bignum_c_len--;
}
return bignum_a;
}
BigNum Pow(BigNum bignum, int n) //大整数乘方运算
{
int i;
BigNum bignum_res = Init("1");
for (i = 0; i < n; i++)
{
bignum_res = Mul(bignum_res, bignum);
}
return bignum_res;
}
void Print_16(BigNum bignum) //打印十六进制大整数
{
int i;
int nFlag = 0;
for (i = SIZE - 1; i >= 0; i--)
{
if (nFlag == 0 && bignum.data[i] == 0) //前面是的不打印
{
continue;
}
else
{
nFlag++;
if (nFlag == 1) //首位的不打印，如x01，只打印
{
printf("%x", bignum.data[i]);
}
else //后面的要打印
{
printf("%02x", bignum.data[i]);
}
}
}
if (nFlag == 0)
{
printf("0");
}
printf(" ");
}
void Print_10(BigNum bignum) //打印十进制大整数
{
int data_10[SIZE_10];
int i = 0;
int j;
while (Cmp(bignum, Init("0")) == 1)
{
data_10[i] = (Mod(bignum, Init("a"))).data[0];
bignum = (Div(bignum, Init("a")));
i++;
}
for (j = i - 1; j >= 0; j--)
{
printf("%d", data_10[j]);
}
printf(" ");
}
//////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////
void GoldNum(int precision)
{
BigNum b1 = Init("2");
BigNum b2 = Init("3");
BigNum b3 = Init("a");
BigNum temp1;
BigNum temp2 = b1;
int count = 1000;
int i = 0;
b3 = Pow(b3, precision);
for(;i < count; i++)
{
temp1 = b2;
b2 = Add(b1, b2);
b1 = temp1;
temp1 = Mul(b1, b3);
temp1 = Div(temp1, b2);
if(!Cmp(temp1, temp2))
{
Print_10(temp1);
break;
}
temp2 = temp1;
}
}
int main()
{
GoldNum(100);
return 0;
}
这里面的大数加减乘除是直接用这里面的http://wenku..com/view/f7d13dd2195f312b3169a5d7.html
我只是写了void GoldNum(int precision)函数，precision参数是精度大小
Ⅳ 用c++怎么编黄金分割法的程序
用pow函数
pow(x,y)表示x的y次方
Ⅳ C语言编程，用黄金分割法求f（a）=a*a-7*a+10的最优解。设初始值a0=0，初始步长h=1,取迭代精度=0.35。急…
#include <stdio.h>
#include <math.h>
float GetEquation(float x)
{
return x*x-7*x+10;
}
void main()
{
float a=0;
float b=6;
float result = 0;
do {
float c=a+0.618*(b-a);
float d=a+b-c;
if(fabs(GetEquation(c)) < fabs(GetEquation(d)))
{
a=d;
result = c;
}
else
{
b=c;
result = d;
}
} while(fabs(GetEquation(result)) > 0.01);
printf("f(a)=a*a-7*a+10\n");
printf("a=%f\n",result);
}