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黄金连分数（注意：大数）

• 1.转为求斐波那契数列的n和n+1项
• 2. n取多少？再增加n，小数点后的101位没有变化。
• 3.不能用C语言定义的整数型直接运算，而是要手工地写大数加法和除法（减法）
  大数的计算

vector是一种顺序容器，事实上和数组差不多，但它比数组更优越。一般来说数组不能动态拓展，因此在程序运行的时候不是浪费内存，就是造成越界。而vector正好弥补了这个缺陷，它的特征是相当于可分配拓展的数组，它的随机访问快，在中间插入和删除慢，但在末端插入和删除快，而且如果你用.at()访问的话，也可以做越界检查。

v.push_back(t) 在数组的最后添加一个值为t的数据

v.size() 当前使用数据的大小

v.pop_back(); // 弹出容器中最后一个元素（容器必须非空）

v.back(); // 返回容器中最后一个元素的引用

/*这里专门就算法中的大数问题进行一个统一归纳*/ #include
   
    #include
    
     #include
     
      using namespace std; //结果支持的最大位数//这个可以依据具体需求调整const static int M = 2000;  int numA[M];int numB[M];  //使用string重置numAvoid resetNumA(string numAStr){
   	memset(numA,0,M*sizeof(int)); 	//将字符串的每一位都转换成数字传入数组	for (int i = 0; i < numAStr.length(); i++)	{
   		numA[i] = numAStr[numAStr.length()-i-1] - '0';	}} //使用string重置numBvoid resetNumB(string numBStr){
   	memset(numB,0,M*sizeof(int)); 	//将字符串的每一位都转换成数字传入数组	for (int i = 0; i < numBStr.length(); i++)	{
   		numB[i] = numBStr[numBStr.length()-i-1] - '0';	}}  //将数组转换为字符串，用于输出string getNumString(int* num){
   	string numString;	bool isBegin = false;	for (int i = M-1; i >= 0 ; i--)	{
   		if(num[i] != 0)		{
   			isBegin = true;		} 		if (isBegin)		{
   			numString += num[i] +'0';		}	}	return numString;} //判断两个数字哪个大int compare(string numAStr,string numBStr){
   	if (numAStr.length() > numBStr.length())	{
   		return 1;	}	else if (numAStr.length() < numBStr.length())	{
   		return -1;	}	else	{
   		for (int i = 0; i < numAStr.length(); i++)		{
   			if (numAStr[i]>numBStr[i])			{
   				return 1;			}						if (numAStr[i]
      
       9)		{
   			numA[i] -= 10;			numA[i+1]++;		}	} 	return getNumString(numA);} //减法string minus(string numAStr, string numBStr){
   	bool isNegative = false; 	//如果numA比numB小	//则结果为负数	//调换位置进行计算	if (compare(numAStr,numBStr)==-1)	{
   		isNegative = true;		string temp = numAStr;		numAStr = numBStr;		numBStr = temp;	}	else if (compare(numAStr,numBStr)==0)	{
   		return "0";	} 	resetNumA(numAStr);	resetNumB(numBStr);   	for (int i = 0; i < M; i++)	{
   		//减数小于被减数就借位		if (numA[i]
       
         nums;	for (int i = 0; i < numBStr.length(); i++)	{
   		//初始化一个临时数据来保存被乘数与乘数的某一位相乘的结果		int temp[M];		memset(temp,0,M*sizeof(int));  		for (int j = i; j < numAStr.length()+i; j++)		{
   			temp[j] += numA[j-i]*numB[i]%10; 			temp[j+1] = numA[j-i]*numB[i]/10; 			//如果大于9，那么就做进位处理			if (temp[j]>9)			{
   				temp[j]-=10;				temp[j+1]++;			}		}		nums.push_back(getNumString(temp));	} 	//每位相乘的结果再用加法加起来	string result = nums[0];	for (int i = 1; i < nums.size(); i++)	{
   		result = plus(result,nums[i]);	} 	return result;}   //除,结果精确到个位string div(string numAStr, string numBStr){
   	resetNumA(numAStr);	resetNumB(numBStr); 	string result;	string left; 	if (compare(numAStr,numBStr)==-1)	{
   		return "0";	} 	//标记第一个不为0的位数的出现	bool flag = false;	for (int i = 0; i < numAStr.length(); i++)	{
   		left +=numAStr[i]; 		//余数比除数大		if (compare(left,numBStr)==1)		{
   			flag = true; 			int count = 1;			string temp = numBStr; 			while (true)			{
   				//每循环一次加上一个余数				temp = plus(temp,numBStr); 				//余数仍然大于除数，继续累加				if (compare(left,temp)==1)				{
   					count++;				}				//余数小于除数				//可以计算结果				else if (compare(left,temp)==-1)				{
   					result += count + '0';					left = minus(left, minus(temp,numBStr));					break;				}				//此时余数刚好是除数的倍数				else if (compare(left,temp) == 0)				{
   					count ++;					result += count + '0';					left = "";					break;				}			}		}		//刚好除尽		else if(compare(left,numBStr)==0)		{
   			flag = true;			result +="1";			left = "";		}		//余数比除数小，跳到下一位		else if(flag)		{
   			result +="0";		}  	} 	return result;} void printTitle(){
   	cout << endl;	cout << "输入1：加法" << endl;	cout << "输入2：减法" << endl;	cout << "输入3：乘法" << endl;	cout << "输入4：除法" << endl;	cout << "选择 : ";} int main(){
    	string numAStr,numBStr;	string operation;	string result; 	printTitle(); 	while (cin >> operation)	{
   		cout << "输入两个数字： ";		cin >> numAStr >> numBStr; 		if(operation == "1")		{
   			result = plus(numAStr,numBStr);		}		else if(operation == "2")		{
   			result = minus(numAStr,numBStr);		}		else if(operation == "3")		{
   			result = mul(numAStr,numBStr);		}		else 		{
   			result = div(numAStr,numBStr);		} 		cout << "结果为：" << result << endl; 		printTitle();	}}
       
      
     
    
   

黄金连分数：

#include 
   
    #include 
    
     //大数乘10void mul10(unsigned char* a) {
   	short i = 0;	for (; i < 99; i++)		a[i] = a[i + 1];	a[99] = 0;}//大数乘个位数ivoid mul_i(unsigned char* a, unsigned char* dest, unsigned char i) {
   	short v = 0;	unsigned char all, num = 0, point;	if (i == 0) {
   		memset(dest, 0, 100);		return;	}	for (v = 99; v >= 0; v--) {
   		all = a[v] * i + num;		num = all / 10;		point = all - num * 10;		dest[v] = point;	}}//大数相减void sub(unsigned char* a, unsigned char* b) {
   	short i = 99;	unsigned char flag = 0;	for (; i >= 0; i--) {
   		if (a[i] - flag < b[i]) {
   			a[i] = a[i] + 10 - flag - b[i];			flag = 1;		}		else {
   			a[i] = a[i] - flag - b[i];			flag = 0;		}	}}//大数相加void add(unsigned char* a, unsigned char* b) {
   	short i = 99, all;	for (; i > 0; i--) {
   		all = a[i] + b[i];		if (all >= 10) {
   			a[i - 1]++;			a[i] = all - 10;		}		else			a[i] = all;	}	a[0] = (a[0] + b[0]) % 10;}//大数相除第一个小数unsigned char div_a_b(unsigned char* a, unsigned char* b) {
   	unsigned char i = 1;	unsigned char tmp[100];	unsigned char tmp1[100] = {
    0 };	memcpy(tmp, a, 100);	mul10(tmp);	for (; i <= 10; i++) {
   		mul_i(b, tmp1, i);		if (memcmp(tmp1, tmp, 100) > 0) {
   			mul10(a);			memset(tmp1, 0, 100);			mul_i(b, tmp1, i - 1);			sub(a, tmp1);			return i - 1;		}		memset(tmp1, 0, 100);	}}//大数是否为零bool isZero(unsigned char* a) {
   	short i = 99;	for (; i >=0; i--)		if (a[i] != 0)			return false;	return true;}//保留100位小数相除void div100(unsigned char* a, unsigned char* b, unsigned char* res) {
   	unsigned char i = 0;	unsigned char tmp[100];	memcpy(tmp, a, 100);	for (; i < 101; i++) {
   		res[i] = div_a_b(tmp, b);		if (isZero(tmp))			return;	}}//信息打印void printRes(unsigned char* a, unsigned char* b, unsigned char* res) {
   	static int count = 0;	int i = 0;	bool flag = false;		printf("%d: ", count++);	for (; i < 100; i++) {
   		if (!flag&&a[i])			flag = true;		if (flag)			printf("%d", a[i]);	}	printf("/");	flag = false;	for (i = 0; i < 100; i++) {
   		if (!flag&&b[i])			flag = true;		if (flag)			printf("%d", b[i]);	}	printf("\n");	printf("0.");	for (i = 0; i < 101; i++)		printf("%d", res[i]);	printf("\n");}int main(){
   	short index;	//使用数组作为大数，res存放结果小数bufen	unsigned char a[100] = {
    0 };	unsigned char b[100] = {
    0 };	unsigned char tmp[100] = {
    0 };	unsigned char pre[101] = {
    0 };	unsigned char res[101] = {
    0 };	a[99] = 1;	b[99] = 2;	do {
   		memcpy(pre, res, 101);		memset(res, 0, 101);		div100(a, b, res);		printRes(a, b, res);		memcpy(tmp, a, 100);		memcpy(a, b, 100);		add(b, tmp);	} while (memcmp(pre, res, 101));	if (res[100] >= 5)		res[99]++;	index = 99;	while (res[index--] == 10) {
   		res[index + 1] = 0;		res[index]++;	}	printf("最终结果为:\n0.");	for (index = 0; index < 100; index++)		printf("%d", res[index]);	printf("\n");    return 0;}
    
   

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