【蓝桥杯】黄金连分数（注意：大数）！！！

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。
常用操作：

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numA
void 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numB
void 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]<numBStr[i])
			{
				return -1;
			}
		}
		return 0;
	}
}
 
//加法
string plus(string numAStr, string numBStr)
{
	resetNumA(numAStr);
	resetNumB(numBStr);
 
	for (int i = 0; i < M; i++)
	{
		//结果保存在numA中
		numA[i] += numB[i];
 
		//数字大于9则进位
		if(numA[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]<numB[i])
		{
			numA[i] = numA[i]+10-numB[i]; 
			numA[i+1]--;
		}
		else
		{
			numA[i] -= numB[i];
		}
	}
	if (isNegative)
	{
		return "-"+  getNumString(numA);
	}
	else
	{
		return getNumString(numA);
	}
	
}
 
//乘法
 
string mul(string numAStr, string numBStr)
{
	resetNumA(numAStr);
	resetNumB(numBStr);
 
	vector<string> 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10
void 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i
void 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;
}