c++高精度小数减法问题
程序主要要求 主程序读入两个字符串类型的大数(小数)，通过函数实现小数的高精度减法。求完整的代码和注释，比较急，谢谢
你这是什么编辑器啊。连作用域都不分，我把变量名都改了，代码如下：
#include <iostream>
using namespace std;
#define MAX_NUMBER_LEN 1024
bool isGreaterThan(const int* pNum1, const int* pNum2)
{
if (pNum1 == NULL || pNum2 == NULL)
{
cout << "您输入的数字有误" << endl;
return false;
}
for(int nIndx = 0; nIndx < MAX_NUMBER_LEN; ++nIndx)
{
if (pNum1[nIndx] > pNum2[nIndx])
return true;
}
return false;
}
int main()
{
char szNum1[1024] = {0};
char szNum2[1024] = {0};
int arNumber1[1024] = {0};
int arNumber2[1024] = {0};
cout << "请输入两个小数(小数位数不要超过500位)，空格分隔：" << endl;
cin >> szNum1;
cin >> szNum2;
// 把第一个字符串转化成数组
int nCount = 0;
int nNumLen = strlen(szNum1);
bool bFindDot = false;
for (int nIndex = 0, pos = nIndex + 1; nIndex < nNumLen; ++nIndex)
{
if (szNum1[nIndex] == '.')
{
bFindDot = true;
}
else
{
if(szNum1[nIndex] < '0' || szNum1[nIndex] > '9')
{
cout << "请输入不是正确的数字" << endl;
return 0;
}
else
{
arNumber1[pos] = szNum1[nIndex] - '0';
}
pos++;
}
if (!bFindDot)
nCount++;
}
arNumber1[0] = nCount;
// 把第二个字符串转化成数组
nCount = 0;
nNumLen = strlen(szNum2);
bFindDot = false;
for (int nIndex1 = 0, pos = nIndex1 + 1; nIndex1 < nNumLen; ++nIndex1)
{
if (szNum2[nIndex1] == '.')
{
bFindDot = true;
}
else
{
if(szNum2[nIndex1] < '0' || szNum2[nIndex1] > '9')
{
cout << "请输入不是正确的数字" << endl;
break;
}
else
{
arNumber2[pos] = szNum2[nIndex1] - '0';
}
pos++;
}
if (!bFindDot)
nCount++;
}
arNumber2[0] = nCount;
bool isGreater = isGreaterThan(arNumber1, arNumber2);
// 比较小数前的位数
if (arNumber1[0] != arNumber2[0])
{
int nshift = abs(arNumber1[0] - arNumber2[0]);
int *pNum;
int nTotal;
if(isGreater)
{
pNum = arNumber2;
nTotal = strstr(szNum2, ".") ? strlen(szNum2) - 1 : strlen(szNum2);
}
else
{
pNum = arNumber1;
nTotal = strstr(szNum1, ".") ? strlen(szNum1) - 1 : strlen(szNum1);
}
for(int nIndex2 = nTotal; nIndex2 >= 0; --nIndex2)
{
pNum[nIndex2 + nshift] = pNum[nIndex2];
}
for(int nIndex2 = nshift; nIndex2 >= 1; --nIndex2)
{
pNum[nIndex2] = 0;
}
}
// 做简单减法
int nlen1 = strstr(szNum1, ".") ? strlen(szNum1) - 1 : strlen(szNum1);
int nlen2 = strstr(szNum2, ".") ? strlen(szNum2) - 1 : strlen(szNum2);
int forcount= nlen1 > nlen2 ?nlen1 : nlen2;
if (isGreater)
{
for (int nIndex3 = 1; nIndex3 <= forcount; nIndex3++)
{
arNumber1[nIndex3] = arNumber1[nIndex3] - arNumber2[nIndex3];
}
}
else
{
for (int nIndex4 = 1; nIndex4 <= forcount; nIndex4++)
{
arNumber1[nIndex4] = arNumber2[nIndex4] - arNumber1[nIndex4];
}
}
// 处理减法中的负数
arNumber1[0] = arNumber1[0] > arNumber2[0] ? arNumber1[0] : arNumber2[0];
for ( int nIndex5 = forcount; nIndex5 > 0; --nIndex5)
{
if (arNumber1[nIndex5] < 0)
{
arNumber1[nIndex5] += 10;
arNumber1[nIndex5 - 1] -= 1;
}
}
cout << "结果为" << endl;
if (!isGreater)
cout << "-";
//输出结果
bool zerooutpput = false;
bool putpucount = 0;
for( int nIndex6 = 1; nIndex6 <= forcount; nIndex6++ )
{
if (arNumber1[nIndex6] != 0 || nIndex6 > arNumber1[0])
{
zerooutpput = true;
cout << arNumber1[nIndex6];
putpucount++;
}
else if (zerooutpput)
{
cout << arNumber1[nIndex6];
putpucount++;
}
if (nIndex6 == arNumber1[0])
{
if (putpucount == 0)
cout << "0";
if (nIndex6 != forcount)
cout << '.';
}
}
system("pause");
return 0;
}
这个和BigInteger的加法减法是一样的原理，只不过小数减法要考虑小数点对齐，用STL中的deque容器实现应该比较简单，直接用C语言的数组也能做处理。你百度一下"C++ BigInteger"应该能找到示例代码
写的有点仓促，凑活着看
#include <iostream>
using namespace std;
#define MAX_NUMBER_LEN 1024
bool isGreaterThan(const int* pNum1, const int* pNum2)
{
if (pNum1 == NULL || pNum2 == NULL)
{
cout << "您输入的数字有误" << endl;
return false;
}
for(int nIndx = 0; nIndx < MAX_NUMBER_LEN; ++nIndx)
{
if (pNum1[nIndx] > pNum2[nIndx])
return true;
}
return false;
}
int main()
{
char szNum1[1024] = {0};
char szNum2[1024] = {0};
int arNumber1[1024] = {0};
int arNumber2[1024] = {0};
cout << "请输入两个小数(小数位数不要超过500位)，空格分隔：" << endl;
cin >> szNum1;
cin >> szNum2;
// 把第一个字符串转化成数组
int nCount = 0;
int nNumLen = strlen(szNum1);
bool bFindDot = false;
for (int nIndex = 0, pos = nIndex + 1; nIndex < nNumLen; ++nIndex)
{
if (szNum1[nIndex] == '.')
{
bFindDot = true;
}
else
{
if(szNum1[nIndex] < '0' || szNum1[nIndex] > '9')
{
cout << "请输入不是正确的数字" << endl;
break;
}
else
{
arNumber1[pos] = szNum1[nIndex] - '0';
}
pos++;
}
if (!bFindDot)
nCount++;
}
arNumber1[0] = nCount;
// 把第二个字符串转化成数组
nCount = 0;
nNumLen = strlen(szNum2);
bFindDot = false;
for (int nIndex = 0, pos = nIndex + 1; nIndex < nNumLen; ++nIndex)
{
if (szNum2[nIndex] == '.')
{
bFindDot = true;
}
else
{
if(szNum2[nIndex] < '0' || szNum2[nIndex] > '9')
{
cout << "请输入不是正确的数字" << endl;
break;
}
else
{
arNumber2[pos] = szNum2[nIndex] - '0';
}
pos++;
}
if (!bFindDot)
nCount++;
}
arNumber2[0] = nCount;
bool isGreater = isGreaterThan(arNumber1, arNumber2);
// 比较小数前的位数
if (arNumber1[0] != arNumber2[0])
{
int nshift = abs(arNumber1[0] - arNumber2[0]);
int *pNum;
int nTotal;
if(isGreater)
{
pNum = arNumber2;
nTotal = strlen(szNum2) -1;
}
else
{
pNum = arNumber1;
nTotal = strlen(szNum1) -1;
}
for(int nIndex = nTotal; nIndex >= nshift; --nIndex)
{
pNum[nIndex + nshift] = pNum[nIndex];
}
for(int nIndex = nshift; nIndex >= 1; --nIndex)
{
pNum[nIndex] = 0;
}
}
// 做简单减法
int forcount= strlen(szNum1) > strlen(szNum2) ? (strlen(szNum1) - 1) : (strlen(szNum2) - 1);
if (isGreater)
{
for (int nIndex = 1; nIndex <= forcount; nIndex++)
{
arNumber1[nIndex] = arNumber1[nIndex] - arNumber2[nIndex];
}
}
else
{
for (int nIndex = 1; nIndex <= forcount; nIndex++)
{
arNumber1[nIndex] = arNumber2[nIndex] - arNumber1[nIndex];
}
}
// 处理减法中的负数
arNumber1[0] = arNumber1[0] > arNumber2[0] ? arNumber1[0] : arNumber2[0];
for ( int nIndex = forcount; nIndex > 0; --nIndex)
{
if (arNumber1[nIndex] < 0)
{
arNumber1[nIndex] += 10;
arNumber1[nIndex - 1] -= 1;
}
}
cout << "结果为" << endl;
if (!isGreater)
cout << "-";
//输出结果
bool zerooutpput = false;
bool putpucount = 0;
for( int nIndex = 1; nIndex <= forcount; nIndex++ )
{
if (arNumber1[nIndex] != 0 || nIndex > arNumber1[0])
{
zerooutpput = true;
cout << arNumber1[nIndex];
putpucount++;
}
else if (zerooutpput)
{
cout << arNumber1[nIndex];
putpucount++;
}
if (nIndex == arNumber1[0])
{
if (putpucount == 0)
cout << "0";
cout << '.';
}
}
system("pause");
return 0;
}
有些情况没有考虑到，可以发现了改一下
这是运行截图