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#include <stdio.h>
#include <string.h>
int judge_value(char c)
{
int x;
if(c >= '0' && c <= '9')
x = c - '0';
else if(c >= 'A' && c <= 'F')
x = c - 'A' + 10;
else if(c >= 'a' && c <= 'f')
x = c - 'a' + 10;
else
{
printf("charactor error!\n");
return -1;
}
return x;
}
void printf_array(char a[])
{
int i = 0;
while(a[i] != '\0')
printf("%c", a[i++]);
}
char * get_result(char *res, char *dest)
{
char *pres = res;
char *pdest = dest;
if(res == NULL || *res == '\0')
{
perror("NULL parameter");
return NULL;
}
while(*pres)
pres++;
pres--;
while(pres >= res)
{
if(*pres == '0')
{
pres--;
}
else
break;
}
while(pres >= res)
{
*pdest++ = *pres--;
}
return dest;
}
int main()
{
char tmp1[64] = "0";
char tmp2[64] = "0";
char result[64] = "\0";
//char *s = "ABCDEF456789123";
char s[64] = {};
char *ps , *pr = result;
long count = 1, i = 0;
int x = 0, k = 0, j = 0, is_overflow = 0;
gets(s);
ps = s;
while(*ps)
ps++;
ps--;
memset(tmp1, '0', sizeof(tmp1)-1);
tmp1[255] = '\0';
memset(tmp2, '0', sizeof(tmp2)-1);
tmp2[255] = '\0';
while(ps >= s)
{
x = judge_value(*ps); //字符‘1’ 转换成 1
i = x * count;
do
{
if(i/10 > 0)
tmp1[k++] = i%10+'0';
else if(i/10 == 0)
{
tmp1[k] = i+'0';
break;
}
}while((i=i/10) > 0); //把算得的数按位存放在tmp1数组中
while(j<=k)
{
if(is_overflow)
{
tmp2[j] = (tmp1[j] -'0') + (tmp2[j] -'0') + 1 + '0'; //tmp1和tmp2 按位相加, 并存放在tmp2数组中
is_overflow = 0;
}
else
tmp2[j] = (tmp1[j] - '0') + (tmp2[j]-'0') + '0'; //tmp1和tmp2 按位相加, 并存放在tmp2数组中
if(tmp2[j] > '9')//such as 7+ 8 = 15 //数据溢出处理
{
tmp2[j] = ((tmp2[j]-'0')%10) + '0';
is_overflow = 1;
}
j++;
}
if(is_overflow)
tmp2[j] = '1';
ps--;
count *= 16;
memset(tmp1, 0, sizeof(tmp1)-1);
k = 0;
i = j = x = 0;
is_overflow = 0;
}
get_result(tmp2, result); // 把tmp2中的数据 倒置, 就得到所求的十进制数字
printf("result:");
printf_array(result);
printf("\n");
return 0;
}
分享代码, 快乐无限~#include <iostream>
#include <string>
using namespace std;
inline int compare(string str1,string str2) {//相等返回0,大于返回1,小于返回-1
if (str1.size()>str2.size()) return 1; //长度长的整数大于长度小的整数
else if (str1.size()<str2.size()) return -1;
else return str1.compare(str2); //若长度相等,则头到尾按位比较
}
string SUB_INT(string str1,string str2);
string ADD_INT(string str1,string str2) {//高精度加法
int sign=1; //sign 为符号位
string str;
if (str1[0]=='-') {
if (str2[0]=='-') {
sign=-1;
str=ADD_INT(str1.erase(0,1),str2.erase(0,1));
} else {
str=SUB_INT(str2,str1.erase(0,1));
}
} else {
if (str2[0]=='-') {
str=SUB_INT(str1,str2.erase(0,1));
} else { //把两个整数对齐,短整数前面加0补齐
string::size_type L1,L2;
int i;
L1=str1.size();
L2=str2.size();
if (L1<L2) {
for (i=1;i<=L2-L1;i++) str1="0"+str1;
} else {
for (i=1;i<=L1-L2;i++) str2="0"+str2;
}
int int1=0,int2=0; //int2 记录进位
for (i=str1.size()-1;i>=0;i--) {
int1=(int(str1[i])-'0'+int(str2[i])-'0'+int2)%10;
int2=(int(str1[i])-'0'+int(str2[i])-'0'+int2)/10;
str=char(int1+'0')+str;
}
if (int2!=0) str=char(int2+'0')+str;
}
}
//运算后处理符号位
if ((sign==-1)&&(str[0]!='0')) str="-"+str;
return str;
}
string SUB_INT(string str1,string str2) {//高精度减法
int sign=1; //sign 为符号位
string str;
int i,j;
if (str2[0]=='-') {
str=ADD_INT(str1,str2.erase(0,1));
} else {
int res=compare(str1,str2);
if (res==0) return "0";
if (res<0) {
sign=-1;
string temp =str1;
str1=str2;
str2=temp;
}
string::size_type tempint;
tempint=str1.size()-str2.size();
for (i=str2.size()-1;i>=0;i--) {
if (str1[i+tempint]<str2[i]) {
j=1;
while (1) {//zhao4zhong1添加
if (str1[i+tempint-j]=='0') {
str1[i+tempint-j]='9';
j++;
} else {
str1[i+tempint-j]=char(int(str1[i+tempint-j])-1);
break;
}
}
str=char(str1[i+tempint]-str2[i]+':')+str;
} else {
str=char(str1[i+tempint]-str2[i]+'0')+str;
}
}
for (i=tempint-1;i>=0;i--) str=str1[i]+str;
}
//去除结果中多余的前导0
str.erase(0,str.find_first_not_of('0'));
if (str.empty()) str="0";
if ((sign==-1) && (str[0]!='0')) str ="-"+str;
return str;
}
string MUL_INT(string str1,string str2) {//高精度乘法
int sign=1; //sign 为符号位
string str;
if (str1[0]=='-') {
sign*=-1;
str1 =str1.erase(0,1);
}
if (str2[0]=='-') {
sign*=-1;
str2 =str2.erase(0,1);
}
int i,j;
string::size_type L1,L2;
L1=str1.size();
L2=str2.size();
for (i=L2-1;i>=0;i--) { //模拟手工乘法竖式
string tempstr;
int int1=0,int2=0,int3=int(str2[i])-'0';
if (int3!=0) {
for (j=1;j<=(int)(L2-1-i);j++) tempstr="0"+tempstr;
for (j=L1-1;j>=0;j--) {
int1=(int3*(int(str1[j])-'0')+int2)%10;
int2=(int3*(int(str1[j])-'0')+int2)/10;
tempstr=char(int1+'0')+tempstr;
}
if (int2!=0) tempstr=char(int2+'0')+tempstr;
}
str=ADD_INT(str,tempstr);
}
//去除结果中的前导0
str.erase(0,str.find_first_not_of('0'));
if (str.empty()) str="0";
if ((sign==-1) && (str[0]!='0')) str="-"+str;
return str;
}
string DIVIDE_INT(string str1,string str2,int flag) {//高精度除法。flag==1时,返回商; flag==0时,返回余数
string quotient,residue; //定义商和余数
int sign1=1,sign2=1;
if (str2 == "0") { //判断除数是否为0
quotient= "ERROR!";
residue = "ERROR!";
if (flag==1) return quotient;
else return residue ;
}
if (str1=="0") { //判断被除数是否为0
quotient="0";
residue ="0";
}
if (str1[0]=='-') {
str1 = str1.erase(0,1);
sign1 *= -1;
sign2 = -1;
}
if (str2[0]=='-') {
str2 = str2.erase(0,1);
sign1 *= -1;
}
int res=compare(str1,str2);
if (res<0) {
quotient="0";
residue =str1;
} else if (res == 0) {
quotient="1";
residue ="0";
} else {
string::size_type L1,L2;
L1=str1.size();
L2=str2.size();
string tempstr;
tempstr.append(str1,0,L2-1);
for (int i=L2-1;i<L1;i++) { //模拟手工除法竖式
tempstr=tempstr+str1[i];
tempstr.erase(0,tempstr.find_first_not_of('0'));//zhao4zhong1添加
if (tempstr.empty()) tempstr="0";//zhao4zhong1添加
for (char ch='9';ch>='0';ch--) { //试商
string str;
str=str+ch;
if (compare(MUL_INT(str2,str),tempstr)<=0) {
quotient=quotient+ch;
tempstr =SUB_INT(tempstr,MUL_INT(str2,str));
break;
}
}
}
residue=tempstr;
}
//去除结果中的前导0
quotient.erase(0,quotient.find_first_not_of('0'));
if (quotient.empty()) quotient="0";
if ((sign1==-1)&&(quotient[0]!='0')) quotient="-"+quotient;
if ((sign2==-1)&&(residue [0]!='0')) residue ="-"+residue ;
if (flag==1) return quotient;
else return residue ;
}
string DIV_INT(string str1,string str2) {//高精度除法,返回商
return DIVIDE_INT(str1,str2,1);
}
string MOD_INT(string str1,string str2) {//高精度除法,返回余数
return DIVIDE_INT(str1,str2,0);
}
int main() {
char ch;
string s1,s2,res;
while (cin>>s1>>ch>>s2) {
switch (ch) {
case '+':res=ADD_INT(s1,s2);break;
case '-':res=SUB_INT(s1,s2);break;
case '*':res=MUL_INT(s1,s2);break;
case '/':res=DIV_INT(s1,s2);break;
case '%':res=MOD_INT(s1,s2);break;
default : break;
}
cout<<res<<endl;
}
return(0);
}