长整数的加减法，求助！

[qg159357qg]

用字符串实现长整数的存储，考虑输入输出方向，＋－号的解决方法，求代码，求方法，求解释！

[gongheguoyingpai11]

给我发邮件给你我之前写的[e-mail]

[wxwlll]

初步认为用字符串模拟，循环位移操作

[cliffbaby]

void Add(char *str1, char *str2, char *str3)
{// str3 = str1 + str2;
     int i, j, i1, i2, tmp, carry;
int len1 = strlen(str1), len2 = strlen(str2);
char ch;
i1 = len1-1; i2 = len2-1;
j = carry = 0;
for( ; i1 >= 0 && i2 >= 0; ++j, --i1, --i2 ){
tmp = str1[i1]-'0'+str2[i2]-'0'+carry;
carry = tmp/10;
str3[j] = tmp%10+'0';
}
while( i1 >= 0 ){
tmp = str1[i1--]-'0'+carry;
carry = tmp/10;
str3[j++] = tmp%10+'0';
}
while( i2 >= 0 ){
tmp = str2[i2--]-'0'+carry;
carry = tmp/10;
str3[j++] = tmp%10+'0';
}
if( carry ) str3[j++] = carry+'0';
str3[j] = '\0';
}
for( i=0, --j; i < j; ++i, --j ){
ch = str3[i]; str3[i] = str3[j]; str3[j] = ch;
}
void Minus(char *str1, char *str2, char *str3)
{// str3 = str1-str2 (str1 > str2)
int i, j, i1, i2, tmp, carry;
int len1 = strlen(str1), len2 = strlen(str2);
char ch;
i1 = len1-1; i2 = len2-1;
j = carry = 0;
while( i2 >= 0 ){
tmp = str1[i1]-str2[i2]-carry;
if( tmp < 0 ) {
str3[j] = tmp+10+'0'; carry = 1;
}
else {
str3[j] = tmp+'0'; carry = 0;
}
--i1; --i2; ++j;
}
while( i1 >= 0 ){
tmp = str1[i1]-'0'-carry;
if( tmp < 0 ) {
str3[j] = tmp+10+'0'; carry = 1;
}
else{
str3[j] = tmp+'0'; carry = 0;
}
--i1; ++j;
}
--j;
while( str3[j] == '0' && j > 0 ) --j;
str3[++j] = '\0';
for( i=0, --j; i < j; ++i, --j ){
ch = str3[i]; str3[i] = str3[j]; str3[j] = ch;
}
}

[cliffbaby]

一个加一个减，奇怪为什么没有了缩格。

[flowingfirefly]

没看懂题意

[qg159357qg]

2L,能实现负数的加减吗？谢谢你呢，老师让我回家前交，写不到啊

[qg159357qg]

[Quote=引用 3 楼 cliffbaby 的回复:]
一个加一个减，奇怪为什么没有了缩格。
[/Quote]嘿嘿，脑袋不好使！