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大整数加法——带有负数的

#include "iostream"

#include "string"

#include "cstring"

using namespace std;

#define maxlen 2001

int a[maxlen];

int b[maxlen];

int len1, len2, i, j;

int bigger(int a, int b)

{

return a>b?a:b;

}

void Add(int underzero)

{

if(underzero)

cout<<\'-\';

int big = bigger(len1, len2);

for(i=0; i<=big; i++)

{

a[i] += b[i];

if(a[i]>=10)

{

a[i]-=10;

a[i+1]++;

}

}

bool flag = false;

for(i=big; i>=0; i--)

{

if(flag)

cout<<a[i];

else if(a[i])

{

cout<<a[i];

flag = true;

}

}

cout<<endl;

}

 void Minus(int aa[], int bb[], int biglen, int underzero)

{

if(underzero)

cout<<\'-\';

for(i=0; i<=biglen; i++)

{

if(aa[i]>=bb[i])

aa[i] -= bb[i];

else 

{

aa[i] = aa[i]+10-bb[i];

aa[i+1]--;

}

}

bool flag = false;

for(i=biglen; i>=0; i--)

{

if(flag)

cout<<aa[i];

else if(aa[i])

{

cout<<aa[i];

flag = true;

}

}

cout<<endl;

}

int main()

{

string s1, s2;

bool success;

while(cin>>s1>>s2)

{

memset(a, 0, sizeof(a));

memset(b, 0, sizeof(b));

len1 = s1.length();

len2 = s2.length();

for(i=len1-1, j=0; i>0; i--)

a[j++] = s1[i]-\'0\';

if(s1[0]!=\'-\')

a[j] = s1[0]-\'0\';

for(i=len2-1, j=0; i>0; i--)

b[j++] = s2[i]-\'0\';

if(s2[0]!=\'-\')

b[j] = s2[0]-\'0\';

if(s1[0]==\'-\' && s2[0]==\'-\')

Add(1);  //和为负

else if(s1[0]!=\'-\' && s2[0]!=\'-\')

Add(0);  //和为正

else if(s1[0]==\'-\')

{

if(len1-1 > len2) //负数大

Minus(a, b, len1-1, 1);

else if(len1-1 < len2) //正数大

Minus(b, a, len2, 0);

else

{

success = true; 

for(i=len1-2, j=len2-1; j>=0; j--, i--) //这里的len-2是个关键,len-1是长度,之所以再减去1,是要减去负号的位置

if(a[i]>b[j]) //负数大

{

success = false;

Minus(a, b, len2, 1);

break;

}

else if(a[i]<b[j]) //正数大

{

success = false;

Minus(b, a, len2, 0);

break;

}

}

if(success) //相等

cout<<"0"<<endl;

}

}

else if(s2[0]==\'-\')

{

if(len2-1 > len1) //负数大

Minus(b, a, len2-1, 1);

else if(len2-1 < len1) //正数大

Minus(a, b, len1, 0);

else

{

success = true;

for(i=len2-2, j=len1-1; j>=0; j--, i--) //这里的len-2是个关键,len-1是长度,之所以再减去1,是要减去负号的位置

{

if(a[j]>b[i]) //正数大

{

success = false;

Minus(a, b, len1, 0);

break;

}

else if(a[j]<b[i]) //负数大

{

success = false;

Minus(b, a, len1, 1);

break;

}

}

if(success) //相等

cout<<"0"<<endl;

}

}

}

return 0;

}

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