# tctony

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# A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9348    Accepted Submission(s): 1557

Problem Description
I have a very simple problem for you. Given two integers A and B, your job is to calculate the Sum of A + B.

Input
The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line consists of two positive integers, A and B. Notice that the integers are very large, that means you should not process them by using 32-bit integer. You may assume the length of each integer will not exceed 1000.

Output
For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line is the an equation "A + B = Sum", Sum means the result of A + B. Note there are some spaces int the equation. Output a blank line between two test cases.

Sample Input
```2
1 2
112233445566778899 998877665544332211```

Sample Output
```Case 1:
1 + 2 = 3
Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110```

#include<iostream>
using namespace std;

#define Base 10
#define MaxLen  1000

struct BigInt {

int len;

int data[MaxLen];

BigInt() :len(
0{}
BigInt(
const BigInt& s) :len(s.len){
memcpy(
this->data,s.data,len*sizeof(*data));
}

BigInt(
int s) :len(0){

for(;s>0;s=s/Base)
data[len
++]=s%Base;
}

BigInt
& operator = (const BigInt& s){

this->len=s.len;
memcpy(
this->data,s.data,len*sizeof(*data));

return *this;
}

int& operator [](int index)return data[index]; }

int operator [](int index) const return data[index]; }
}
;

BigInt
operator + (BigInt & a, BigInt & b){
BigInt c;

int i,carry=0;
c.len
=a.len>b.len?a.len:b.len;

for(i=0;i<c.len||carry>0;++i){

if(i<a.len)    carry+=a[i];

if(i<b.len)    carry+=b[i];
c[i]
=carry%Base;
carry
/=Base;
}

c.len
=i;

return c;
}

istream
& operator >> (istream & is, BigInt & s)
{

int i;

char ch[MaxLen];

is>>ch;
s.len
=strlen(ch);

for(i=s.len-1;i>-1;--i)
s[i]
=ch[s.len-1-i]-'0';

return is;
}

ostream
& operator << (ostream & os, BigInt & s)
{

int i;

for(i=s.len-1;i>-1;--i)
os
<<s[i];

return os;
}

int main () {

int n,i;
BigInt a,b,c;

cin
>>n;

for(i=1;i<n;++i){
cin
>>a>>b;
c
=a+b;
cout
<<"Case "<<i<<":"<<endl;
cout
<<a<<" + "<<b<<" = "<<c<<endl<<endl;
}

cin
>>a>>b;
c
=a+b;
cout
<<"Case "<<i<<":"<<endl;
cout
<<a<<" + "<<b<<" = "<<c<<endl;

return 0;
}

posted on 2007-11-28 18:57 tctony 阅读(725) 评论(1)  编辑 收藏 引用

### 评论

# re: HDU 1002 2011-05-17 21:12 bell