# 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(711) 评论(1)  编辑 收藏 引用 ### 评论

# re: HDU 1002 2011-05-17 21:12 bell