HDOJ 1002 的几种方法
1 #include <iostream>
2 #include <string>
3
4 usingnamespace std;
5
6 string LSum ( string, string );
7
8 int main()
9 {
10 string s_1, s_2;
11 int count_inside =0, count_outside;
12 cin >> count_outside ;
13 while ( count_inside != count_outside && cin >> s_1 >> s_2 )
14 {
15 cout <<"Case "<<++count_inside <<":"<< endl;
16 cout << s_1 <<" + "<< s_2 <<" = "<< LSum ( s_1, s_2 ) << endl;
17 if ( count_inside != count_outside ) cout << endl;
18 }
19 return0;
20 }
21
22 string LSum ( string A, string B )
23 {
24 string::size_type i = A.size(), j = B.size();
25 int c =0;
26 string Ans;
27 while ( i || j )
28 {
29 int tmp = c;
30 if ( i ) tmp += A[--i] -48;
31 if ( j ) tmp += B[--j] -48;
32 if ( ( i +1 ) || ( j +1 ) )
33 Ans += ( tmp %10+48 );
34 c = tmp /10;
35 }
36 if ( c ) Ans += c +48;
37 string AnsCopy ( Ans );
38 string::size_type AnsLength = Ans.size();
39 for ( int k = AnsLength; k !=0; --k )
40 Ans[AnsLength-k] = AnsCopy[k-1];
41 return Ans;
42 }
这个是我自己的，没有加工，看完string那一部分后写的。
下面这个是网上找到的，很多个链接都是这个：
#include <iostream>
#include <string.h>
using namespace std;
void add ( char a[], char b[] )
{
char sum[1010] = {' '};
int flg = 0;
int temp = 0;
int len_a = strlen ( a );
int len_b = strlen ( b );
int i = len_a;
int j = len_b;
for ( ; i > 0; i-- )
{
if ( j > 0 )
{
temp = a[i-1] + b[j-1] + flg - 96;
j--;
}
else temp = a[i-1] + flg - 48;
if ( temp >= 10 )
{
flg = 1;
}
else flg = 0;
temp = temp % 10;
sum[i] = temp + 48;
}
if ( flg == 1 ) sum[0] = 49;
i = 0;
while ( i <= len_a )
{
if ( sum[i] != ' ' ) cout << sum[i];
i++;
}
cout << endl;
}
int main()
{
int N;
while ( cin >> N )
{
for ( int i = 1; i <= N; i++ )
{
char a[1000];
char b[1000];
cin >> a;
cin >> b;
int len_a = strlen ( a );
int len_b = strlen ( b );
cout << "Case " << i << ":\n" << a << " + " << b << " = ";
if ( len_a >= len_b )
{
add ( a, b );
}
else add ( b, a );
if ( i != N ) cout << endl;
}
}
return 0;
}
我看了这道题的统计，代码最少的只有380B左右，很想看到代码，可惜看不到，也没在网上找到。
作为一个新手，希望大家能多多指点，谢谢！
