杭电水题 1002 大数加法

杭电acm1002
A + B Problem II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 60478 Accepted Submission(s): 11083

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
注意输出时的格式问题
代码一：
#include <iostream>
#include <string.h>
using namespace std;

int main()
{
char str1[1000],str2[1000],str3[1000],str4[1000];
char sum1[1000],sum2[1000];
int n;
int p = 1;
int temp;
int flag = 0;
int len1 = 0;
int len2 = 0;
cin>>n;
while(n--)
{
flag = 0;
memset(str1,'0',sizeof(str1));
memset(str2,'0',sizeof(str2));
memset(str3,'0',sizeof(str3));
memset(str4,'0',sizeof(str4));
memset(sum1,'0',sizeof(sum1));
memset(sum2,'\0',sizeof(sum2));
cin>>str1;
cin>>str2;
len1 = strlen(str1);
len2 = strlen(str2);
for(int i = 0;i < len1;i++)
{
str3[i] = str1[len1-1-i];
}
for(int j = 0;j < len2;j++)
{
str4[j] = str2[len2-1-j];
}

if(len1 > len2)
{
for(int i = 0;i < len1;i++)
{
temp = sum1[i] - '0';
sum1[i] = (temp + (str3[i] -'0') + (str4[i] - '0'))%10 + '0';
if(((temp + str3[i] -'0' + str4[i] - '0') > 10)||((temp + str3[i] -'0' + str4[i] - '0') == 10))
{
sum1[i+1]++;
}
}
flag = 0;
if(sum1[len1] != '0')
flag = 1;
for(int m = 0;m < len1 + flag;m++)
{
sum2[m] = sum1[len1 + flag - 1 - m];
}
}
else
{
for(int i = 0;i < len2;i++)
{
temp = sum1[i] -'0';
sum1[i] = (temp + (str3[i] -'0') + (str4[i] - '0'))%10 + '0';
if(((temp + str3[i] -'0' + str4[i] - '0') > 10)||((temp + str3[i] -'0' + str4[i] - '0') == 10))
{
sum1[i+1]++;
}
}
flag = 0;
if(sum1[len2] != '0')
flag = 1;
for(int m = 0;m < len2 + flag;m++)
{
sum2[m] = sum1[len2 + flag - 1 - m];
}
}
if(n == 0)
{
cout<<"Case "<<p++<<":"<<endl;
cout<<str1<<" + "<<str2<<" = "<<sum2<<endl;
}
else
{
cout<<"Case "<<p++<<":"<<endl;
cout<<str1<<" + "<<str2<<" = "<<sum2<<endl<<endl;
}
}
return 0;
}
代码2：
#include<stdio.h>
#include<string.h>
int shu(char a)
{
return (a-'0');
}
int main(){
char a[1000],b[1000];
int num[1001];
int n,i,j=1,al,bl,k,t;
scanf("%d",&n);
while(n--)
{
getchar();
if(j!=1)
printf("\n");
scanf("%s",a);
al=strlen(a);
scanf("%s",b);
bl=strlen(b);
k=(al>bl)?al:bl;
for(i=0;i<=k;i++)
num[i]=0;
t=k;
for(k;al>0&&bl>0;k--)
{
num[k]+=shu(a[--al])+shu(b[--bl]);

if(num[k]/10)
{
num[k-1]++;
num[k]%=10;
}
}
while(al>0)
{
num[k--]+=shu(a[--al]);
if(num[k+1]/10)
{
num[k]++;
num[k+1]%=10;
}
}
while(bl>0)
{
num[k--]+=shu(b[--bl]);
if(num[k+1]/10)
{
num[k]++;
num[k+1]%=10;
}
}

printf("Case %d:\n",j++);
printf("%s + %s = ",a,b);
for(i=0;i<=t;i++)
{
if(i==0&&num[i]==0)
i++;
printf("%d",num[i]);
}
printf("\n");
}
return 0;
}