A + B Problem II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 204234 Accepted Submission(s): 39250
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.
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.
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
/*
1002 A + B Problem II
大数加法？
*/
#include<iostream>
#include<cstring>
using namespace std;
char a[1001],b[1001];
int c[1001];
int main(){
int i,j,n,k,z,len1,len2,t,num;
cin>>n;
num=1;
while(num<=n)
{
scanf("%s %s",a,b);
len1=strlen(a)-1;
len2=strlen(b)-1;
k=0;
z=0;
for(;len1>=0&&len2>=0;len1--,len2--)
{
t=(a[len1]-'0')+(b[len2]-'0')+z;
c[k++]=t%10;
z=t/10;
}
while(len1>=0)
{
t=a[len1]-'0'+z;
c[k++]=t%10;
z=t/10;
len1--;
}
while(len2>=0)
{
t=b[len2]-'0'+z;
c[k++]=t%10;
z=t/10;
len2--;
}
printf("Case %d:\n",num++);
printf("%s + %s = ",a,b);
for(i=k-1;i>=0;i--)
{
cout<<c[i];
}
printf("\n");
if(num<=n) printf("\n");
}
return 0;
}