A + B Problem II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 315214 Accepted Submission(s): 61139
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
1 + 2 = 3
Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110
112233445566778899 + 998877665544332211 = 1111111111111111110
Author
Ignatius.L
#include<iostream>
#include<string>
#include<algorithm>
using namespace std;
int main() {
int t,i,k;
cin>>t;
for(k=; k<=t; k++) {
string a,b,c,m,n;
int sum,add;
cin>>a>>b;
m=a;
n=b;
int lena=a.length();
int lenb=b.length();
reverse(a.begin(),a.end());//反转字符串
reverse(b.begin(),b.end());
add=;
//模拟加法运算
for(i=; i<lena||i<lenb; i++) {
if(i<lena&&i<lenb)
sum=a[i]-''+b[i]-''+add;
else if(i<lena)
sum=a[i]-''+add;
else if(i<lenb)
sum=b[i]-''+add;
add=;
if(sum>) {
add=;
sum-=;
}
c+=sum+'';
}
if(add)
c+=add+'';
reverse(c.begin(),c.end());//还原
cout<<"Case "<<k<<":"<<endl;
cout<<m<<" + "<<n<<" = ";
cout<<c<<endl;
if(k!=t)
cout<<endl;
}
return ;
}