A + B Problem II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 399645 Accepted Submission(s): 77352
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
用JAVA
import java.math.BigInteger;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n;
BigInteger a,b;
n=in.nextInt();
int i=0;
while(n>0) {
n--;
a=in.nextBigInteger();
b=in.nextBigInteger();
i++;
System.out.println("Case" + " " + i + ":");
if(n>0) {
System.out.println(a + " + " + b + " = "+a.add(b));
System.out.println();
}
else {
System.out.println(a + " + " + b + " = "+a.add(b));
}
}
}
}
C++:
要审清题意，注意输出格式！
#include<string>
#include <cstdio>
#include<iostream>
using namespace std;
string add(string a,string b)
{
int len1=a.length();
int len2=b.length();
if(len1>len2)
{
for(int i=;i<=len1-len2;i++)
b=""+b;
}
else
{
for(int i=;i<=len2-len1;i++)
a=""+a;
}
int len=a.length();
int cf=,t;
string str;
for(int i=len-;i>=;i--)
{
t=a[i]-''+b[i]-''+cf;
cf=t/;
t%=;
str=char(t+'')+str;
}
if(cf!=)
str=char(cf+'')+str;
return str;
}
int main()
{
int t;
cin>>t;
int k=;
while(t--)
{
k++;
string a,b;
cin>>a>>b;
string str;
str=add(a,b);
cout<<"Case "<<k<<":"<<endl;
cout<<a<<" + "<<b<<" = "<<str<<endl;
if(t)
cout<<endl;
}
return ;
}