题解报告:hdu 1002 A + B Problem II(大数加法)
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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
解题思路:大数加法,简单模拟竖式运算或者直接调用java中BigInteger大数类的add()方法即可。
AC之C++代码:
1 #include<bits/stdc++.h> 2 using namespace std; 3 int main(){ 4 string a,b,r;int t,tmp;bool flag; 5 while(cin>>t){ 6 flag=true; 7 for(int i=1;i<=t;++i){ 8 cin>>a>>b;tmp=0;r=""; 9 if(!flag)cout<<endl; 10 cout<<"Case "<<i<<": "<<a<<" + "<<b<<" = "; 11 if(a.size()>b.size())swap(a,b); 12 for(int i=a.size()-1,j=b.size()-1;~j;--i,--j){ 13 int now=(i>=0?a[i]-‘0‘:0)+b[j]-‘0‘+tmp; 14 tmp=now>9?1:0;now%=10; 15 r+=‘0‘+now; 16 } 17 if(tmp)r+=‘1‘; 18 for(int i=r.size()-1;~i;--i)cout<<r[i]; 19 cout<<endl;flag=false; 20 } 21 } 22 return 0; 23 }
AC之java代码:
1 import java.util.Scanner; 2 import java.math.BigInteger; 3 public class Main { 4 public static void main(String[] args) { 5 Scanner scan = new Scanner(System.in); 6 int t=scan.nextInt(); 7 for(int i=1;i<=t;++i){ 8 BigInteger a = scan.nextBigInteger(); 9 BigInteger b = scan.nextBigInteger(); 10 System.out.println("Case "+i+":"); 11 System.out.println(a+" + "+b+" = "+a.add(b)); 12 if(i!=t)System.out.println(); 13 } 14 } 15 }
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