题解报告：hdu 1002 A + B Problem II（大数加法）

发布时间：2020-12-14 03:49:48 所属栏目：大数据来源：网络整理

导读：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 li

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

1 2

112233445566778899 998877665544332211

Sample Output

Case 1:

1 + 2 = 3

Case 2:

112233445566778899 + 998877665544332211 = 2222222222222221110

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<<":n"<<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 }

（编辑：李大同）

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