【hdoj 1102】A + B Problem II

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Tiramicrystal

最新推荐文章于 2024-02-08 13:38:17 发布

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分类专栏： c/c++刷题算法

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本文介绍了一个经典的大数加法问题“A+B Problem II”，通过使用字符串来存储大数，并实现了一个从低位到高位进行逐位相加的算法。该算法能够有效处理长度不超过1000的大整数相加的问题。

A + B Problem II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 263357 Accepted Submission(s): 50986

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

题意：大树加法，用字符串存储数据，然后从低到高相加，进位标志flag

code：

#include<iostream>
#include<string>
#include<fstream>
#define N 1000
using namespace std;
string bigmult(string a,string b)
{
     int la,lb,lc,i,j,flag;
     string c="";
     char t;
     i=la=a.length()-1;
     j=lb=b.length()-1;
     flag=0;
     while(i>=0&&j>=0)
     {
         t=a[i]+b[j]+flag-'0';
         if(t>'9')
         {
             t=t-10;flag=1;
         }
         else
            flag=0;
         c+=t;
         i--;
         j--;
     }
    while(i>=0)
    {
         t=a[i]+flag;
         if(t>'9')
         {
             t=t-10;
             flag=1;
         }
         else
            flag=0;
         c+=t;
         i--;
    }
     while(j>=0)
     {
         t=b[j]+flag;
         if(t>'9')
         {
             t=t-10;
             flag=1;
         }
         else
            flag=0;
         c+=t;
         j--;
     }
     if(flag)
        c+=(flag+'0');
     lc=c.length();
     for(i=0,j=lc-1;i<j;i++,j--)
     {
         t=c[i];
         c[i]=c[j];
         c[j]=t;
     }
     return c;
 }

string ans[N];

int main()
{
     int T,k=1;
     string a,b;
     scanf("%d",&T);
     while(T--)
     {
         cin>>a>>b;
         cout<<"Case "<<k++<<":"<<endl;
         cout<<a<<" + "<<b<<" = "<<bigmult(a,b)<<endl;
         if(T)
         cout<<endl;
     }
     return 0;
}