大数计算------杭电HDOJ 1002(简单题)

发布时间：2020-12-14 03:40:28 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

   
   
    
    2
1 2
112233445566778899 998877665544332211

Sample Output

   
   
    
    Case 1:
1 + 2 = 3

Case 2:
112233445566778899 + 998877665544332211 = 2222222222222221110

题意很简单：这里就不多说。

普通int定义数据范围：2147483648~2147483648，就算是__int64 也是有限的，那怎么办呢？我们用数组来计算这道题：

代码如下：

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<math.h>
#include<iostream>
using namespace std;
#define max 10006
char liu[max][max],xp[max][max],a[max][max];
int d[max];
int main()
{
    int i,j,k;
    int t,f;
    int len1,len2,len3;
    int b;
    cin>>t;
    getchar();
    for(i=0;i<t;i++)
    {
        cin>>liu[i]>>xp[i];
        //printf("%sn",liu[i]);
        //printf("%sn",xp[i]);
    }
    for(f=0;f<t;f++)
    {
        memset(d,max);
        len1=strlen(liu[f]);
        len2=strlen(xp[f]);
        b=0;
        for(i=len1-1,j=len2-1;i>=0&&j>=0;i--,j--)
        {
            d[b+1]=(d[b]+(liu[f][i]-'0')+(xp[f][j]-'0'))/10;
            d[b]=(d[b]+(liu[f][i]-'0')+(xp[f][j]-'0'))%10;
            b=b+1;
           }
         if(len1<len2)
         {
               for(i=len2-len1-1;i>=0;i--)
               {
                   d[b+1]=(d[b]+(xp[f][i]-'0'))/10;
                d[b]=(d[b]+(xp[f][i]-'0'))%10;
                   b=b+1;
               }
            }
         else if(len1>len2)
           {
             for(i=len1-len2-1;i>=0;i--)
            {
                 d[b+1]=(d[b]+(liu[f][i]-'0'))/10;
                d[b]=(d[b]+(liu[f][i]-'0'))%10;
                   b=b+1;
            }
        }
           printf("Case %d:n",f+1);
           printf("%s + %s = ",liu[f],xp[f]);
          for(j=max-1;j>=0;j--)
        {
            if(d[j]) break;
        }
           for(i=j;i>=0;i--)
           {
             printf("%d",d[i]);
        }
           printf("n");
           if(f!=t-1) printf("n");
    }
    return 0;
}

实在惭愧，当初不会这道题的解法，活生生的错了40次；

能力有限，代码也不懂得优化，大家要是有什么优化的建议，欢迎提出，大家一起学习进步，谢谢；

（编辑：李大同）

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