码迷,mamicode.com
首页 > 其他好文 > 详细

hdu1002 A + B Problem II(大数题)

时间:2014-12-16 00:50:15      阅读:186      评论:0      收藏:0      [点我收藏+]

标签:des   style   blog   http   io   ar   color   os   sp   

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1002

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 230247    Accepted Submission(s): 44185


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

 

题目大意:题意很容易理解,具体就不解释了,主要就是要解决大数的问题。

题目思路:如果会java的话,可以轻松AC。其他的小伙伴们只能用最笨的方法解决。我们用一个数字将数字倒过来存下,无论是乘法还是加法,这是最好的解决办法。

下面附上两个代码。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 
 5 using namespace std;
 6 
 7 int main ()
 8 {
 9     char a[8000],b[8000];
10     int na[8000],nb[8000],sum[8000],pre,flag=1;
11     int t;
12     scanf("%d",&t);
13         while (t--)
14         {
15             memset(sum,0,sizeof(sum));
16             memset(na,0,sizeof(na));
17             memset(nb,0,sizeof(nb));
18             scanf("%s%s",a,b);
19             pre=0;
20             int lena=strlen(a);
21             int lenb=strlen(b);
22             for (int i=0; i<lena; i++)
23                 na[lena-1-i]=a[i]-0;
24             for (int j=0; j<lenb; j++)
25                 nb[lenb-1-j]=b[j]-0;
26             int lenx=lena>lenb?lena:lenb;
27             for (int k=0; k<lenx; k++)
28             {
29                 sum[k]=na[k]+nb[k]+pre/10;
30                 pre=sum[k];
31             }
32             while (pre>9)
33             {
34                 sum[lenx]=pre/10%10;
35                 lenx++;
36                 pre/=10;
37             }
38             printf ("Case %d:\n",flag++);
39             printf ("%s + %s = ",a,b);
40             for (int i=lenx-1; i>=0; i--)
41             {
42                 printf ("%d",sum[i]%10);
43             }
44             printf ("\n");
45             if (t)
46                 printf ("\n");
47         }
48 
49     return 0;
50 }

java代码。

 1 import java.util.*;
 2 import java.math.*;
 3 public class Main {
 4     public static void main(String[] args) {
 5         Scanner sc=new Scanner (System.in);
 6         int l=sc.nextInt();
 7         for(int i=1;i<=l;i++){
 8             if(i!=1) System.out.println();
 9             BigInteger a,b;
10             a=sc.nextBigInteger();
11             b=sc.nextBigInteger();
12             System.out.println("Case "+i+":");
13             System.out.println(a+" + "+b+" = "+a.add(b));
14         }
15     }
16 }

 

hdu1002 A + B Problem II(大数题)

标签:des   style   blog   http   io   ar   color   os   sp   

原文地址:http://www.cnblogs.com/qq-star/p/4166105.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!