杭电A+B problem1002 （大数）

发布时间：2020-12-14 03:07:26 所属栏目：大数据来源：网络整理

导读：大数的问题真是一让人头疼的问题。用数组存数，我这个初学者学着好困难。打了好长时间才把代码整好，纪念一下。 ? ? ? ? ? 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

大数的问题真是一让人头疼的问题。用数组存数，我这个初学者学着好困难。

打了好长时间才把代码整好，纪念一下。

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

代码：

#include<stdio.h> #include<string.h> #define MAX 1000 int a1[MAX+10]; int a2[MAX+10]; char b1[MAX+10]; char b2[MAX+10]; int main(void) { ??? int n,lenb1,lenb2,i,j,t,k=1; ??? scanf("%d",&n); ??? getchar(); ??? //scanf("%s %s",b1,b2); ??? while(n--) ??? { ??????? scanf("%s %s",b2); ??????? getchar(); ??????? //lenb1=strlen(b1); ??????? //lenb2=strlen(b2); ??????? memset(a1,sizeof(a1)); ??????? memset(a2,sizeof(a2)); ??????? lenb1=strlen(b1); ??????? for(j=0,i=lenb1-1;i>=0;i--) ??????? { ?????????? a1[j++]=b1[i]-'0'; ??????? } ??????? lenb2=strlen(b2); ??????? for(j=0,i=lenb2-1;i>=0;i--) ??????? { ??????????? a2[j++]=b2[i]-'0'; ??????? } ??????? if(lenb1<lenb2) ??????? { ???????????? t=lenb1;lenb1=lenb2;lenb2=t; ??????? } ??????? for(i=0;i<lenb1;i++) ??????? { ?????????? a1[i]+=a2[i]; ?????????? if(a1[i]>=10) ?????????? { ????????????? a1[i]-=10; ????????????? a1[i+1]++; ?????????? } ??????? } ??????? printf("Case %d:n",k++); ??????? //k++; ??????? printf("%s + %s = ",b2); ??????? if(a1[lenb1]!=0) ??????????? printf("%d",a1[lenb1]); ??????? for(i=lenb1-1;i>=0;i--) ??????????? printf("%d",a1[i]); ??????? printf("n"); ??????? if(n) ??????????? printf("n"); ??? } ??? return 0; }

（编辑：李大同）

【声明】本站内容均来自网络，其相关言论仅代表作者个人观点，不代表本站立场。若无意侵犯到您的权利，请及时与联系站长删除相关内容!