A + B Problem II
http://acm.hdu.edu.cn/showproblem.php?pid=1002
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 137421 Accepted Submission(s): 26074
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
Author
Ignatius.L
#include<iostream> #include<cstring> using namespace std; int s1[1005],s2[1005],ans[1005]; int main(){ int t; char str1[1005],str2[1005]; cin>>t; int cases=0; while(t--){ cin>>str1>>str2; if(cases) cout<<endl; int i,j; int len1=strlen(str1),len2=strlen(str2); memset(s1,0,sizeof(s1)); memset(s2,0,sizeof(s2)); memset(ans,0,sizeof(ans)); int si=0,sj=0; for(i=len1-1;i>=0;i--) s1[si++]=str1[i]-'0'; for(j=len2-1;j>=0;j--) s2[sj++]=str2[j]-'0'; for(i=0;i<1005;i++){ ans[i]+=s1[i]+s2[i]; if(ans[i]>=10){ ans[i+1]+=ans[i]/10; ans[i]%=10; } } for(j=1004;j>=0;j--) if(ans[j]!=0) break; cout<<"Case "<<++cases<<":"<<endl; if(j==-1){ cout<<"0 + 0 = 0"<<endl; continue; } cout<<str1<<" + "<<str2<<" = "; for(i=j;i>=0;i--) cout<<ans[i]; cout<<endl; } return 0; }