HDU 1002 A + B Problem II
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A + B Problem II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 303792 Accepted Submission(s): 58634
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
21 2112233445566778899 998877665544332211
Sample Output
Case 1:1 + 2 = 3Case 2:112233445566778899 + 998877665544332211 = 1111111111111111110
Author
Ignatius.L
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#include <stdio.h>#include<iostream>#include<string.h>#include<string>#include<cmath>#include <stdlib.h>using namespace std;string s1,s2;string sum(string s1,string s2){if(s1.length()<s2.length()){string temp=s1;s1=s2;s2=temp;}int i,j;for(i=s1.length()-1,j=s2.length()-1;i>=0;i--,j--){s1[i]=char(s1[i]+(j>=0?s2[j]-'0':0));if(s1[i]-'0'>=10){s1[i]=char((s1[i]-'0')%10+'0');if(i) s1[i-1]++;else s1='1'+s1;}}return s1;}int main(){int t1,t2,k=0;cin>>t1;t2=t1;while(t1--){k++;cin>>s1>>s2;cout<<"Case "<<k<<":"<<endl;cout<<s1<<" + "<<s2<<" = "<<sum(s1,s2)<<endl;if(k<t2)cout<<endl;s1[0]=0,s2[0]=0;}return 0;}
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