A + B Problem II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 124589 Accepted Submission(s): 23993
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
1 #include2 using namespace std; 3 4 void sum(char (&a)[1001],char (&b)[1001]); 5 int main() 6 { 7 int T, count_T; 8 char a[1001], b[1001]; 9 cin >> T;10 for(count_T=1; count_T <= T; count_T++) 11 {12 cin >> a >> b;13 14 if(count_T==1) 15 {16 cout << "Case " << count_T << ":" << endl;17 }18 else 19 {20 cout << "\nCase " << count_T << ":" << endl;21 }22 23 cout << a << " + " << b << " = ";24 25 sum(a,b);26 }27 return(0);28 }29 void sum(char (&a)[1001],char (&b)[1001])30 {31 char c[1002]; 32 int la, lb;33 la=strlen(a)-1;34 lb=strlen(b)-1;35 int i;36 int flag=0;37 for(i=0;la>=0 && lb>=0;la--,lb--,i++)38 {39 flag+=(a[la]-'0')+(b[lb]-'0');40 c[i]=flag%10;41 flag/=10;42 }43 for(la=la;la>=0;la--,i++)44 {45 flag+=(a[la]-'0');46 c[i]=flag%10;47 flag/=10;48 }49 for(lb=lb;lb>=0;lb--,i++)50 {51 flag+=(b[lb]-'0');52 c[i]=flag%10;53 flag/=10;54 }55 if(flag)56 {57 c[i]=flag;58 i++;59 }60 for(i--;i>=0;i--)61 {62 cout<<(char)(c[i]+'0');63 }64 cout<