杭电ACM 1002

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 306310    Accepted Submission(s): 59185

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
#include<iostream>#define MAX 1000using namespace std;int main(){    int A[MAX],B[MAX];    char a[MAX]={0},b[MAX]={0};    int T,cont=1;;    cin>>T;    while(T--){        cin>>a>>b;        int alen=strlen(a);        int blen=strlen(b);        int c=0;//代表进位        int sum;//代表和        int i,j;        for(i=0;i<MAX;i++){            A[i]=0;            B[i]=0;        }        for(i=0;i<alen;i++){            A[i]=a[alen-1-i]-'0';        }        for(j=0;j<blen;j++){            B[j]=b[blen-1-j]-'0';        }        cout<<"Case "<<cont++<<":"<<endl;        for(i=alen-1;i>=0;i--)            cout<<A[i];        cout<<" + ";        for(j=blen-1;j>=0;j--){            cout<<B[j];        }        cout<<" = ";        int count=alen-blen>0?alen:blen;            for(int k=0;k<count;k++){            sum=A[k]+B[k]+c;            c=sum/10;///求进位            A[k]=sum%10;//求低位        }        A[count]=c;///最高位为进位        count= c==1?count:count-1;            for(int n=count;n>=0;n--){             cout<<A[n];        }                  cout<<endl;        if(0!=T)            cout<<endl;    }        return 0;}