超出long long長度的大數相加

存：字串 轉換數字：a[i] - ‘0’ 難點：進位的標記，定義整型變數temp標記進位值

HDU 1002

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

AC一例

#include <bits/stdc++.h>
using namespace std;
int main(){
    string a,b;
    int t,c[1001],p=0;
    while(cin>>t){
        for(int i=1;i<=t;i++){
            cin>>a>>b;
            int alen,blen,clen=0,temp=0;
            alen = a.length()-1;
            blen = b.length()-1;
            while(alen>=0 && blen>=0){
                c[clen++] = (temp + a[alen] - '0' + b[blen] - '0') % 10;
                temp = (temp + a[alen] - '0' + b[blen] - '0') / 10;
                alen--;
                blen--;
            }
            if(alen > blen)
                while(alen >= 0){
                    c[clen++] = (temp + a[alen] - '0') % 10;
                    temp = (temp + a[alen] - '0') / 10;
                    alen--;
                }
            else if(blen > alen)
                while(blen >= 0){
                    c[clen++] = (temp + b[blen] - '0') % 10;
                    temp = (temp + b[blen] - '0') / 10;
                    blen--;
                }
            c[clen] = temp;
            if(p)
                cout<<endl;
            p++;
            cout<<"Case "<<i<<':'<<endl;
            cout<<a<<" + "<<b<<" = ";
            if(c[clen]!=0)
                cout<<c[clen];
            for(--clen;clen>=0;clen--)
                cout<<c[clen];
            cout<<endl;
        }
    }
    return 0;
}