HDU 1002 A + B Problem II（大整数相加）

A + B Problem II

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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 

 

题目大意：

大整数相加。

解题思路：

先把短的补齐，从最后一位开始计算，不进为就直接放进容器，进为把取余的放进容器，然后前一位加一。

代码：

#include<iostream>#include<string>#include<cstdio>#include<vector>using namespace std;int t;string str1,str2;vector <char> v;void solve(){    string temp;    int a,l2;    if(str1.length()<str2.length()){        temp=str1;str1=str2;str2=temp;        l2=str2.length();    }    for(int i=0;i<str1.length()-l2;i++){        str2.insert(0,1,'0');    }    for(int i=0;i<str1.length();i++){        a=str1[str1.length()-i-1]+str2[str2.length()-i-1]-2*'0';        if(a>=10){            v.push_back(a%10+'0');            if(str1.length()-i-1==0){                v.push_back('1');break;            }            str1[str1.length()-i-2]=(char)(str1[str1.length()-i-2]+1);        }        else v.push_back((char)(a+'0'));    }     vector<char>::iterator it=v.end();     it--;     while(it!=v.begin()){        if(*it=='0')            v.erase(it);        else break;        it--;     }    for(int i=v.size()-1;i>=0;i--){       cout<<v[i];    }    cout<<endl;}int main(){    int casen=0;    scanf("%d",&t);    while(t-->0){        cin>>str1>>str2;        printf("Case %d:\n%s + %s = ",++casen,str1.c_str(),str2.c_str());        v.clear();        solve();        if(t!=0)            cout<<endl;    }    return 0;}