A + B Problem II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3820 Accepted Submission(s): 1351 

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 <stdio.h>#include <string.h>#include<iostream>using namespace std;int main(){    int i,n,j,t,xt,zt;    char z[10000],x[10000],v[10000]={0};    char a;    cin>>n;    int tt=n;    j=1;    while(n--)    {        cin>>z>>x;        cout<<"Case "<<j<<":"<<endl<<z<<" + "<<x<<" = ";        zt=strlen(z);        xt=strlen(x);        for(i=0;i<zt/2;i++)//将字符串倒置  从后往前        {            a=z[i];            z[i]=z[zt-1-i];            z[zt-1-i]=a;        }        for(i=0;i<xt/2;i++)        {            a=x[i];            x[i]=x[xt-1-i];            x[xt-1-i]=a;        }        for(i=0;i<zt;i++)//转化为数字        {            z[i]-='0';        }        for(i=0;i<xt;i++)        {            x[i]-='0';        }        if(zt>xt)            t=zt;        else            t=xt;        for(i=0;i<t;i++)        {            if(i<zt)v[i]+=z[i];            if(i<xt)v[i]+=x[i];            if(v[i]>=10)            {                v[i]=v[i]%10;v[i+1]++;//是否需要进位            }        }        if(v[t]!=0)             t=t+1;        for(i=t-1;i>=0;i--)           {                 printf("%d",v[i]);                v[i]=0;            }        cout<<endl;        if(j!=tt)cout<<endl;        j++;    }    return 0;}