HD OJ 1002Problem 大数相加

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 139440    Accepted Submission(s): 26450

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

 

分析：

该题为大数相加，将数字放在字串中进行运算即可。

代码：

#include <iostream>using namespace std;int main(int argc, const char * argv[]){    char a[1001],b[1001],c[2000];    int x,y,m,z,n;    cin>>n;getchar();    for(int j=0;j<n;j++){z=0;cout<<"Case "<<j+1<<":"<<endl;    for(x=1,a[0]='0';(a[x]=getchar())>='0'&&a[x]<='9';x++){cout<<a[x];};    cout<<" + ";x--;    for(y=1,b[0]='0';(b[y]=getchar())>='0'&&b[y]<='9';y++){cout<<b[y];};    cout<<" = ";y--;    for(m=0;x>0||y>0;m++)    {c[m]=(z=(a[x]-'0'+z+b[y]-'0'))%10+'0';z/=10;     x>0?x--:x=0;y>0?y--:y=0;    }if(z!=0){c[m]=z+'0';m++;}    for(m-=1;m>=0;m--)        cout<<c[m];    if(j!=n-1)cout<<endl<<endl;    }    return 0;}