hdu 1002 高精度加法

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 276647    Accepted Submission(s): 53400

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

 

Author

Ignatius.L

 

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#include<iostream>#include<algorithm>#include<cstring>#include<cstdio>using namespace std;int ans[1003];char a[1003],b[1003];int len_a=0,len_b=0,mlen=0;void init (){memset (ans,0,sizeof ans);/*for (int i=0;i<=1003;i++)ans[i]=0;*/scanf("%s",a);scanf("%s",b);len_a=strlen (a); len_b=strlen(b);}void add (){int p=0;for (int i=len_a-1;i>=0;i--)ans[p++]+=(a[i]-'0');p=0;for (int i=len_b-1;i>=0;i--)ans[p++]+=(b[i]-'0');}void produce_ans(){    mlen=max(len_a,len_b);//前面声明了mlen做全局变量，后面不用定义局部变量mlen。//    cout<<mlen<<endl;for (int i=0;i<mlen;i++){if (i==mlen-1&&ans[i]>=10)mlen++;if (ans[i]>=10){    int x=ans[i]/10;ans[i]%=10;ans[i+1]+=x;//写++是可以的，因为最多只可能进一位}}}void output(){    printf("%s + %s = ",a,b);for (int i=mlen-1;i>=0;i--)//不是i++，是i--！！！cout<<ans[i];cout<<endl;}int main (){int n;cin>>n;for (int i=1;i<=n;i++){    if(i>1) putchar('\n');    printf("Case %d:\n",i);init ();add();produce_ans();output();//"1 + 2 = "}return 0;}