HDU1002 : A + B Problem II

A + B Problem II

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 367113 Accepted Submission(s): 71500

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

2
1 2
112233445566778899 998877665544332211

Sample Output

Case 1:
1 + 2 = 3
Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110

代码如下，部分代码直接用了老师的课件上的，更正了错误。

#include<iostream>#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>using namespace std;int m3[100002];int f = 1;void Reverse(char *word,int len)   // 反转数字 {                              char temp;    int i, j;    for (j = 0, i = len - 1; j < i; --i, ++j) {        temp = word[i];        word[i] = word[j];        word[j] = temp;    }}int check(int a[],int num)      //归整{  int k=0,len=num;    while(a[len-1]==0&&len>1) len--;    //去掉前导0    for(k=0; k<len; k++)        if(a[k]>=10)        {          a[k+1]=a[k+1]+ a[k]/10;  a[k]=a[k] % 10;        }    if (a[k]!=0) len=k+1;  //确定数组最终长度    return len;}int addition(int m3[], char m1[], int num1, char m2[], int num2){    int i,len1,len2,len;    len1=num1;    len2=num2;    len=(len1>=len2)?len1:len2;     //定位数    for(i=0; i<=len; i++) m3[i]=0;   //初始化    for (i=len1; i<len + 1; i++) m1[i]='0'; //缺位前导补0    m1[i] = '\0';    for (i=len2; i<len + 1; i++)    m2[i]='0';     m2[i] = '\0';    Reverse(m1,len1);    Reverse(m2,len2);    for (i=0; i<=len; i++)        m3[i]=(int)(m1[i]-'0'+m2[i]-'0');   //加法    len=check(m3,len);                  return len;}int Fun(char a[]){   //确定位数    int i = 0;    for(i = 0; a[i] != '\0'; i++);    return i;  //精确位数}int main(){    char num1[100001],num2[100001];    int m,len1,len2,len;    cin >> m;    while(m--){        cin >> num1 >> num2;        len1 = Fun(num1);        len2 = Fun(num2);        if(f != 1){            cout << endl;        }        cout << "Case " << f << ":" << endl;        cout << num1 << " + " << num2 << " = ";        len = addition(m3,num1,len1,num2,len2);        for(int i = len -1; i >= 0; i--){            cout << m3[i];        }        cout << endl ;        f++;    }    return 0;}