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] = ' ';
for (i=len2; i<len + 1; i++) m2[i]='0';
m2[i] = ' ';
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] != ' '; 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;
}