HDU - 1002 A + B Problem II

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

题解：很明显是大数相加，但是它全部都是正数那更容易了，输入之后全部反序变成数字，两个数组个个相加，把进位记录，差不多就是这样，不过要注意换行之类的

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

int a[1010],b[1010],sum[1010];
int len_a,len_b,len_sum;
void changeInt(char str[],int num[],int *l){
*l = strlen(str) - 1;
int index = 0;
for(int i = *l;i >= 0 ;i--){
num[i] = str[index++] - '0';
}
}

void add(int a[],int b[],int s[]){
int c = 0;
int m = 0;
while(len_a>=m || len_b>=m){
if(len_a<m)a[m] = 0;
if(len_b<m)b[m] = 0;
s[m] = a[m]+b[m]+c;
c = s[m]/10;
if(s[m]>9){
s[m]=s[m]%10;
}
m++;
}
if(c) s[m] = c;
len_sum = m - 1 + c;
}

void write(int num[],int l){
for(int i = l;i >= 0;i--)printf("%d",num[i]);
}

int main(){
int T;cin>>T;int count = 0;
for(int i = 0; i < T ; i++){
char m[1010],n[1010];
if(count != T && count)printf("\n");
cin>>m>>n;
changeInt(m,a,&len_a);changeInt(n,b,&len_b);
add(a,b,sum);
printf("Case %d:\n",++count);
write(a,len_a);
printf(" + ");
write(b,len_b);
printf(" = ");
write(sum,len_sum);
printf("\n");
}
return 0;
}