之前总是在查阅别人的文档，看着其他人的博客，自己心里总有一份冲动，想记录一下自己学习的经历。学习算法有一段时间了，于是想从算法开始自己的博客生涯O(∩_∩)O~~
今天在网上看了一道大数相加（高精度）的题目，题目很简单，但是体现了算法编程的细心之处。
题目：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
/*注意哦这里是两个test之间会有一个空行*/
Case 2:
112233445566778899 + 998877665544332211 = 1111111111111111110
/*在最后一个test输出之后是不需要空行的*/
1 #include <iostream> 2 #include <stdio.h> 3 #include <string.h> 4 using namespace std; 5 6 int main() 7 { 8 int n; 9 cin>>n; 10 int t = 0; 11 while(n--) 12 { 13 t++; 14 char a[1005]; 15 char b[1005]; 16 char c[1005] = {'0'}; //c数组用来存放两数相加之和，需要初始化一下 17 cin>>a; //cin会忽略回车、空格、tab键 18 cin>>b; 19 int k,sum; 20 k = strlen(a)>strlen(b)?strlen(a):strlen(b); //k 取两个长整数中较长的一个 21 a[k+1]=' '; // 对长整数之后的一位赋值' ' 结束标志 22 sum = 0; //累加器 23 for(int i = strlen(a)-1,j=strlen(b)-1;j>=0||i>=0;i--,j--,k--) 24 { 25 if(i>=0) sum+=a[i]-'0'; 26 if(j>=0) sum+=b[j]-'0'; 27 c[k] = sum%10 + '0'; 28 sum /= 10; 29 } 30 if(sum!=0) 31 c[0] = sum + '0'; 32 else 33 strcpy(c,&c[1]); 34 printf("Case %d: ",t); 35 printf("%s + %s = %s ",a,b,c); 36 if(n!=0) //最后一个test 不需要输出空行 37 printf(" "); 38 } 39 return 0; 40 }