A + B Problem II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 347161 Accepted Submission(s): 67385
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
Author
Ignatius.L
分析：高精度计算，大数相加！模版在博客中已给出，翻翻看，按照模版写就行了，要注意细节，空格的输出，因为这个PE了2次！
下面给出AC代码：
1 #include <bits/stdc++.h> 2 using namespace std; 3 int main() 4 { 5 char a1[1005],b1[1005]; 6 int a[1005],b[1005],c[1005];//a，b，c分别存储加数，加数，结果 7 int x,i,j,n; 8 int lena,lenb,lenc; 9 while(scanf("%d",&n)!=EOF) 10 { 11 for(j=1;j<=n;j++) 12 { 13 memset(a,0,sizeof(a));//数组a清零 14 memset(b,0,sizeof(b));//数组b清零 15 memset(c,0,sizeof(c));//数组c清零 16 scanf("%s%s",&a1,&b1); 17 lena=strlen(a1); 18 lenb=strlen(b1); 19 for(i=0;i<=lena;i++) 20 a[lena-i]=a1[i]-'0';//将数串a1转化为数组a，并倒序存储 21 for(i=0;i<=lenb;i++) 22 b[lenb-i]=b1[i]-'0';//将数串b1转化为数组a，并倒序存储 23 x=0;//x是进位 24 lenc=1;//lenc表示第几位 25 while(lenc<=lena||lenc<=lenb) 26 { 27 c[lenc]=a[lenc]+b[lenc]+x;//第lenc位相加并加上次的进位 28 x=c[lenc]/10;//向高位进位 29 c[lenc]%=10;//存储第lenc位的值 30 lenc++;//位置下标变量 31 } 32 c[lenc]=x; 33 if(c[lenc]==0)//处理最高进位 34 lenc--; 35 printf("Case %d:\n",j);//格式要求吧！学着点 36 printf("%s + %s = ",a1,b1);//这也是格式要求吧！学着点 37 for(i=lenc;i>=1;i--) 38 printf("%d",c[i]); 39 printf("\n"); 40 if(j!=n)//对于2组之间加空行的情况 41 printf("\n"); 42 } 43 } 44 return 0; 45 }
java写法大数，真是有毒！
1 import java.math.BigInteger; 2 import java.util.Scanner; 3 4 public class Main { 5 6 /** 7 * @param args 8 */ 9 public static void main(String[] args) 10 { 11 // TODO Auto-generated method stub 12 //System.out.println("Hello World!"); 13 Scanner in=new Scanner(System.in); 14 while(in.hasNextInt()) 15 { 16 // int []arr=new int[3]; 17 int n; 18 n=in.nextInt(); 19 for(int i=1;i<=n;i++) 20 { 21 BigInteger a,b; 22 a=in.nextBigInteger(); 23 b=in.nextBigInteger(); 24 if(i<n) 25 { 26 System.out.println("Case "+i+":"); 27 System.out.print(a+" + "+b+" = "); 28 System.out.println(a.add(b)); 29 System.out.println(); 30 } 31 else 32 { 33 System.out.println("Case "+i+":"); 34 System.out.print(a+" + "+b+" = "); 35 System.out.println(a.add(b)); 36 } 37 } 38 } 39 } 40 }
