A + B Problem II
Time
Limit: 2000/1000 MS (Java/Others) Memory Limit:
65536/32768 K (Java/Others)
Total Submission(s):
201400 Accepted Submission(s):
38599
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
大数问题，总觉得自己写的麻烦了，不解释。。。Java 和 C ~~~~
Java 代码：
1 import java.math.BigDecimal;
2 import java.util.*;
3
4 public class Main {
5 public static void main(String[] args) {
6 Scanner cin = new Scanner(System.in);
7 int n = cin.nextInt();
8 for (int i = 1; i <= n; ++i) {
9
10 BigDecimal a1, b1, c1;
11 a1 = cin.nextBigDecimal();
12 b1 = cin.nextBigDecimal();
13 c1 = a1.add(b1);
14 System.out.println("Case " + i + ":");
15 System.out.println(a1 + " + " + b1 + " = " + c1);
16 if (i < n)
17 System.out.println();
18 }
19 }
20 }
C代码：
1 #include <cstdio>
2 #include <iostream>
3 #include <string.h>
4 using namespace std;
5
6 #define N 1000
7 int main()
8 {
9 int T, f;
10 int m, n, temp;
11 char A[N], B[N];
12 int sum[N+1], t[N];
13
14 cin>>T;
15 f=T;
16
17 while(T--)
18 {
19 scanf("%s",A);
20 scanf("%s",B);
21
22 m = strlen(A);
23 n = strlen(B);
24
25 if(m==n) {
26
27 for(int i = m-1; i>=0; i--) t[i]=A[i]+B[i]-‘0‘-‘0‘;
28
29 for(int i = m-1; i>=0; i--) {
30
31 temp = t[i]/10;
32 sum[i+1] = t[i]%10;
33 if(temp && i) t[i-1] += temp;
34 if(i==0) sum[0] = temp;
35 }
36
37 printf("Case %d:\n",f-T);
38 printf("%s + %s = ",A,B);
39
40 if(sum[0]) printf("%d", sum[0]);
41
42 for(int i = 1; i < m; i++) printf("%d", sum[i]);
43
44 printf("%d\n",sum[m]);
45 }
46 else if(m>n) {
47
48 for(int i = m-1; i>=0; i--) t[i]=A[i]-‘0‘;
49 for(int i = m-1; n>=m-i; i--) t[i]+=B[n-m+i]-‘0‘;
50
51 for(int i = m-1; i>=0; i--) {
52
53 temp = t[i]/10;
54 sum[i+1] = t[i]%10;
55 if(temp && i) t[i-1] += temp;
56 if(i==0) sum[0] = temp;
57 }
58
59 printf("Case %d:\n",f-T);
60 printf("%s + %s = ",A,B);
61
62 if(sum[0]) printf("%d", sum[0]);
63
64 for(int i = 1; i < m; i++) printf("%d", sum[i]);
65
66 printf("%d\n",sum[m]);
67
68 }
69 else {
70
71 for(int i = n-1; i>=0; i--) t[i]=B[i]-‘0‘;
72 for(int i = n-1; m>=n-i; i--) t[i]+=A[m-n+i]-‘0‘;
73
74 for(int i = n-1; i>=0; i--) {
75
76 temp = t[i]/10;
77 sum[i+1] = t[i]%10;
78 if(temp && i) t[i-1] += temp;
79 if(i==0) sum[0] = temp;
80 }
81
82 printf("Case %d:\n",f-T);
83 printf("%s + %s = ",A,B);
84
85 if(sum[0]) printf("%d", sum[0]);
86
87 for(int i = 1; i < n; i++) printf("%d", sum[i]);
88
89 printf("%d\n",sum[n]);
90
91 }
92 if(T) printf("\n");
93 }
94 return 0;
95 }