POJ 1503 Integer Inquiry
Integer Inquiry
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 23072 | Accepted: 8987 |
Description
One of the first users of BIT's new supercomputer was Chip Diller. He extended his exploration of powers of 3 to go from 0 to 333 and he explored taking various sums of those numbers.
``This supercomputer is great,'' remarked Chip. ``I only wish Timothy were here to see these results.'' (Chip moved to a new apartment, once one became available on the third floor of the Lemon Sky apartments on Third Street.)
``This supercomputer is great,'' remarked Chip. ``I only wish Timothy were here to see these results.'' (Chip moved to a new apartment, once one became available on the third floor of the Lemon Sky apartments on Third Street.)
Input
The
input will consist of at most 100 lines of text, each of which contains a
single VeryLongInteger. Each VeryLongInteger will be 100 or fewer
characters in length, and will only contain digits (no VeryLongInteger
will be negative).
The final input line will contain a single zero on a line by itself.
The final input line will contain a single zero on a line by itself.
Output
Your program should output the sum of the VeryLongIntegers given in the input.
Sample Input
123456789012345678901234567890
123456789012345678901234567890
123456789012345678901234567890
0
Sample Output
370370367037037036703703703670
Source
只有正数的高精度加法。用字符串就好。
1 #include <iostream>
2 #include <cstring>
3 #include <cstdlib>
4 #include <cstdio>
5 #include <algorithm>
6
7 using namespace std;
8
9 int result[105];
10
11 void add(string line)
12 {
13 int current = 104;
14 for(int i(line.size()-1);i >= 0;--i)
15 {
16 result[current] += (line[i]-'0');
17 if(result[current] >= 10)
18 {
19 result[current-1]++;
20 result[current] -= 10;
21 }
22 current--;
23 }
24 if(result[current] >= 10)
25 {
26 result[current-1]++;
27 result[current] -= 10;
28 }
29 }
30
31 int main(void)
32 {
33 string line;
34 memset(result,0,sizeof(int)*105);
35
36 while(cin >> line && line != "0")
37 {
38 add(line);
39 }
40
41 bool flag = false;
42 for(int i = 0;i !=105;++i)
43 {
44 if(result[i] == 0 && !flag)
45 continue;
46 else
47 {
48 flag = true;
49 printf("%d",result[i]);
50 }
51 }
52 printf("\n");
53
54 return 0;
55 }
