【原】 POJ 1503 Integer Inquiry 大整数加法 解题报告
http://poj.org/problem?id=1503
方法:
很简单,模拟手动加法过程, 结果中的前置0要消去
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.)
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.
Output
Your program should output the sum of the VeryLongIntegers given in the input.
Sample Input
123456789012345678901234567890
123456789012345678901234567890
123456789012345678901234567890
0
Sample Output
370370367037037036703703703670
1: #include <iostream>
2: #include <string>
3: #include <fstream>
4:
5: using namespace std ;
6:
7: string sum_bignum(const string& lhs,const string& rhs)
8: {
9: int len1 = lhs.size() ;
10: int len2 = rhs.size() ;
11: int lenResult = len1>len2 ? len1+1 : len2+1 ;
12:
13: int i ;
14: int lval,rval,carry = 0 ;
15: int tmpresult ;
16: int lhsEnd = len1-1 ;
17: int rhsEnd = len2-1 ;
18: int resultEnd = lenResult-1 ;
19: string result(lenResult,'0') ;
20: for( i=resultEnd ; i>=0 ; --i )
21: {
22: //*** 如果不判断直接赋值,则会出错,例如999+9
23: //对结果的每一位都进行计算***
24: if(lhsEnd>=0)
25: lval = lhs[lhsEnd--] - '0' ;
26: else
27: lval = 0 ;
28:
29: if(rhsEnd>=0)
30: rval = rhs[rhsEnd--] - '0' ;
31: else
32: rval = 0 ;
33:
34: tmpresult = lval+rval+carry ;
35: result[i] = tmpresult%10 + '0' ;
36: carry = tmpresult/10 ;
37: }
38:
39: return result.substr( result.find_first_not_of('0') ) ;
40: }
41:
42: void run1503()
43: {
44: ifstream in("in.txt") ;
45: string lhs,rhs ;
46: getline(in,lhs);
47: while(getline(in,rhs) && rhs!="0")
48: lhs = sum_bignum(lhs,rhs);
49: cout<<lhs<<endl;
50: }
【推荐】编程新体验,更懂你的AI,立即体验豆包MarsCode编程助手
【推荐】凌霞软件回馈社区,博客园 & 1Panel & Halo 联合会员上线
【推荐】抖音旗下AI助手豆包,你的智能百科全书,全免费不限次数
【推荐】轻量又高性能的 SSH 工具 IShell:AI 加持,快人一步