POJ 1503 Integer Inquiry

发布时间：2020-12-14 02:53:43 所属栏目：大数据来源：网络整理

导读：来源:http://poj.org/problem?id=1503 Integer Inquiry Time Limit: ?1000MS ? Memory Limit: ?10000K Total Submissions: ?30480 ? Accepted: ?11870 Description One of the first users of BIT's new supercomputer was Chip Diller. He extended his exp

来源:http://poj.org/problem?id=1503

Integer Inquiry

Time Limit:?1000MS	?	Memory Limit:?10000K
Total Submissions:?30480	?	Accepted:?11870

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

Source

East Central North America 1996

题意: 输入n行正整数数以0结束，输出这n行数的和。

题解: 简单大数加法，我用压位高精度写的~~

AC代码:

#include<iostream>
#include<string>
#include<cmath>
#include<cstring>
#define carry 100000000
#define maxn 115
using namespace std;
string str,sum;
void str2num(string &x,int *num){
	int len = x.size(),pos = 0;
	for (int i = 1; i <= len / 8; i++){
		string temp = x.substr(len - 8 * i,8);
		for (int i = 0; i<8; i++)
			num[pos] += (temp[7 - i] - '0')*pow(10,i);
		pos++;
	}
	int left = len % 8;
	if (left){
		for (int i = 0; i < left; i++)
			num[pos] += (x[left - i-1] - '0')*pow(10,i);
		pos++;
	}
}
void num2str(int x,string &str){
	int temp[10] = { 0 },pos=0;
	while (x){
		temp[pos++] = x % 10;
		x /= 10;
	}
	for (int i = 7; i >= 0; i--)
		str += temp[i] + '0';
}
void Add(string x,string & y)
{
	string rest = "";
	int num1[maxn] = { 0 },num2[maxn] = { 0 };
	int len = x.size() > y.size() ? x.size() : y.size();
	str2num(x,num1);
//	cout << "num1[0]=" << num1[0] << endl;
	str2num(y,num2);
	for (int i = 0; i <= len/8; i++){
		num1[i] += num2[i];
		if (num1[i] >= carry){
			num1[i + 1] += num1[i] / carry;
			num1[i] %= carry;
		}
	}
	int k;
	for (k = len; k>0 && !num1[k]; k--);
	for (; k>= 0; k--) num2str(num1[k],rest);
	y = rest;
}
int main()
{
	cin.sync_with_stdio(false);
	sum = "0";
	while (cin >> str,str != "0"){
	Add(str,sum);
	}
	int i;
	for (i = 0; i <sum.size();i++)
	if (sum[i] != '0')break;
	sum = sum.substr(i,sum.size() - i+1);
	cout << sum << endl;
	return 0;
}

（编辑：李大同）

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