【算法】大数乘法

POJ：1001

http://poj.org/problem?id=1001

===============我是分割线=================

/*
*Copyright: CheerM
*Author: CheerM
*Date: 2016-11-14
*Description: 实现 底数B为6位宽浮点数 指数E为（0，25】的数，求值
*/

#include <iostream>
#include <string>
#include <vector>
#include <iomanip>

using namespace std;

const int max_ = 10000;//保留四位数

/*
*Function:       multiply
*Description:    被乘数任意长vector<int>& multiplicand, 乘数为定长，不超过6位的int multiplier，相乘的值保存在vector<int>& multiplicand中
*parameter：         vector<int>& multiplicand 表示任意长的被乘数
*                 int multiplier 表示定长不超过6位数的乘数
*Return:         void
*/
void multiply(vector<int>& multiplicand, int multiplier) {
    int carry = 0;
    for (int i = 0; i < multiplicand.size(); i++) {
        int t1 = multiplicand[i] * multiplier + carry;
        multiplicand[i] = t1 % max_;
        carry = t1 / max_;
    }
    if (carry != 0) {
        while (carry) {
            multiplicand.push_back(carry % max_);
            carry /= max_;
        }
    }
}

int main() {
    string base;//底数
    int exponentiation, point, mark;//指数，小数点后有效位数，‘.’标识符
    while (cin >> base >> exponentiation) {
        point = mark = 0;

        //记录小数点后数字位数，忽略最右的连续0
        for (int i = base.size() - 1; i >= 0; i--) {
            if (base[i] == '0' && point == 0)
                continue;
            else if (base[i] == '.') {
                mark = 1;
                break;
            }
            else
                point++;
        }
        if (!mark)point = 0;

        //把底数base从字符串转化为整型，e.g. 12.345 -> int:12345 point=3, 1.0100 -> int:101 point=2
        int tempBase = 0, tmp = point;//tempBase是底数的int形式
        mark = 0;
        for (int i = 0; i < base.size() && tmp >= 0; i++) {
            if (base[i] >= '0' && base[i] <= '9') {
                tempBase = tempBase * 10 + base[i] - '0';
            }
            else if (base[i] == '.') {
                mark = 1;
            }

            if (mark == 1) {
                tmp--;
            }
        }

        //初始化被乘数为1，循环相乘
        vector<int> result;//vector来存储超长int型整数，因为乘数有可能是6位数，而int最大值可以10位，所以result的每一个单位用来存储4位结果，从低到高
        result.push_back(1);
        point *= exponentiation;
        while (exponentiation--) {
            multiply(result, tempBase);
        }

        //把vector转为string
        string ss;
        for (int i = 0; i < result.size(); i++) {
            int tt = result[i];
            for (int j = 0; j < 4; j++) {
                ss.push_back(tt % 10 + '0');
                tt /= 10;
                if (ss.size() == point) ss.push_back('.');
                if (i == result.size() - 1 && tt == 0) break;
            }
        }
        while (ss.size() < point) {//补0
            ss.push_back('0');
            if (ss.size() == point) ss.push_back('.');
        }

        //输出
        for (int i = ss.size() - 1; i >= 0; i--)
            cout << ss[i];
        cout << endl;
    }

    system("pause");
    return 0;
}