poj1001

简单题

先把小数点去掉（乘以10的若干次幂），按整数计算，然后再计算有多少小数位即可。

View Code

#include <iostream>

#include <string>

using namespace std;



const int w = 100000, maxn = 300, ww = 5;



string st, ansst;

int pointpos, n, fac;

long long ans[maxn];



void init()

{

    int i, x = 0;



    pointpos = 0;

    for (i = 0; i < 6; i++)

    {

        if (st[i] == '.')

        {

            pointpos = 6 - i - 1;

            continue;

        }

        x *= 10;

        x += st[i] - '0';

    }

    fac = x;

    cin >> n;

}



void mul(long long *ans, int fac)

{

    int i;

    long long jw = 0, ls;



    for (i = 1; i <= ans[0]; i++)

    {

        ls = jw + ans[i] * fac;

        ans[i] = ls % w;

        jw = ls / w;

    }

    if (jw > 0)

    {

        ans[0]++;

        ans[ans[0]] = jw;

    }

}



void work()

{

    int i;



    ans[0] = 1;

    ans[1] = 1;

    for (i = 0; i < n; i++)

        mul(ans, fac);

}



string tostring(long long x)

{

    string re;

    int i, j;



    j = w;

    for (i = 0; i < ww; i++)

    {

        j /= 10;

        re += char('0' + x / j);

        x %= j;

    }

    return re;

}



void print()

{

    int pp, i;



    ansst = "";

    for (i = ans[0]; i > 0; i--)

        ansst += tostring(ans[i]);

    while (ansst[0] == '0')

        ansst.erase(0, 1);

    pp = pointpos * n;

    if (pp > ansst.length())

        for (i = ansst.length(); i < pp; i++)

            ansst.insert(0, "0");

    if (pp > 0)

    {

        ansst.insert(ansst.length() - pp, ".");

        while (ansst[ansst.length() - 1] == '0')

            ansst.erase(ansst.length() - 1, 1);

    }

    if (ansst[ansst.length() - 1] == '.')

        ansst.erase(ansst.length() - 1, 1);

    cout << ansst << endl;

}



int main()

{

    //freopen("t.txt", "r", stdin);

    while (cin >> st)

    {

        init();

        work();

        print();

    }

}