poj1019

规律题

View Code

#include <iostream>
using namespace std;

long long        mi[20], len[50], eachlen[50];

long long    make(long long num, long long n)
{
    long long        i, x, goal;

    i = 1;
    while (true)
    {
        if (n > i * (mi[i] - mi[i - 1]))
            n -= i * (mi[i] - mi[i - 1]);
        else
            break;
        i++;
    }
    x = n / i;
    if (n % i == 0)
        x -= 1;
    n -= x * i;
    goal = x + mi[i - 1];
    n = i - n + 1;
    for (i = 0; i < n - 1; i++)
        goal /= 10;
    return goal % 10;
}

long long    work(long long n)
{
    long long        i, start, each, l, r, mid, num;
    
    i = 0;
    while (n > len[i])
        i++;
    i--;
    start = i;
    n -= len[i];
    each = eachlen[i];
    l = 0;
    r = mi[start + 1] - mi[start];
    while (l < r)
    {
        mid = (l + r) / 2;
        if ((l + r) % 2 != 0)
            mid +=1;
        if ((each + start + 1 + each + mid * (start + 1)) * mid / 2 < n)
            l = mid;
        else
            r = mid - 1;
    }
    mid = l;
    n -= (each + start + 1 + each + mid * (start + 1)) * mid / 2;
    num = mi[start] + mid;
    return make(num, n);
}

int main()
{
    long long        i, t, n;

    //freopen("t.txt", "r", stdin);
    mi[0] = 1;
    for (i = 1; i <= 5; i++)
        mi[i] = mi[i - 1] * 10;
    memset(len, 0, sizeof(len));
    memset(eachlen, 0, sizeof(eachlen));
    for (i = 1; i <= 5; i++)
    {
        len[i] = len[i - 1];
        len[i] += eachlen[i - 1] * (mi[i] - mi[i - 1]);
        len[i] += (i + i * (mi[i] - mi[i - 1])) * (mi[i] - mi[i - 1]) / 2;
        eachlen[i] += eachlen[i - 1];
        eachlen[i] += i * (mi[i] - mi[i - 1]);
    }
    cin >> t;
    while (t--)
    {
        cin >> n;
        cout << work(n) << endl;
    }
    return 0;
}