HDU 3215 The first place of 2^n（数论-其他）

The first place of 2^n

Problem Description

LMY and YY are mathematics and number theory lovers. They like to find and solve interesting mathematic problems together. One day LMY calculates 2n one by one, n=0, 1, 2,… and writes the results on a sheet of paper: 1，2，4，8，16，32，64，128，256，512，1024，……

LMY discovers that for every consecutive 3 or 4 results, there must be one among them whose first digit is 1, and comes to the conclusion that the first digit of 2n isn’t evenly distributed between 1 and 9, and the number of 1s exceeds those of others. YY now intends to use statistics to prove LMY’s discovery.

 

Input

Input consists of one or more lines, each line describing one test case: an integer N, where 0≤N≤10000.

End of input is indicated by a line consisting of -1.

 

Output

For each test case, output a single line. Each line contains nine integers. The i^th integer represents the number of js satisfying the condition that 2^j begins with i (0≤j≤N).

 

Sample Input

01310-1

 

Sample Output

1 0 0 0 0 0 0 0 01 1 0 0 0 0 0 0 01 1 0 1 0 0 0 1 04 2 1 1 1 1 0 1 0

 

Source

2009 Shanghai Network Contest Host by DHU

 

Recommend

zhuweicong

题目大意：

题目大意就是计算2^0到2^n这n个数首位为1的次数，2的次数，...9的次数。

解题思路：

log10，纯数学。注意精度误差。

X = pow(10, i * log10(2.0) - int(i * log10(2.0)));

自己想为什么。

参考代码：

#include <iostream>#include <cstring>#include <cmath>using namespace std;const int MAXN = 10010;const double eps = 1e-10;int first[MAXN], cnt[10], n;void init() {    for (int i = 0; i <= 10000; i++) {        first[i] = int(pow(10, i * log10(2.0) - int(i * log10(2.0))) + eps);    }}void solve() {    memset(cnt, 0, sizeof(cnt));    for (int i = 0; i <= n; i++) {        cnt[first[i]]++;    }    for (int i = 1; i <= 9; i++) {        if (i > 1) cout << " ";        cout << cnt[i];    }    cout << endl;}int main() {    ios::sync_with_stdio(false);    init();    while (cin >> n && n != -1) {        solve();    }    return 0;}