HDU 3215 The first place of 2^n (number theory-water problem)

The first place of 2^n
Problem Descriptionlmy 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, every consecutive 3 or 4 results, there must be one among them whose first digit are 1, and comes to The conclusion that first digit of 2n ' t evenly distributed between 1 and 9, and the number of 1s exceeds those of Others. YY now intends to the use of statistics to prove Lmy ' s discovery.
Inputinput 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.

Outputfor each test case, output a single line. Each line contains nine integers. The ith integer represents the number of JS satisfying the condition that 2j 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

Source2009 Shanghai Network Contest Host by Dhu
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To the effect of the topic ":

Mom, this is our big Donghua 09 out of the question Ah, good, is actually good water ah.

The main problem is to calculate 2^0 to 2^n this n number of the first 1 times, 2 times, ... 9 of the Times.

Problem Solving Ideas:

I'm not going to tell you that LOG10 will find what you're looking for.

Problem Solving Code:

#include <iostream> #include <cstdio> #include <cmath>using namespace Std;const int Maxn=10010;const Double lg2=log10 (2.0); int a[maxn];void ini () {    a[0]=1,a[1]=2,a[2]=4,a[3]=8;    for (int i=4;i<maxn;i++) {        double x=i*lg2-int (i*lg2+1e-7);        A[i]=pow (10.0,x);    }    for (int i=0;i<20;i++) cout<< "2^" <<i<< ":" <<a[i]<<endl;} int main () {    ini ();    int n;    while (scanf ("%d", &n)!=eof && n!=-1) {        int cnt[10]={0};        for (int i=0;i<=n;i++) {            cnt[a[i]]++;        }        printf ("%d", cnt[1]);        for (int i=2;i<=9;i++) {            printf ("%d", Cnt[i]);        }        printf ("\ n");    }    return 0;}

HDU 3215 The first place of 2^n (number theory-water problem)