HDU 3215 The first place of 2^n（计算2^0到2^n首位出现的1,2,3...9的个数）

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

 

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题目大意：

计算2^0到2^n首位出现的1,2,3...9的个数。

解题思路：利用log10计算整理一下，根据小数部分=浮点数-整数。

#include<iostream>#include<cstdio>#include<cmath>using namespace std;const int maxN=11000;const double lg2=log10(2);int a[maxN],n;void initial(){     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]=(int)(pow(10.0,x)+1e-7);    }    //for(int i=0;i<20;i++) cout<<"2^"<<i<<" :"<<a[i]<<endl;}int main(){    initial();    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<10;i++){            printf(" %d",cnt[i]);        }        printf("\n");    }    return 0;}