Foj 1559 Count Zeros

Problem 1559 Count Zeros

Accept: 143Submit: 416
Time Limit: 1000 mSecMemory Limit : 32768 KB

Problem Description

As is known to us all, any positive decimal integer can be expressed into a unique binary digit. For example the decimal number 13 can be expressed in the form of binary digit as 1101, your task is to count the total number of 0s in the binary digit of 1..2ⁿ. For example, if n = 2, then 2ⁿ = 4. Binary digits of 1,2,3,4 are 1,10,11,100. So the number of 0s is 3.

Input

There are multiple test cases. Each test case contains a non-negative integer n (n<=5000)

Output

For each test case, print the number of 0s which the binary digits of 1..2ⁿ contains. The result may be vary large, so output it mod 2007.

Sample Input

2

3

Sample Output

3

8

推公式,

t[n]=3*t[n-1]-2*t[n-2]+2^(n-2)-1

汗死,

把t[1]写成0了, wa了无数次.......

orz...

orz...

orz...

#include<stdio.h> int t[5001]; void init() { int k=4,tmp; t[1]=1;t[2]=3;t[3]=8; for(int i=4;i<=5000;i++) { tmp=3*t[i-1]-2*t[i-2]+k-1; if(tmp<=0) tmp=3*(t[i-1]+2007)-2*t[i-2]+k-1; t[i]=tmp%2007; k*=2; k%=2007; } } int main() { int n; init(); while(scanf("%d",&n)!=EOF) printf("%d/n",t[n]); return 0; }

这是暴力的Java代码

import java.math.BigInteger; import java.util.*; public class Main { static BigInteger[] big = new BigInteger[5001]; public static void main(String[] args) { init(); Scanner in = new Scanner(System.in); int n; while(in.hasNext()) { n=in.nextInt(); System.out.println(big[n].mod(new BigInteger("2007"))); } } public static void init() { BigInteger tmp1,tmp2; big[0]=new BigInteger("0"); big[1]=new BigInteger("1"); big[2]=new BigInteger("3"); big[3]=new BigInteger("8"); int k=4; for(int i=4;i<=5000;i++) { tmp1=big[i-1].multiply(new BigInteger("3")); tmp2=big[i-2].multiply(new BigInteger("2")); tmp1=tmp1.subtract(tmp2); tmp1=tmp1.add(new BigInteger(k+"")); big[i]=tmp1.subtract(new BigInteger("1")); k*=2; k%=2007; } } }