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2013
11-10

POJ 2231 Moo Volume [解题报告] Java

Moo Volume

问题描述 :

Farmer John has received a noise complaint from his neighbor, Farmer Bob, stating that his cows are making too much noise.

FJ’s N cows (1 <= N <= 10,000) all graze at various locations on a long one-dimensional pasture. The cows are very chatty animals. Every pair of cows simultaneously carries on a conversation (so every cow is simultaneously MOOing at all of the N-1 other cows). When cow i MOOs at cow j, the volume of this MOO must be equal to the distance between i and j, in order for j to be able to hear the MOO at all. Please help FJ compute the total volume of sound being generated by all N*(N-1) simultaneous MOOing sessions.

输入:

* Line 1: N

* Lines 2..N+1: The location of each cow (in the range 0..1,000,000,000).

输出:

There are five cows at locations 1, 5, 3, 2, and 4.

样例输入:

5
1
5
3
2
4

样例输出:

40

温馨提示:

INPUT DETAILS:

There are five cows at locations 1, 5, 3, 2, and 4.

OUTPUT DETAILS:

Cow at 1 contributes 1+2+3+4=10, cow at 5 contributes 4+3+2+1=10, cow at 3 contributes 2+1+1+2=6, cow at 2 contributes 1+1+2+3=7, and cow at 4 contributes 3+2+1+1=7. The total volume is (10+10+6+7+7) = 40.

解题代码:

import java.io.BufferedInputStream;
import java.math.BigInteger;
import java.util.Arrays;
import java.util.Scanner;

public class Main {

	public static BigInteger getVolume(int[] a, int n) {
		BigInteger sum = new BigInteger("0");
		BigInteger s = new BigInteger("0");
		int i = 0;
		for (; i < n - 1; i++) {
			s = s.add(new BigInteger(String
				.valueOf((Math.abs(a[0] - a[i + 1])))));
		}
		sum = new BigInteger(String.valueOf(s)).add(sum);
		i = 1;
		while (i < n) {
		s = s.subtract(new BigInteger(String.valueOf(new BigInteger(String
			.valueOf((n - 2 * i))).multiply(new BigInteger(String
			.valueOf((Math.abs(a[i] - a[i - 1]))))))));
		sum = sum.add(new BigInteger(String.valueOf(s)));
			i++;
		}
		return sum;
	}

	public static void main(String[] args) {

		Scanner sc = new Scanner(new BufferedInputStream(System.in));
		int n = sc.nextInt();
		int[] a = new int[n];
		for (int i = 0; i < n; i++) {
			a[i] = sc.nextInt();
		}
		
		Arrays.sort(a);
		System.out.println(Main.getVolume(a, n));
	}

}

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  2. /*
    * =====================================================================================
    *
    * Filename: 1366.cc
    *
    * Description:
    *
    * Version: 1.0
    * Created: 2014年01月06日 14时52分14秒
    * Revision: none
    * Compiler: gcc
    *
    * Author: Wenxian Ni (Hello World~), [email protected]
    * Organization: AMS/ICT
    *
    * =====================================================================================
    */

    #include
    #include

    using namespace std;

    int main()
    {
    stack st;
    int n,i,j;
    int test;
    int a[100001];
    int b[100001];
    while(cin>>n)
    {
    for(i=1;i>a[i];
    for(i=1;i>b[i];
    //st.clear();
    while(!st.empty())
    st.pop();
    i = 1;
    j = 1;

    while(in)
    break;
    }
    while(!st.empty()&&st.top()==b[j])
    {
    st.pop();
    j++;
    }
    }
    if(st.empty())
    cout<<"YES"<<endl;
    else
    cout<<"NO"<<endl;
    }
    return 0;
    }