UVa1210 - Sum of Consecutive Prime Numbers(欧拉筛法即线性筛法)

Some positive integers can be represented by a sum of one or more consecutive prime numbers. How many such representations does a given positive integer have? For example, the integer 53 has two representations
5 + 7 + 11 + 13 + 17 and 53. The integer
41 has three representations 2 + 3 + 5 + 7 + 11 + 13,
11 + 13 + 17, and 41. The integer 3 has only one representation, which is 3. The integer 20 has no such representations. Note that summands must be consecutive prime numbers, so neither
7 + 13 nor 3 + 5 + 5 + 7 is a valid representation for the integer 20. Your mission is to write a program that reports the number of representations for the given positive integer.

Input 

The input is a sequence of positive integers each in a separate line. The integers are between 2 and 10000, inclusive. The end of the input is indicated by a zero.

Output 

The output should be composed of lines each corresponding to an input line except the last zero. An output line includes the number of representations for the input integer as the sum of one or more consecutive prime numbers. No other characters should be
inserted in the output.

Sample Input 

2
3
17
41
20
666
12
53
0

Sample Output 

1
1
2
3
0
0
1
2

用欧拉筛法（即线性筛法）求出10000内的素数，然后用数组f[i]表示从第1个素数到第i个素数的和，用两重循环判断f[i] - f[j]是否等于n
import java.io.FileInputStream;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.io.OutputStreamWriter;
import java.io.StreamTokenizer;
import java.util.ArrayList;

public class Main implements Runnable
{
private static final boolean DEBUG = false;
private static final int N = 10001;
private BufferedReader cin;
private PrintWriter cout;
private StreamTokenizer tokenizer;
private boolean[] vis;
private ArrayList<Integer> vPrime = new ArrayList<Integer>();
private int[] f;
private int n;

private void preprocess()
{
vis = new boolean
;
vis[0] = vis[1] = true;

for (int i = 2; i < N; i++) {
if (!vis[i]) {
vPrime.add(i);
}

int size = vPrime.size();
for (int j = 0; j < size && i * vPrime.get(j) < N; j++) {
int num = vPrime.get(j);
vis[i * num] = true;

if (i % num == 0) break;
}
}

int size = vPrime.size();
f = new int[size + 1];

f[0] = 0;
for (int i = 0; i < size; i++) {
f[i + 1] = f[i] + vPrime.get(i);
}
}

private void init()
{
try {
if (DEBUG) {
cin = new BufferedReader(new InputStreamReader(
new FileInputStream("d:\\OJ\\uva_in.txt")));
} else {
cin = new BufferedReader(new InputStreamReader(System.in));
}

tokenizer = new StreamTokenizer(cin);

cout = new PrintWriter(new OutputStreamWriter(System.out));
preprocess();

} catch (Exception e) {
e.printStackTrace();
}
}

private String next()
{
try {
tokenizer.nextToken();
if (tokenizer.ttype == StreamTokenizer.TT_EOF) return null;
else if (tokenizer.ttype == StreamTokenizer.TT_NUMBER) {
return String.valueOf((int)tokenizer.nval);
} else return tokenizer.sval;
} catch (Exception e) {
e.printStackTrace();
return null;
}
}

private boolean input()
{
n = Integer.parseInt(next());
if (n == 0) return false;

return true;
}

private void solve()
{
int size = f.length;

int ans = 0;
for (int i = 0; i < size; i++) {
for (int j = i + 1; j < size; j++) {
if (f[j] - f[i] == n) ans++;
}
}

cout.println(ans);
cout.flush();
}

public void run()
{
init();
while (input()) {
solve();
}
}

public static void main(String[] args)
{
new Thread(new Main()).start();
}
}