ZOJ 1442 Dinner Is Ready 容斥原理 + java大数

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/**
 *
 * @author Liu
 */
import java.util.*; 
import java.math.*; //大数头文件
public class Main {
    static final int maxn = 15;
    static int[] be = new int[maxn];
    static int[] en = new int[maxn];
    static int n, m;
    static BigInteger ans;
    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        int t = input.nextInt();
        while ((t--) > 0) { //返回值必须bool
            n = input.nextInt();
            m = input.nextInt();
            int sum = 0;
            for (int i = 1; i <= n; ++i) {
                be[i] = input.nextInt();
                en[i] = input.nextInt();
                sum += be[i];
            }
            ans = BigInteger.ZERO; //清0
            dfs(1, sum, 0);
            System.out.println(ans);
        }
    }
    public static BigInteger C(int n, int m) { //这个数字很大
        if (n < m) return BigInteger.ZERO;
        if (n == m) return BigInteger.ONE;
        BigInteger ans = BigInteger.ONE;
        int mx = Math.max(n - m, m); //调用最大值
        int mi = n - mx;
        for (int i = 1; i <= mi; ++i) {
            ans = ans.multiply(BigInteger.valueOf(mx + i)); //转换成大数的方法 
            ans = ans.divide(BigInteger.valueOf(i)); //记得接收返回值
        }
        return ans;
    }
    public static void dfs(int cur, int tot, int has) {
        if (cur == n + 1) {
            if (has % 2 == 1) {
                ans = ans.subtract(C(m - tot + n - 1, n - 1));
            } else ans = ans.add(C(m - tot + n - 1, n - 1));
            return;
        }
        dfs(cur + 1, tot - be[cur] + en[cur] + 1, has + 1);
        dfs(cur + 1, tot, has);
    }
}