对于Java中的大量输入,解析很长时间失败

Question

我正在尝试将一个非常大的输入读取为String,然后将其转换long为如下所示:[该程序适用于短输入]

输入是两个int用空格分隔的,例如:"1248614876148768372675689568324619856329856295619253291561358926935829358293587932857923857934572895729511 413241"

我的代码:

import java.io.*;
import java.math.*;
import java.util.*;
public class Solution {
    public static void main(String args[] ) throws Exception {
        Solution obj = new Solution();
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int T = Integer.parseInt(br.readLine());
        String input[]=new String[T];
        for (int i=0;i<T;i++) {
            input[i] = br.readLine();
            }
        for (int i=0;i<T;i++){
            StringTokenizer st = new StringTokenizer(input[i]," ");
                BigInteger N = new BigInteger(st.nextToken());
                BigInteger P = new BigInteger(st.nextToken());
                System.out.println(obj.result(N,P));
            }
        }
    }

    public BigInteger result(BigInteger N, BigInteger P){
        BigInteger temp=1;
        BigInteger c=0;
        for (BigInteger i=0;i<=N;i++){
            //System.out.println(nck(N,i));
            if ((nck(N,i)%P) ==0)
                c++;
        }
    return c;
    }

    public BigInteger nck(BigInteger N, BigInteger k){
        if (k==0)
            return 1;
        else {
            BigInteger temp=1;
            BigInteger y=1;
            BigInteger z=N;
                while(k>=1){
                temp=temp*z/y;
                y++;
                z--;
                k--;
            }
        return temp;
        }
    }
}

我得到了一个 java.lang.NumberFormatException

Answer 1

你不能将这个字符串解析为long,它太大了(大于Long.MAX_VALUE),你需要BigInteger:

BigInteger bi = new BigInteger(st.nextToken());

编辑后:

不要试图迭代一个bigInteger:如果它太长而不适合长,那么循环对你来说太长了.将它与合理的限制进行比较,如果它更小,则将其作为int并输入循环:

    BigInteger MAX = new BigInteger("1000000");
    if (bi.compareTo(MAX)<0) {
        int N = bi.intValue();
        for (int i=0; i<N; i++) {
            Test...
        }
    }