java - 使用按位移位运算符进行相乘，得到 TLE

Question

Question

Given N and M, write an equation using left shift operators whose result will be equal to the product N * M.

Input : First line has 0 < T ≤ 50000 denoting number of test cases.
Next T lines have two integers 0 < N, M ≤ 10¹⁶.

Output : For each test case print an equation for N * M resembling

(N << p1) + (N << p2)+ ...+(N << pk) where p1 ≥ p2 ≥ ... ≥ pk and k is minimum.

SAMPLE INPUT    SAMPLE OUTPUT

2
2 1             (2<<0)
2 3             (2<<1) + (2<<0)

Time Limit: 1.0 sec

我的解决方案第一种方法

int dig = (int)(Math.floor(Math.log10(m)/Math.log10(2))+1);
boolean flag = false;
for(long i = dig; i>=0; --i) {       
      if(((m>>(i-1l)) & 1l) == 1l) {
           if(flag)
               System.out.print(" + ("+n+ "<<"+(i-1)+")");
           else {
               System.out.print("("+n+"<<"+(i-1)+")");
               flag = true; 
                }
             } 
         }

第二种方法

boolean[] arr = new boolean[dig];
        int i = dig-1;
        while(m > 0) {
            if((m&1) == 1 ) {
                arr[i] = true;
            }
            i--;
            m = m>>1l;
        }
        int j = dig-1;
        for( i = 0; i < dig; ++i) {

            if(arr[i]) {
               if(flag) 
               System.out.print(" + ("+n+"<<"+j+")");
               else {
                   System.out.print("("+n+"<<"+j+")");
                   flag = true;
               }
            }

            j--;
        }

在这两种情况下，我在 8 题中得到了 5 题，其余 3 题都是 TLE，为什么？

Answer 1

实际上，我并没有看到您的两种方法中存在任何阻止最多 57 位数字的数万个乘积在一秒钟内表示为 String 的情况:
TLE 可能是由于 System.out.print() 花费了过多的时间。
也就是说，使用像这样的实用程序

   /** builds <code>n * m</code> in the form
    * <code>(n&lt;&lt;p1) + (n&lt;&lt;p2) + ... + (n&lt;&lt;pk)</code>
    * using left shift.
    * @param n  (0 < multiplicand <= 10**16)
    * @param m  0 < multiplier <= 10**16
    * @return a verbose <code>String</code> for <code>n * m</code>
    */
    static String verboseBinaryProduct(Object n, long m) {
        int shift = Long.SIZE - Long.numberOfLeadingZeros(m) - 1;
        final long highest = 1 << shift;
        final StringBuilder binary = new StringBuilder(42);
        final String chatter = ") + (" + n + "<<";
        final long rest = highest - 1;
        while (true) {
            if (0 != (highest & m))
                binary.append(chatter).append(shift);
            if (0 == (rest & m)) {
                binary.append(')');
                return binary.substring(4);
            }
            m <<= 1;
            shift -= 1;
        }
    }

和System.out.println(verboseBinaryProduct(n, m));。