(高精度+找规律)
题意:给定
思路:没思路就打表,打表千万不能打错!!!(打错了就会像我一样找规律找个半个小时都没找到,要注意
代码:
import java.math.BigDecimal;
import java.math.BigInteger;
import java.util.*;
public class Main {
public static double eps = 1e-6;
public static BigInteger[] a = new BigInteger[80];
public static BigInteger sqrt(BigInteger x) {
BigInteger div = BigInteger.ZERO.setBit(x.bitLength()/2);
BigInteger div2 = div;
// Loop until we hit the same value twice in a row, or wind
// up alternating.
for(;;) {
BigInteger y = div.add(x.divide(div)).shiftRight(1);
if (y.equals(div) || y.equals(div2))
return y;
div2 = div;
div = y;
}
}
public static void print_table(String mx_num) {
for(BigDecimal i=BigDecimal.valueOf(3); i.compareTo(new BigDecimal(mx_num))<=0; i=i.add(BigDecimal.ONE)) {
// p = 3*i / 2
BigDecimal p = BigDecimal.valueOf(3).multiply(i).divide(BigDecimal.valueOf(2)) ;
// area2_dec = p*(p-i+1)*(p-i)*(p-i-1)
BigDecimal x1 = p.subtract(i).add(BigDecimal.ONE);
BigDecimal x2 = p.subtract(i);
BigDecimal x3 = p.subtract(i).subtract(BigDecimal.ONE);
BigDecimal area2_dec = p.multiply(x1).multiply(x2).multiply(x3);
BigInteger area2_int = area2_dec.setScale(0, BigDecimal.ROUND_HALF_DOWN).toBigInteger();
// if(fabs(area2_int - area2_dec) > eps) continue ;
if((new BigDecimal(area2_int).subtract(area2_dec).abs()).compareTo(BigDecimal.valueOf(eps)) > 0) continue ;
BigInteger area_int = sqrt(area2_int);
// if(area_int * area_int != area2_int) continue ;
if(area_int.multiply(area_int).compareTo(area2_int) != 0) continue ;
System.out.println(i);
}
}
public static void pre_treat() {
a[0] = BigInteger.valueOf(2);
a[1] = BigInteger.valueOf(4);
for(int i=2; i<60; i++) {
a[i] = a[i-1].multiply(new BigInteger("4")).subtract(a[i-2]);
//System.out.println(a[i]);
}
}
public static void main(String[] args) {
//print_table("10000000");
pre_treat();
Scanner cin = new Scanner(System.in);
int T;
T = cin.nextInt();
while((T--) > 0) {
BigInteger n = cin.nextBigInteger();
for(int i=1; i<60; i++){
if(a[i].compareTo(n) >= 0) {
System.out.println(a[i]);
break ;
}
}
}
cin.close();
}
}