https://nanti.jisuanke.com/t/38222
In ACM labs, there are only few members who have girlfriends. And these people can make FFF Party angry easily. One day, FFF Party prepared a function Fand a set S.
There are several unequal positive integers fi in the set S.
They calculated the value of W=∑f∈SF(F(f)) , and tried to cause W damage to those who have girlfriends. Suppose you are not a single dog and have been attacked. Now, give you the value of W and please write a program to calculate these fi in set S.
Input
The first line consists of a single integer T denoting the number of test cases.
TT lines follow, with an integer in each, denoting the result of W.
Output
For each test case, print a sequence of space-separated integers fi satisfying the equation.
If there are more than one answers, please print the lexicographically smallest one.
If there’s no such sequence, print -1 instead.
Constraints
1≤T≤10
1≤W≤10100,000
样例输入
2 1 3
样例输出
1 1 2 3
题意:问你斐波那契套斐波那契能否用这些数加起来组成W
解法：因为斐波那契套斐波那契的第29项就到了10的10万次方，那么只要矩阵快速幂预处理处斐波那契套斐波那契的第1-29项
从大到小遍历W能减则减就行 W<=10时要特殊判断，因为要输出字典序最小
所以比如3 可以是1 2 3 也可以是1 4，但字典序小是1 2 3
import java.math.BigInteger;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
// TODO 自动生成的方法存根
Scanner cin = new Scanner(System.in);
int H[] = new int [40];
BigInteger Ans[]=new BigInteger [40];
H[1]=H[2]=1;
for(int i=3;i<=29;i++)
{
H[i]=H[i-1]+H[i-2];
}
for(int i=1;i<=29;i++)
{
int n = H[i];
n -= 1;//
BigInteger a[][] = { { BigInteger.valueOf(1), BigInteger.valueOf(1) }, { BigInteger.valueOf(1), BigInteger.valueOf(0) } };
BigInteger b[][] = { { BigInteger.valueOf(1), BigInteger.valueOf(0) }, { BigInteger.valueOf(0), BigInteger.valueOf(1) } };// 单位矩阵
while (n > 0) {
if (n % 2 == 1) {
b = q(a, b);
}
a = q(a, a);
n /= 2;
}
Ans[i]=b[0][0];
//System.out.println(i+" "+Ans[i]);
}
int T=cin.nextInt();
for(int t=1;t<=T;t++)
{
BigInteger s;
s=cin.nextBigInteger();
int num[]=new int [100];
int cnt=0;
for(int i=29;i>=6;i--)
{
if(s.compareTo(Ans[i])==1||s.compareTo(Ans[i])==0)
{
s=s.subtract(Ans[i]);
num[++cnt]=i;
}
}
if(s.compareTo(BigInteger.valueOf(10))==1)
{
System.out.println("-1");
continue;
}
if(s.equals(BigInteger.valueOf(0)))
{
for(int i=cnt;i>=1;i--)
{
if(i==cnt)
System.out.print(num[i]);
else
System.out.print(" "+num[i]);
}
System.out.println();
}
else
{
if(s.equals(BigInteger.valueOf(1)))
{
System.out.print("1");
}
else if(s.equals(BigInteger.valueOf(2)))
{
System.out.print("1 2");
}
else if(s.equals(BigInteger.valueOf(3)))
{
System.out.print("1 2 3");
}
else if(s.equals(BigInteger.valueOf(4)))
{
System.out.print("1 2 4");
}
else if(s.equals(BigInteger.valueOf(5)))
{
System.out.print("1 2 3 4");
}
else if(s.equals(BigInteger.valueOf(6)))
{
System.out.print("1 5");
}
else if(s.equals(BigInteger.valueOf(7)))
{
System.out.print("1 2 5");
}
else if(s.equals(BigInteger.valueOf(8)))
{
System.out.print("1 2 3 5");
}
else if(s.equals(BigInteger.valueOf(9)))
{
System.out.print("1 2 4 5");
}
else if(s.equals(BigInteger.valueOf(10)))
{
System.out.print("1 2 3 4 5");
}
for(int i=cnt;i>=1;i--)
{
System.out.print(" "+num[i]);
}
System.out.println();
}
}
}
static BigInteger[][] q(BigInteger[][] a, BigInteger[][] b) {//
BigInteger value1 = (a[0][0].multiply( b[0][0] ) ).add( a[0][1] .multiply(b[1][0]));// 左上
BigInteger value2 = (a[0][0].multiply( b[0][1] ) ).add( a[0][1] .multiply(b[1][1]));// 左上
BigInteger value3 =(a[1][0].multiply( b[0][0] ) ).add( a[1][1] .multiply(b[1][0]));// 左上
BigInteger value4 = (a[1][0].multiply( b[0][1] ) ).add( a[1][1] .multiply(b[1][1]));// 左上
BigInteger c[][] = new BigInteger[2][2];
c[0][0] = value1 ;
c[0][1] = value2 ;
c[1][0] = value3;
c[1][1] = value4;
return c;
}
}