B Fibonacci

斐波那契数求和表示

最新推荐文章于 2024-12-30 16:53:07 发布

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Fibonacci

Time Limit: 2000MS Memory limit: 131072K

题目描述

Fibonacci numbers are well-known as follow:

Now given an integer N, please find out whether N can be represented as the sum of several Fibonacci numbers in such a way that the sum does not include any two consecutive Fibonacci numbers.

输入

Multiple test cases, the first line is an integer T (T<=10000), indicating the number of test cases.

Each test case is a line with an integer N (1<=N<=10⁹).

输出

One line per case. If the answer don’t exist, output “-1” (without quotes). Otherwise, your answer should be formatted as “N=f1+f2+…+fn”. N indicates the given number and f1, f2, … , fn indicating the Fibonacci numbers in ascending order. If there are multiple ways, you can output any of them.

示例输入

示例输出

5=5

6=1+5

7=2+5

100=3+8+89

题意：给出一个数，这个数是斐波那契数列中的数的和，按斐波那契数列的顺序输出，同时这些数不可以相邻。

因为数据最大是 10^9，打个斐波那契数列数列的表，就45个，然后暴力搜索，每个数都可以表示成斐波那契数的和，所以不存在输出 -1 的情况。

当时在做题的时候一点儿也没想到，然后队友用搜索做的，而且当时以为任一种顺序都可以，提交WA后改了好几次都没想到哪错了，幸亏最后又看了一遍题，过了，以后一定要仔细读完题，代码如下：

#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <string.h>
using namespace std;
int a[46];
int result[46];
int vis[47];


int main(){
	int T, i, N;
	a[1] = 1;
	a[2] = 2;
	for(i = 3; i < 46; ++i)
		a[i] = a[i - 1] + a[i - 2];	
	cin >> T;
	
	while(T--){
		cin >> N;
		cout << N << "=";
		memset(vis, 0, sizeof(vis));
		memset(result, 0, sizeof(result));
		int k = 1;
		for(i = 45; i > 0; --i){
			if(N >= a[i] && vis[i - 1] == 0 && vis[i + 1] == 0){
				vis[i] = 1;
				N -= a[i];
				result[k++] = a[i];
				if(N == 0)
					break;
			}
		}
		
		for(i = --k; i > 1; --i){
			cout << result[i] << "+";
		}
		cout << result[1] << endl;
	} 
	return 0;
}

Fibonacci

Time Limit: 2000MS Memory limit: 131072K

题目描述

Fibonacci numbers are well-known as follow:

Now given an integer N, please find out whether N can be represented as the sum of several Fibonacci numbers in such a way that the sum does not include any two consecutive Fibonacci numbers.