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Commit c6e0817

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feat: add solutions to lc problem: No.1414 (doocs#4273)
No.1414.Find the Minimum Number of Fibonacci Numbers Whose Sum Is K
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‎solution/0800-0899/0830.Positions of Large Groups/README.md‎

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<strong>输出:</strong>[]
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</pre>
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<p><strong>提示:</strong></p>
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<ul>

‎solution/1400-1499/1414.Find the Minimum Number of Fibonacci Numbers Whose Sum Is K/README.md‎

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<!-- solution:start -->
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### 方法一
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### 方法一:贪心
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我们可以每次贪心地选取一个不超过 $k$ 的最大的斐波那契数,然后将 $k$ 减去该数,答案加一,一直循环,直到 $k = 0$ 为止。
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由于每次贪心地选取了最大的不超过 $k$ 的斐波那契数,假设为 $b,ドル前一个数为 $a,ドル后一个数为 $c$。将 $k$ 减去 $b,ドル得到的结果,一定小于 $a,ドル也即意味着,我们选取了 $b$ 之后,一定不会选到 $a$。这是因为,如果能选上 $a,ドル那么我们在前面就可以贪心地选上 $b$ 的下一个斐波那契数 $c,ドル这不符合我们的假设。因此,我们在选取 $b$ 之后,可以贪心地减小斐波那契数。
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时间复杂度 $O(\log k),ドル空间复杂度 $O(1)$。
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<!-- tabs:start -->
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```python
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class Solution:
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def findMinFibonacciNumbers(self, k: int) -> int:
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def dfs(k):
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if k < 2:
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return k
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a = b = 1
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while b <= k:
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a, b = b, a + b
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return 1 + dfs(k - a)
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return dfs(k)
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a = b = 1
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while b <= k:
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a, b = b, a + b
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ans = 0
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while k:
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if k >= b:
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k -= b
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ans += 1
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a, b = b - a, a
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return ans
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```
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#### Java
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```java
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class Solution {
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public int findMinFibonacciNumbers(int k) {
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if (k < 2) {
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return k;
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}
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int a = 1, b = 1;
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while (b <= k) {
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b = a + b;
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a = b - a;
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int c = a + b;
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a = b;
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b = c;
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}
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return 1 + findMinFibonacciNumbers(k - a);
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int ans = 0;
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while (k > 0) {
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if (k >= b) {
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k -= b;
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++ans;
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}
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int c = b - a;
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b = a;
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a = c;
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}
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return ans;
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}
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}
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```
@@ -113,80 +127,100 @@ class Solution {
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class Solution {
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public:
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int findMinFibonacciNumbers(int k) {
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if (k < 2) return k;
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int a = 1, b = 1;
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while (b <= k) {
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b = a + b;
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a = b - a;
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int c = a + b;
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a = b;
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b = c;
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}
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return 1 + findMinFibonacciNumbers(k - a);
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int ans = 0;
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while (k > 0) {
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if (k >= b) {
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k -= b;
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++ans;
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}
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int c = b - a;
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b = a;
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a = c;
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}
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return ans;
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}
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};
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```
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#### Go
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```go
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func findMinFibonacciNumbers(k int) int {
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if k < 2 {
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return k
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}
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func findMinFibonacciNumbers(k int) (ans int) {
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a, b := 1, 1
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for b <= k {
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a, b = b, a+b
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c := a + b
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a = b
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b = c
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}
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return 1 + findMinFibonacciNumbers(k-a)
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for k > 0 {
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if k >= b {
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k -= b
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ans++
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}
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c := b - a
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b = a
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a = c
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}
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return
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}
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```
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#### TypeScript
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```ts
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const arr = [
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1836311903, 1134903170, 701408733, 433494437, 267914296, 165580141, 102334155, 63245986,
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39088169, 24157817, 14930352, 9227465, 5702887, 3524578, 2178309, 1346269, 832040, 514229,
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317811, 196418, 121393, 75025, 46368, 28657, 17711, 10946, 6765, 4181, 2584, 1597, 987, 610,
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377, 233, 144, 89, 55, 34, 21, 13, 8, 5, 3, 2, 1,
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];
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function findMinFibonacciNumbers(k: number): number {
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let res = 0;
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for (const num of arr) {
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if (k >= num) {
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k -= num;
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res++;
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if (k === 0) {
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break;
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}
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let [a, b] = [1, 1];
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while (b <= k) {
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let c = a + b;
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a = b;
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b = c;
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}
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let ans = 0;
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while (k > 0) {
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if (k >= b) {
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k -= b;
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ans++;
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}
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let c = b - a;
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b = a;
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a = c;
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}
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return res;
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return ans;
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}
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```
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#### Rust
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```rust
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const FIB: [i32; 45] = [
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1836311903, 1134903170, 701408733, 433494437, 267914296, 165580141, 102334155, 63245986,
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39088169, 24157817, 14930352, 9227465, 5702887, 3524578, 2178309, 1346269, 832040, 514229,
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317811, 196418, 121393, 75025, 46368, 28657, 17711, 10946, 6765, 4181, 2584, 1597, 987, 610,
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377, 233, 144, 89, 55, 34, 21, 13, 8, 5, 3, 2, 1,
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];
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impl Solution {
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pub fn find_min_fibonacci_numbers(mut k: i32) -> i32 {
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let mut res = 0;
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for &i in FIB.into_iter() {
181-
if k >= i {
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k -= i;
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res += 1;
184-
if k == 0 {
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break;
186-
}
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let mut a = 1;
206+
let mut b = 1;
207+
while b <= k {
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let c = a + b;
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a = b;
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b = c;
211+
}
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213+
let mut ans = 0;
214+
while k > 0 {
215+
if k >= b {
216+
k -= b;
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ans += 1;
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}
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let c = b - a;
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b = a;
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a = c;
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}
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res
223+
ans
190224
}
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}
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```

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