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feat: add solutions to lc problem: No.2197 (#2640)
No.2198.Number of Single Divisor Triplets
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‎solution/2100-2199/2198.Number of Single Divisor Triplets/README.md

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@@ -58,68 +58,69 @@
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## 解法
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### 方法一
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### 方法一：计数 + 枚举
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我们注意到，数组 $\textit{nums}$ 的元素的范围是 $[1, 100]$，因此我们可以枚举三个数 $a, b, c$，其中 $a, b, c \in [1, 100]$，然后判断 $a + b + c$ 是否只能被 $a, b, c$ 中的一个数整除。如果是，则我们可以计算出以 $a, b, c$ 为元素的单因数三元组的个数。具体计算方法如下：
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- 如果 $a = b$，那么以 $a, b, c$ 为元素的单因数三元组的个数为 $x \times (x - 1) \times z$，其中 $x$, $y$, $z$ 分别表示 $a$, $b$, $c$ 在数组 $\textit{nums}$ 中出现的次数。
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- 如果 $a = c$，那么以 $a, b, c$ 为元素的单因数三元组的个数为 $x \times (x - 1) \times y$。
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- 如果 $b = c$，那么以 $a, b, c$ 为元素的单因数三元组的个数为 $x \times y \times (y - 1)$。
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- 如果 $a, b, c$ 互不相等，那么以 $a, b, c$ 为元素的单因数三元组的个数为 $x \times y \times z$。
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最后，我们将所有的单因数三元组的个数相加即可。
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时间复杂度 $O(M^3)$，空间复杂度 $O(M)$。其中 $M$ 为数组 $\textit{nums}$ 中元素的取值范围。
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<!-- tabs:start -->
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```python
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class Solution:
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def singleDivisorTriplet(self, nums: List[int]) -> int:
68-
def check(a, b, c):
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s = a + b + c
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return sum(s % x == 0 for x in [a, b, c]) == 1
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72-
counter = Counter(nums)
79+
cnt = Counter(nums)
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ans = 0
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for a, cnt1 in counter.items():
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for b, cnt2 in counter.items():
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for c, cnt3 in counter.items():
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if check(a, b, c):
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for a, x in cnt.items():
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for b, y in cnt.items():
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for c, z in cnt.items():
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s = a + b + c
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if sum(s % v == 0 for v in (a, b, c)) == 1:
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if a == b:
79-
ans += cnt1 * (cnt1 - 1) * cnt3
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ans += x * (x - 1) * z
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elif a == c:
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ans += cnt1 * (cnt1 - 1) * cnt2
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ans += x * (x - 1) * y
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elif b == c:
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ans += cnt1 * cnt2 * (cnt2 - 1)
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ans += x * y * (y - 1)
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else:
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ans += cnt1 * cnt2 * cnt3
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ans += x * y * z
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return ans
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```
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```java
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class Solution {
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public long singleDivisorTriplet(int[] nums) {
92-
int[] counter = new int[101];
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int[] cnt = new int[101];
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for (int x : nums) {
94-
++counter[x];
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++cnt[x];
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}
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long ans = 0;
97-
for (int i = 1; i <= 100; ++i) {
98-
for (int j = 1; j <= 100; ++j) {
99-
for (int k = 1; k <= 100; ++k) {
100-
int cnt1 = counter[i], cnt2 = counter[j], cnt3 = counter[k];
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int s = i + j + k;
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int cnt = 0;
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if (s % i == 0) {
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++cnt;
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}
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if (s % j == 0) {
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++cnt;
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}
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if (s % k == 0) {
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++cnt;
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}
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if (cnt != 1) {
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continue;
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}
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if (i == j) {
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ans += (long) cnt1 * (cnt1 - 1) * cnt3;
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} else if (i == k) {
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ans += (long) cnt1 * (cnt1 - 1) * cnt2;
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} else if (j == k) {
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ans += (long) cnt1 * cnt2 * (cnt2 - 1);
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} else {
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ans += (long) cnt1 * cnt2 * cnt3;
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for (int a = 1; a <= 100; ++a) {
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for (int b = 1; b <= 100; ++b) {
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for (int c = 1; c <= 100; ++c) {
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int s = a + b + c;
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int x = cnt[a], y = cnt[b], z = cnt[c];
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int t = 0;
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t += s % a == 0 ? 1 : 0;
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t += s % b == 0 ? 1 : 0;
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t += s % c == 0 ? 1 : 0;
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if (t == 1) {
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if (a == b) {
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ans += 1L * x * (x - 1) * z;
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} else if (a == c) {
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ans += 1L * x * (x - 1) * y;
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} else if (b == c) {
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ans += 1L * x * y * (y - 1);
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} else {
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ans += 1L * x * y * z;
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}
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}
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}
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}
@@ -133,24 +134,28 @@ class Solution {
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class Solution {
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public:
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long long singleDivisorTriplet(vector<int>& nums) {
136-
vector<int> counter(101);
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for (int& x : nums) ++counter[x];
137+
int cnt[101]{};
138+
for (int x : nums) {
139+
++cnt[x];
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}
138141
long long ans = 0;
139-
for (int i = 1; i <= 100; ++i) {
140-
for (int j = 1; j <= 100; ++j) {
141-
for (int k = 1; k <= 100; ++k) {
142-
int cnt1 = counter[i], cnt2 = counter[j], cnt3 = counter[k];
143-
int s = i + j + k;
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int cnt = (s % i == 0) + (s % j == 0) + (s % k == 0);
145-
if (cnt != 1) continue;
146-
if (i == j)
147-
ans += 1ll * cnt1 * (cnt1 - 1) * cnt3;
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else if (i == k)
149-
ans += 1ll * cnt1 * (cnt1 - 1) * cnt2;
150-
else if (j == k)
151-
ans += 1ll * cnt1 * cnt2 * (cnt2 - 1);
152-
else
153-
ans += 1ll * cnt1 * cnt2 * cnt3;
142+
for (int a = 1; a <= 100; ++a) {
143+
for (int b = 1; b <= 100; ++b) {
144+
for (int c = 1; c <= 100; ++c) {
145+
int s = a + b + c;
146+
int x = cnt[a], y = cnt[b], z = cnt[c];
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int t = (s % a == 0) + (s % b == 0) + (s % c == 0);
148+
if (t == 1) {
149+
if (a == b) {
150+
ans += 1LL * x * (x - 1) * z;
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} else if (a == c) {
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ans += 1LL * x * (x - 1) * y;
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} else if (b == c) {
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ans += 1LL * x * y * (y - 1);
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} else {
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ans += 1LL * x * y * z;
157+
}
158+
}
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}
155160
}
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}
@@ -160,45 +165,68 @@ public:
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```
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```go
163-
func singleDivisorTriplet(nums []int) int64 {
164-
counter := make([]int, 101)
168+
func singleDivisorTriplet(nums []int) (ans int64) {
169+
cnt := [101]int{}
165170
for _, x := range nums {
166-
counter[x]++
171+
cnt[x]++
167172
}
168-
var ans int64
169-
check := func(a, b, c int) bool {
170-
s := a + b + c
171-
cnt := 0
172-
if s%a == 0 {
173-
cnt++
173+
f := func(a, b int) int {
174+
if a%b == 0 {
175+
return 1
174176
}
175-
if s%b == 0 {
176-
cnt++
177-
}
178-
if s%c == 0 {
179-
cnt++
180-
}
181-
return cnt == 1
177+
return 0
182178
}
183-
for i := 1; i <= 100; i++ {
184-
for j := 1; j <= 100; j++ {
185-
for k := 1; k <= 100; k++ {
186-
if check(i, j, k) {
187-
cnt1, cnt2, cnt3 := counter[i], counter[j], counter[k]
188-
if i == j {
189-
ans += int64(cnt1 * (cnt1 - 1) * cnt3)
190-
} else if i == k {
191-
ans += int64(cnt1 * (cnt1 - 1) * cnt2)
192-
} else if j == k {
193-
ans += int64(cnt1 * cnt2 * (cnt2 - 1))
179+
for a := 1; a <= 100; a++ {
180+
for b := 1; b <= 100; b++ {
181+
for c := 1; c <= 100; c++ {
182+
s := a + b + c
183+
t := f(s, a) + f(s, b) + f(s, c)
184+
if t == 1 {
185+
if a == b {
186+
ans += int64(cnt[a] * (cnt[a] - 1) * cnt[c])
187+
} else if a == c {
188+
ans += int64(cnt[a] * (cnt[a] - 1) * cnt[b])
189+
} else if b == c {
190+
ans += int64(cnt[b] * (cnt[b] - 1) * cnt[a])
194191
} else {
195-
ans += int64(cnt1 * cnt2 * cnt3)
192+
ans += int64(cnt[a] * cnt[b] * cnt[c])
196193
}
197194
}
198195
}
199196
}
200197
}
201-
return ans
198+
return
199+
}
200+
```
201+
202+
```ts
203+
function singleDivisorTriplet(nums: number[]): number {
204+
const cnt: number[] = Array(101).fill(0);
205+
for (const x of nums) {
206+
++cnt[x];
207+
}
208+
let ans = 0;
209+
const f = (a: number, b: number) => (a % b === 0 ? 1 : 0);
210+
for (let a = 1; a <= 100; ++a) {
211+
for (let b = 1; b <= 100; ++b) {
212+
for (let c = 1; c <= 100; ++c) {
213+
const s = a + b + c;
214+
const t = f(s, a) + f(s, b) + f(s, c);
215+
if (t === 1) {
216+
if (a === b) {
217+
ans += cnt[a] * (cnt[a] - 1) * cnt[c];
218+
} else if (a === c) {
219+
ans += cnt[a] * (cnt[a] - 1) * cnt[b];
220+
} else if (b === c) {
221+
ans += cnt[b] * (cnt[b] - 1) * cnt[a];
222+
} else {
223+
ans += cnt[a] * cnt[b] * cnt[c];
224+
}
225+
}
226+
}
227+
}
228+
}
229+
return ans;
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}
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```
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