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feat: add solutions to lc problem: No.2197 #2640

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204 changes: 116 additions & 88 deletions solution/2100-2199/2198.Number of Single Divisor Triplets/README.md
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Expand Up @@ -58,68 +58,69 @@

## 解法

### 方法一
### 方法一：计数 + 枚举

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：

- 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。
- 如果 $a = c$，那么以 $a, b, c$ 为元素的单因数三元组的个数为 $x \times (x - 1) \times y$。
- 如果 $b = c$，那么以 $a, b, c$ 为元素的单因数三元组的个数为 $x \times y \times (y - 1)$。
- 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 $x \times y \times z$。

最后，我们将所有的单因数三元组的个数相加即可。

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。

<!-- tabs:start -->

```python
class Solution:
def singleDivisorTriplet(self, nums: List[int]) -> int:
def check(a, b, c):
s = a + b + c
return sum(s % x == 0 for x in [a, b, c]) == 1

counter = Counter(nums)
cnt = Counter(nums)
ans = 0
for a, cnt1 in counter.items():
for b, cnt2 in counter.items():
for c, cnt3 in counter.items():
if check(a, b, c):
for a, x in cnt.items():
for b, y in cnt.items():
for c, z in cnt.items():
s = a + b + c
if sum(s % v == 0 for v in (a, b, c)) == 1:
if a == b:
ans += cnt1 * (cnt1 - 1) * cnt3
ans += x * (x - 1) * z
elif a == c:
ans += cnt1 * (cnt1 - 1) * cnt2
ans += x * (x - 1) * y
elif b == c:
ans += cnt1 * cnt2 * (cnt2 - 1)
ans += x * y * (y - 1)
else:
ans += cnt1 * cnt2 * cnt3
ans += x * y * z
return ans
```

```java
class Solution {
public long singleDivisorTriplet(int[] nums) {
int[] counter = new int[101];
int[] cnt = new int[101];
for (int x : nums) {
++counter[x];
++cnt[x];
}
long ans = 0;
for (int i = 1; i <= 100; ++i) {
for (int j = 1; j <= 100; ++j) {
for (int k = 1; k <= 100; ++k) {
int cnt1 = counter[i], cnt2 = counter[j], cnt3 = counter[k];
int s = i + j + k;
int cnt = 0;
if (s % i == 0) {
++cnt;
}
if (s % j == 0) {
++cnt;
}
if (s % k == 0) {
++cnt;
}
if (cnt != 1) {
continue;
}
if (i == j) {
ans += (long) cnt1 * (cnt1 - 1) * cnt3;
} else if (i == k) {
ans += (long) cnt1 * (cnt1 - 1) * cnt2;
} else if (j == k) {
ans += (long) cnt1 * cnt2 * (cnt2 - 1);
} else {
ans += (long) cnt1 * cnt2 * cnt3;
for (int a = 1; a <= 100; ++a) {
for (int b = 1; b <= 100; ++b) {
for (int c = 1; c <= 100; ++c) {
int s = a + b + c;
int x = cnt[a], y = cnt[b], z = cnt[c];
int t = 0;
t += s % a == 0 ? 1 : 0;
t += s % b == 0 ? 1 : 0;
t += s % c == 0 ? 1 : 0;
if (t == 1) {
if (a == b) {
ans += 1L * x * (x - 1) * z;
} else if (a == c) {
ans += 1L * x * (x - 1) * y;
} else if (b == c) {
ans += 1L * x * y * (y - 1);
} else {
ans += 1L * x * y * z;
}
}
}
}
Expand All @@ -133,24 +134,28 @@ class Solution {
class Solution {
public:
long long singleDivisorTriplet(vector<int>& nums) {
vector<int> counter(101);
for (int& x : nums) ++counter[x];
int cnt[101]{};
for (int x : nums) {
++cnt[x];
}
long long ans = 0;
for (int i = 1; i <= 100; ++i) {
for (int j = 1; j <= 100; ++j) {
for (int k = 1; k <= 100; ++k) {
int cnt1 = counter[i], cnt2 = counter[j], cnt3 = counter[k];
int s = i + j + k;
int cnt = (s % i == 0) + (s % j == 0) + (s % k == 0);
if (cnt != 1) continue;
if (i == j)
ans += 1ll * cnt1 * (cnt1 - 1) * cnt3;
else if (i == k)
ans += 1ll * cnt1 * (cnt1 - 1) * cnt2;
else if (j == k)
ans += 1ll * cnt1 * cnt2 * (cnt2 - 1);
else
ans += 1ll * cnt1 * cnt2 * cnt3;
for (int a = 1; a <= 100; ++a) {
for (int b = 1; b <= 100; ++b) {
for (int c = 1; c <= 100; ++c) {
int s = a + b + c;
int x = cnt[a], y = cnt[b], z = cnt[c];
int t = (s % a == 0) + (s % b == 0) + (s % c == 0);
if (t == 1) {
if (a == b) {
ans += 1LL * x * (x - 1) * z;
} else if (a == c) {
ans += 1LL * x * (x - 1) * y;
} else if (b == c) {
ans += 1LL * x * y * (y - 1);
} else {
ans += 1LL * x * y * z;
}
}
}
}
}
Expand All @@ -160,45 +165,68 @@ public:
```

```go
func singleDivisorTriplet(nums []int) int64 {
counter := make([]int, 101)
func singleDivisorTriplet(nums []int) (ans int64) {
cnt := [101]int{}
for _, x := range nums {
counter[x]++
cnt[x]++
}
var ans int64
check := func(a, b, c int) bool {
s := a + b + c
cnt := 0
if s%a == 0 {
cnt++
f := func(a, b int) int {
if a%b == 0 {
return 1
}
if s%b == 0 {
cnt++
}
if s%c == 0 {
cnt++
}
return cnt == 1
return 0
}
for i := 1; i <= 100; i++ {
for j := 1; j <= 100; j++ {
for k := 1; k <= 100; k++ {
if check(i, j, k) {
cnt1, cnt2, cnt3 := counter[i], counter[j], counter[k]
if i == j {
ans += int64(cnt1 * (cnt1 - 1) * cnt3)
} else if i == k {
ans += int64(cnt1 * (cnt1 - 1) * cnt2)
} else if j == k {
ans += int64(cnt1 * cnt2 * (cnt2 - 1))
for a := 1; a <= 100; a++ {
for b := 1; b <= 100; b++ {
for c := 1; c <= 100; c++ {
s := a + b + c
t := f(s, a) + f(s, b) + f(s, c)
if t == 1 {
if a == b {
ans += int64(cnt[a] * (cnt[a] - 1) * cnt[c])
} else if a == c {
ans += int64(cnt[a] * (cnt[a] - 1) * cnt[b])
} else if b == c {
ans += int64(cnt[b] * (cnt[b] - 1) * cnt[a])
} else {
ans += int64(cnt1 * cnt2 * cnt3)
ans += int64(cnt[a] * cnt[b] * cnt[c])
}
}
}
}
}
return ans
return
}
```

```ts
function singleDivisorTriplet(nums: number[]): number {
const cnt: number[] = Array(101).fill(0);
for (const x of nums) {
++cnt[x];
}
let ans = 0;
const f = (a: number, b: number) => (a % b === 0 ? 1 : 0);
for (let a = 1; a <= 100; ++a) {
for (let b = 1; b <= 100; ++b) {
for (let c = 1; c <= 100; ++c) {
const s = a + b + c;
const t = f(s, a) + f(s, b) + f(s, c);
if (t === 1) {
if (a === b) {
ans += cnt[a] * (cnt[a] - 1) * cnt[c];
} else if (a === c) {
ans += cnt[a] * (cnt[a] - 1) * cnt[b];
} else if (b === c) {
ans += cnt[b] * (cnt[b] - 1) * cnt[a];
} else {
ans += cnt[a] * cnt[b] * cnt[c];
}
}
}
}
}
return ans;
}
```

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