feat: add solutions to lc problem: No.3770

solution/3700-3799/3770.Largest Prime from Consecutive Prime Sum/README.md

@@ -75,32 +75,224 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3700-3799/3770.La
 
 <!-- solution:start -->
 
-### 方法一
+### 方法一：预处理 + 二分查找
+
+我们可以预处理出所有小于等于 $5 \times 10^5$ 的质数列表，然后计算从 2 开始的连续质数和，并将这些和中是质数的数存储在一个数组中 $s$ 中。
+
+对于每个查询，我们只需要在数组 $s$ 中使用二分查找找到小于等于 $n$ 的最大值即可。
+
+时间复杂度方面，预处理质数的时间复杂度为 $O(M \log \log M)$，每次查询的时间复杂度为 $(\log k)$，其中 $M$ 是预处理的上限，而 $k$ 是数组 $s$ 的长度，本题中 $k \leq 40$。
 
 <!-- tabs:start -->
 
 #### Python3
 
 ```python
-
+mx = 500000
+is_prime = [True] * (mx + 1)
+is_prime[0] = is_prime[1] = False
+primes = []
+for i in range(2, mx + 1):
+    if is_prime[i]:
+        primes.append(i)
+        for j in range(i * i, mx + 1, i):
+            is_prime[j] = False
+s = [0]
+t = 0
+for x in primes:
+    t += x
+    if t > mx:
+        break
+    if is_prime[t]:
+        s.append(t)
+
+
+class Solution:
+    def largestPrime(self, n: int) -> int:
+        i = bisect_right(s, n) - 1
+        return s[i]
 ```
 
 #### Java
 
 ```java
-
+class Solution {
+    private static final int MX = 500000;
+    private static final boolean[] IS_PRIME = new boolean[MX + 1];
+    private static final List<Integer> PRIMES = new ArrayList<>();
+    private static final List<Integer> S = new ArrayList<>();
+
+    static {
+        Arrays.fill(IS_PRIME, true);
+        IS_PRIME[0] = false;
+        IS_PRIME[1] = false;
+
+        for (int i = 2; i <= MX; i++) {
+            if (IS_PRIME[i]) {
+                PRIMES.add(i);
+                if ((long) i * i <= MX) {
+                    for (int j = i * i; j <= MX; j += i) {
+                        IS_PRIME[j] = false;
+                    }
+                }
+            }
+        }
+
+        S.add(0);
+        int t = 0;
+        for (int x : PRIMES) {
+            t += x;
+            if (t > MX) {
+                break;
+            }
+            if (IS_PRIME[t]) {
+                S.add(t);
+            }
+        }
+    }
+
+    public int largestPrime(int n) {
+        int i = Collections.binarySearch(S, n + 1);
+        if (i < 0) {
+            i = ~i;
+        }
+        return S.get(i - 1);
+    }
+}
 ```
 
 #### C++
 
 ```cpp
-
+static const int MX = 500000;
+static vector<bool> IS_PRIME(MX + 1, true);
+static vector<int> PRIMES;
+static vector<int> S;
+
+auto init = [] {
+    IS_PRIME[0] = false;
+    IS_PRIME[1] = false;
+
+    for (int i = 2; i <= MX; i++) {
+        if (IS_PRIME[i]) {
+            PRIMES.push_back(i);
+            if (1LL * i * i <= MX) {
+                for (int j = i * i; j <= MX; j += i) {
+                    IS_PRIME[j] = false;
+                }
+            }
+        }
+    }
+
+    S.push_back(0);
+    int t = 0;
+    for (int x : PRIMES) {
+        t += x;
+        if (t > MX) break;
+        if (IS_PRIME[t]) {
+            S.push_back(t);
+        }
+    }
+
+    return 0;
+}();
+
+class Solution {
+public:
+    int largestPrime(int n) {
+        auto it = upper_bound(S.begin(), S.end(), n);
+        return *(it - 1);
+    }
+};
 ```
 
 #### Go
 
 ```go
+const MX = 500000
+
+var (
+	isPrime = make([]bool, MX+1)
+	primes  []int
+	S       []int
+)
+
+func init() {
+	for i := range isPrime {
+		isPrime[i] = true
+	}
+	isPrime[0] = false
+	isPrime[1] = false
+
+	for i := 2; i <= MX; i++ {
+		if isPrime[i] {
+			primes = append(primes, i)
+			if i*i <= MX {
+				for j := i * i; j <= MX; j += i {
+					isPrime[j] = false
+				}
+			}
+		}
+	}
+
+	S = append(S, 0)
+	t := 0
+	for _, x := range primes {
+		t += x
+		if t > MX {
+			break
+		}
+		if isPrime[t] {
+			S = append(S, t)
+		}
+	}
+}
+
+func largestPrime(n int) int {
+	i := sort.SearchInts(S, n+1)
+	return S[i-1]
+}
+```
 
+#### TypeScript
+
+```ts
+const MX = 500000;
+
+const isPrime: boolean[] = Array(MX + 1).fill(true);
+isPrime[0] = false;
+isPrime[1] = false;
+
+const primes: number[] = [];
+const s: number[] = [];
+
+(function init() {
+    for (let i = 2; i <= MX; i++) {
+        if (isPrime[i]) {
+            primes.push(i);
+            if (i * i <= MX) {
+                for (let j = i * i; j <= MX; j += i) {
+                    isPrime[j] = false;
+                }
+            }
+        }
+    }
+
+    s.push(0);
+    let t = 0;
+    for (const x of primes) {
+        t += x;
+        if (t > MX) break;
+        if (isPrime[t]) {
+            s.push(t);
+        }
+    }
+})();
+
+function largestPrime(n: number): number {
+    const i = _.sortedIndex(s, n + 1) - 1;
+    return s[i];
+}
 ```
 
 <!-- tabs:end -->