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weekly-contest-183

A

Statement

Metadata

给你一个数组 nums,请你从中抽取一个子序列,满足该子序列的元素之和 严格 大于未包含在该子序列中的各元素之和。

如果存在多个解决方案,只需返回 长度最小 的子序列。如果仍然有多个解决方案,则返回 元素之和最大 的子序列。

与子数组不同的地方在于,「数组的子序列」不强调元素在原数组中的连续性,也就是说,它可以通过从数组中分离一些(也可能不分离)元素得到。

注意,题目数据保证满足所有约束条件的解决方案是 唯一 的。同时,返回的答案应当按 非递增顺序 排列。

 

示例 1:

输入:nums = [4,3,10,9,8]
输出:[10,9] 
解释:子序列 [10,9] 和 [10,8] 是最小的、满足元素之和大于其他各元素之和的子序列。但是 [10,9] 的元素之和最大。 

示例 2:

输入:nums = [4,4,7,6,7]
输出:[7,7,6] 
解释:子序列 [7,7] 的和为 14 ,不严格大于剩下的其他元素之和(14 = 4 + 4 + 6)。因此,[7,6,7] 是满足题意的最小子序列。注意,元素按非递增顺序返回。

示例 3:

输入:nums = [6]
输出:[6]

 

提示:

  • 1 <= nums.length <= 500
  • 1 <= nums[i] <= 100

Metadata

Given the array nums, obtain a subsequence of the array whose sum of elements is strictly greater than the sum of the non included elements in such subsequence. 

If there are multiple solutions, return the subsequence with minimum size and if there still exist multiple solutions, return the subsequence with the maximum total sum of all its elements. A subsequence of an array can be obtained by erasing some (possibly zero) elements from the array. 

Note that the solution with the given constraints is guaranteed to be unique. Also return the answer sorted in non-increasing order.

 

Example 1:


Input: nums = [4,3,10,9,8]
Output: [10,9] 
Explanation: The subsequences [10,9] and [10,8] are minimal such that the sum of their elements is strictly greater than the sum of elements not included, however, the subsequence [10,9] has the maximum total sum of its elements. 

Example 2:


Input: nums = [4,4,7,6,7]
Output: [7,7,6] 
Explanation: The subsequence [7,7] has the sum of its elements equal to 14 which is not strictly greater than the sum of elements not included (14 = 4 + 4 + 6). Therefore, the subsequence [7,6,7] is the minimal satisfying the conditions. Note the subsequence has to returned in non-decreasing order.

Example 3:


Input: nums = [6]
Output: [6]

 

Constraints:

  • 1 <= nums.length <= 500
  • 1 <= nums[i] <= 100

B

Statement

Metadata

给你一个以二进制形式表示的数字 s 。请你返回按下述规则将其减少到 1 所需要的步骤数:

  • 如果当前数字为偶数,则将其除以 2 。

  • 如果当前数字为奇数,则将其加上 1 。

题目保证你总是可以按上述规则将测试用例变为 1 。

 

示例 1:

输入:s = "1101"
输出:6
解释:"1101" 表示十进制数 13 。
Step 1) 13 是奇数,加 1 得到 14 
Step 2) 14 是偶数,除 2 得到 7
Step 3) 7  是奇数,加 1 得到 8
Step 4) 8  是偶数,除 2 得到 4  
Step 5) 4  是偶数,除 2 得到 2 
Step 6) 2  是偶数,除 2 得到 1  

示例 2:

输入:s = "10"
输出:1
解释:"10" 表示十进制数 2 。
Step 1) 2 是偶数,除 2 得到 1 

示例 3:

输入:s = "1"
输出:0

 

提示:

  • 1 <= s.length <= 500
  • s 由字符 '0''1' 组成。
  • s[0] == '1'

Metadata

Given the binary representation of an integer as a string s, return the number of steps to reduce it to 1 under the following rules:

  • If the current number is even, you have to divide it by 2.

  • If the current number is odd, you have to add 1 to it.

It is guaranteed that you can always reach one for all test cases.

 

Example 1:

Input: s = "1101"
Output: 6
Explanation: "1101" corressponds to number 13 in their decimal representation.
Step 1) 13 is odd, add 1 and obtain 14. 
Step 2) 14 is even, divide by 2 and obtain 7.
Step 3) 7 is odd, add 1 and obtain 8.
Step 4) 8 is even, divide by 2 and obtain 4.  
Step 5) 4 is even, divide by 2 and obtain 2. 
Step 6) 2 is even, divide by 2 and obtain 1.  

Example 2:

Input: s = "10"
Output: 1
Explanation: "10" corressponds to number 2 in their decimal representation.
Step 1) 2 is even, divide by 2 and obtain 1.  

Example 3:

Input: s = "1"
Output: 0

 

Constraints:

  • 1 <= s.length <= 500
  • s consists of characters '0' or '1'
  • s[0] == '1'

C

Statement

Metadata

如果字符串中不含有任何 'aaa''bbb''ccc' 这样的字符串作为子串,那么该字符串就是一个「快乐字符串」。

给你三个整数 abc,请你返回 任意一个 满足下列全部条件的字符串 s

  • s 是一个尽可能长的快乐字符串。
  • s最多a 个字母 'a'b 个字母 'b'c 个字母 'c'
  • s 中只含有 'a''b''c' 三种字母。

如果不存在这样的字符串 s ,请返回一个空字符串 ""

 

示例 1:

输入:a = 1, b = 1, c = 7
输出:"ccaccbcc"
解释:"ccbccacc" 也是一种正确答案。

示例 2:

输入:a = 2, b = 2, c = 1
输出:"aabbc"

示例 3:

输入:a = 7, b = 1, c = 0
输出:"aabaa"
解释:这是该测试用例的唯一正确答案。

 

提示:

  • 0 <= a, b, c <= 100
  • a + b + c > 0

Metadata

A string s is called happy if it satisfies the following conditions:

  • s only contains the letters 'a', 'b', and 'c'.
  • s does not contain any of "aaa", "bbb", or "ccc" as a substring.
  • s contains at most a occurrences of the letter 'a'.
  • s contains at most b occurrences of the letter 'b'.
  • s contains at most c occurrences of the letter 'c'.

Given three integers a, b, and c, return the longest possible happy string. If there are multiple longest happy strings, return any of them. If there is no such string, return the empty string "".

A substring is a contiguous sequence of characters within a string.

 

Example 1:

Input: a = 1, b = 1, c = 7
Output: "ccaccbcc"
Explanation: "ccbccacc" would also be a correct answer.

Example 2:

Input: a = 7, b = 1, c = 0
Output: "aabaa"
Explanation: It is the only correct answer in this case.

 

Constraints:

  • 0 <= a, b, c <= 100
  • a + b + c > 0

Solution

#include <bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define SZ(x) int((x).size())
#define endl "\n"
using db = double;
using ll = long long;
using ull = unsigned long long;
using pII = pair<int, int>;
using pLL = pair<ll, ll>;
constexpr int mod = 1e9 + 7;
template <class T1, class T2>
inline void chadd(T1 &x, T2 y, int Mod = mod) {
    x += y;
    while (x >= Mod) x -= Mod;
    while (x < 0) x += Mod;
}
template <class T1, class T2>
inline void chmax(T1 &x, T2 y) {
    if (x < y)
        x = y;
}
template <class T1, class T2>
inline void chmin(T1 &x, T2 y) {
    if (x > y)
        x = y;
}
inline int nextInt() {
    int x;
    cin >> x;
    return x;
}
void rd() {}
template <class T, class... Ts>
void rd(T &arg, Ts &...args) {
    cin >> arg;
    rd(args...);
}
#define dbg(x...)                             \
    do {                                      \
        cout << "\033[32;1m" << #x << " -> "; \
        err(x);                               \
    } while (0)
void err() {
    cout << "\033[39;0m" << endl;
}
template <class T, class... Ts>
void err(const T &arg, const Ts &...args) {
    cout << arg << ' ';
    err(args...);
}
template <template <typename...> class T, typename t, typename... A>
void err(const T<t> &arg, const A &...args) {
    for (auto &v : arg) cout << v << ' ';
    err(args...);
}
void ptt() {
    cout << endl;
}
template <class T, class... Ts>
void ptt(const T &arg, const Ts &...args) {
    cout << ' ' << arg;
    ptt(args...);
}
template <class T, class... Ts>
void pt(const T &arg, const Ts &...args) {
    cout << arg;
    ptt(args...);
}
void pt() {}
template <template <typename...> class T, typename t, typename... A>
void pt(const T<t> &arg, const A &...args) {
    for (int i = 0, sze = arg.size(); i < sze; ++i) cout << arg[i] << " \n"[i == sze - 1];
    pt(args...);
}
inline ll qpow(ll base, ll n) {
    assert(n >= 0);
    ll res = 1;
    while (n) {
        if (n & 1)
            res = res * base % mod;
        base = base * base % mod;
        n >>= 1;
    }
    return res;
}
// head

class Solution {
public:
    bool gao(string &res, int &a, int &b, int &c, char _a, char _b, char _c) {
        //	pt(res, res[-2], res[-3], _a);
        if (SZ(res) > 1 && *(res.end() - 1) == _a && *(res.end() - 2) == _a) {
            if (b + c == 0)
                return false;
            if (b > c) {
                res += _b;
                --b;
            } else {
                res += _c;
                --c;
            }
        } else {
            res += _a;
            --a;
        }
        return true;
    }
    string longestDiverseString(int a, int b, int c) {
        string res = "";
        while (a + b + c > 0) {
            int t = max({a, b, c});
            if (a == t) {
                if (!gao(res, a, b, c, 'a', 'b', 'c'))
                    break;
            } else if (b == t) {
                if (!gao(res, b, a, c, 'b', 'a', 'c'))
                    break;
            } else {
                if (!gao(res, c, a, b, 'c', 'a', 'b'))
                    break;
            }
        }
        return res;
    }
};

#ifdef LOCAL

int main() {
    return 0;
}

#endif

D

Statement

Metadata
  • Link: 石子游戏 III
  • Difficulty: Hard
  • Tag: 数组 数学 动态规划 博弈

Alice 和 Bob 用几堆石子在做游戏。几堆石子排成一行,每堆石子都对应一个得分,由数组 stoneValue 给出。

Alice 和 Bob 轮流取石子,Alice 总是先开始。在每个玩家的回合中,该玩家可以拿走剩下石子中的的前 1、2 或 3 堆石子 。比赛一直持续到所有石头都被拿走。

每个玩家的最终得分为他所拿到的每堆石子的对应得分之和。每个玩家的初始分数都是 0 。比赛的目标是决出最高分,得分最高的选手将会赢得比赛,比赛也可能会出现平局。

假设 Alice 和 Bob 都采取 最优策略 。如果 Alice 赢了就返回 "Alice" Bob 赢了就返回 "Bob",平局(分数相同)返回 "Tie"

 

示例 1:

输入:values = [1,2,3,7]
输出:"Bob"
解释:Alice 总是会输,她的最佳选择是拿走前三堆,得分变成 6 。但是 Bob 的得分为 7,Bob 获胜。

示例 2:

输入:values = [1,2,3,-9]
输出:"Alice"
解释:Alice 要想获胜就必须在第一个回合拿走前三堆石子,给 Bob 留下负分。
如果 Alice 只拿走第一堆,那么她的得分为 1,接下来 Bob 拿走第二、三堆,得分为 5 。之后 Alice 只能拿到分数 -9 的石子堆,输掉比赛。
如果 Alice 拿走前两堆,那么她的得分为 3,接下来 Bob 拿走第三堆,得分为 3 。之后 Alice 只能拿到分数 -9 的石子堆,同样会输掉比赛。
注意,他们都应该采取 最优策略 ,所以在这里 Alice 将选择能够使她获胜的方案。

示例 3:

输入:values = [1,2,3,6]
输出:"Tie"
解释:Alice 无法赢得比赛。如果她决定选择前三堆,她可以以平局结束比赛,否则她就会输。

示例 4:

输入:values = [1,2,3,-1,-2,-3,7]
输出:"Alice"

示例 5:

输入:values = [-1,-2,-3]
输出:"Tie"

 

提示:

  • 1 <= values.length <= 50000
  • -1000 <= values[i] <= 1000

Metadata
  • Link: Stone Game III
  • Difficulty: Hard
  • Tag: Array Math Dynamic Programming Game Theory

Alice and Bob continue their games with piles of stones. There are several stones arranged in a row, and each stone has an associated value which is an integer given in the array stoneValue.

Alice and Bob take turns, with Alice starting first. On each player's turn, that player can take 1, 2, or 3 stones from the first remaining stones in the row.

The score of each player is the sum of the values of the stones taken. The score of each player is 0 initially.

The objective of the game is to end with the highest score, and the winner is the player with the highest score and there could be a tie. The game continues until all the stones have been taken.

Assume Alice and Bob play optimally.

Return "Alice" if Alice will win, "Bob" if Bob will win, or "Tie" if they will end the game with the same score.

 

Example 1:

Input: values = [1,2,3,7]
Output: "Bob"
Explanation: Alice will always lose. Her best move will be to take three piles and the score become 6. Now the score of Bob is 7 and Bob wins.

Example 2:

Input: values = [1,2,3,-9]
Output: "Alice"
Explanation: Alice must choose all the three piles at the first move to win and leave Bob with negative score.
If Alice chooses one pile her score will be 1 and the next move Bob's score becomes 5. In the next move, Alice will take the pile with value = -9 and lose.
If Alice chooses two piles her score will be 3 and the next move Bob's score becomes 3. In the next move, Alice will take the pile with value = -9 and also lose.
Remember that both play optimally so here Alice will choose the scenario that makes her win.

Example 3:

Input: values = [1,2,3,6]
Output: "Tie"
Explanation: Alice cannot win this game. She can end the game in a draw if she decided to choose all the first three piles, otherwise she will lose.

 

Constraints:

  • 1 <= stoneValue.length <= 5 * 104
  • -1000 <= stoneValue[i] <= 1000

Solution

#include <bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define SZ(x) int((x).size())
#define endl "\n"
using db = double;
using ll = long long;
using ull = unsigned long long;
using pII = pair<int, int>;
using pLL = pair<ll, ll>;
constexpr int mod = 1e9 + 7;
template <class T1, class T2>
inline void chadd(T1 &x, T2 y, int Mod = mod) {
    x += y;
    while (x >= Mod) x -= Mod;
    while (x < 0) x += Mod;
}
template <class T1, class T2>
inline void chmax(T1 &x, T2 y) {
    if (x < y)
        x = y;
}
template <class T1, class T2>
inline void chmin(T1 &x, T2 y) {
    if (x > y)
        x = y;
}
inline int nextInt() {
    int x;
    cin >> x;
    return x;
}
void rd() {}
template <class T, class... Ts>
void rd(T &arg, Ts &...args) {
    cin >> arg;
    rd(args...);
}
#define dbg(x...)                             \
    do {                                      \
        cout << "\033[32;1m" << #x << " -> "; \
        err(x);                               \
    } while (0)
void err() {
    cout << "\033[39;0m" << endl;
}
template <class T, class... Ts>
void err(const T &arg, const Ts &...args) {
    cout << arg << ' ';
    err(args...);
}
template <template <typename...> class T, typename t, typename... A>
void err(const T<t> &arg, const A &...args) {
    for (auto &v : arg) cout << v << ' ';
    err(args...);
}
void ptt() {
    cout << endl;
}
template <class T, class... Ts>
void ptt(const T &arg, const Ts &...args) {
    cout << ' ' << arg;
    ptt(args...);
}
template <class T, class... Ts>
void pt(const T &arg, const Ts &...args) {
    cout << arg;
    ptt(args...);
}
void pt() {}
template <template <typename...> class T, typename t, typename... A>
void pt(const T<t> &arg, const A &...args) {
    for (int i = 0, sze = arg.size(); i < sze; ++i) cout << arg[i] << " \n"[i == sze - 1];
    pt(args...);
}
inline ll qpow(ll base, ll n) {
    assert(n >= 0);
    ll res = 1;
    while (n) {
        if (n & 1)
            res = res * base % mod;
        base = base * base % mod;
        n >>= 1;
    }
    return res;
}
// head

const int INF = 0x3f3f3f3f, N = 1e6 + 10;
int n;
vector<int> vec;

int f[N];

int dfs(int cur) {
    if (cur >= n)
        return 0;
    if (f[cur] != -INF)
        return f[cur];
    int tot = 0, Max = -INF;
    for (int i = 0; i < 3 && cur + i < n; ++i) {
        tot += vec[cur + i];
        chmax(Max, tot - dfs(cur + i + 1));
    }
    //	dbg(cur, Max);
    return f[cur] = Max;
}

class Solution {
public:
    string stoneGameIII(vector<int> &stoneValue) {
        vec = stoneValue;
        n = SZ(vec);
        for (int i = 0; i <= n + 5; ++i) f[i] = -INF;
        int t = dfs(0);
        if (t > 0)
            return "Alice";
        else if (t < 0)
            return "Bob";
        else
            return "Tie";
    }
};

#ifdef LOCAL

int main() {
    return 0;
}

#endif

最后更新: January 15, 2023
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