2020-12-05 22:42 已编辑文远知行_产品组_软件开发

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牛客小白月赛30

黑白边

https://ac.nowcoder.com/acm/contest/9667/A

牛客小白月赛30

A. 黑白边

题解

并查集模板题，构成最小生成树，优先选择黑边。

注意不能构成树的情况，吃了一发罚时。

代码

#include <cmath>
#include <cstdio>
#include <iostream>
#include <algorithm>

using namespace std;
#define endl "\n" 
typedef long long ll;
typedef unsigned long long ull;
const ll mod = 1e9+7;

const int maxn = 2e5+55;
struct node {
    int x, y, z;
}num[maxn];
int bin[maxn];
void init(int n) {
    for (int i = 1; i <= n; ++ i) {
        bin[i] = i;
    }
}
int find(int x) {
    return bin[x] = (x == bin[x] ? x : find(bin[x]));
}
int main() {
    int n, m;
    cin >> n >> m;
    init(n);
    for (int i = 0; i < m; ++ i) {
        cin >> num[i].x >> num[i].y >> num[i].z;
    }
    sort(num, num+m, [](node x, node y) {
        return x.z < y.z;
    });
    int ans = 0, cnt = n-1;
    for (int i = 0; i < m; ++ i) {
        int xx = find(num[i].x);
        int yy = find(num[i].y);
        if (xx != yy) {
            if (num[i].z) ans ++;
            cnt --;
            bin[xx] = yy;
        }
    }
    if (cnt) cout << -1 << endl;
    else cout << ans << endl;
    return 0;
}

B. 最好的宝石

题解

线段树，每个节点中存放这个节点所代表的最左区间 l ，最右区间 r ，区间 [l,r] 之间的最大值以及最大值出现的次数。

struct node {
    int l, r, cnt, maxx;
};

注意区间合并就好了吧

tree[step].maxx = max(tree[step<<1].maxx, tree[step<<1|1].maxx);
if (tree[step].maxx == tree[step<<1].maxx) tree[step].cnt += tree[step<<1].cnt;
if (tree[step].maxx == tree[step<<1|1].maxx) tree[step].cnt += tree[step<<1|1].cnt;

更新区间的时候需要注意的是

tree[step].cnt = 0;

代码

#include <queue>
#include <cmath>
#include <cstdio>
#include <string>
#include <iostream>
#include <algorithm>

using namespace std;
#define endl "\n" 
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 2e5+55;
struct node {
    int l, r, cnt, maxx;
}tree[maxn<<2];
int a[maxn];
void Build(int l, int r, int step) {
    tree[step].l = l;
    tree[step].r = r;
    tree[step].cnt = 0;
    tree[step].maxx = 0;
    if (l == r) {
        tree[step].cnt = 1;
        tree[step].maxx = a[l];
        return;
    }
    int mid = (l+r) >> 1;
    Build(l, mid, step<<1);
    Build(mid+1, r, step<<1|1);

    tree[step].maxx = max(tree[step<<1].maxx, tree[step<<1|1].maxx);
    if (tree[step].maxx == tree[step<<1].maxx) tree[step].cnt += tree[step<<1].cnt;
    if (tree[step].maxx == tree[step<<1|1].maxx) tree[step].cnt += tree[step<<1|1].cnt;
}

void Update(int pos, int val, int step) {
    if (tree[step].l > pos || tree[step].r < pos) return;
    if (tree[step].l == tree[step].r) {
        tree[step].cnt = 1;
        tree[step].maxx = val;
        return;
    }
    Update(pos, val, step<<1);
    Update(pos, val, step<<1|1);

    tree[step].maxx = max(tree[step<<1].maxx, tree[step<<1|1].maxx);
    tree[step].cnt = 0;
    if (tree[step].maxx == tree[step<<1].maxx) tree[step].cnt += tree[step<<1].cnt;
    if (tree[step].maxx == tree[step<<1|1].maxx) tree[step].cnt += tree[step<<1|1].cnt;
}

int maxx = 0, cnt = 0;

void Query(int l, int r, int step) {
    if (tree[step].l > r || tree[step].r < l) return;
    if (tree[step].l >= l && tree[step].r <= r) {
        if (maxx == tree[step].maxx) {
            cnt += tree[step].cnt;
        }
        if (maxx < tree[step].maxx) {
            maxx = tree[step].maxx;
            cnt = tree[step].cnt;
        }
        return;
    }
    Query(l, r, step<<1);
    Query(l, r, step<<1|1);
}
string s;
int main() {
    int n, m;
    cin >> n >> m;
    for (int i = 1; i <= n; ++ i) {
        cin >> a[i];
    }
    Build(1, n, 1);
    while (m --) {
        int x, y;
        cin >> s >> x >> y;
        if (s[0] == 'A') {
            maxx = 0, cnt = 0;
            Query(x, y, 1);
            cout << maxx << " " << cnt << endl;
        }
        else {
            Update(x, y, 1);
        }
    }
    return 0;
}

C. 滑板上楼梯

题解

根据贪心思路，优先走 3 台阶，那么最优的话就会形成

31313131.... 等，这样走2 步上 4 台阶，后面的进行特判即可，

代码

#include <cmath>
#include <cstdio>
#include <iostream>
#include <algorithm>

using namespace std;
#define endl "\n" 
typedef long long ll;
typedef unsigned long long ull;
const ll mod = 1e9+7;

int main() {
    ll n;
    cin >> n;
    ll x = n/4*2;
    n %= 4;
    if (n == 3) x ++;
    else x += n;
    cout << x << endl;
    return 0;
}

D. GCD

题解

考虑一下最复杂的情况，根据抽屉原则，把这个区间之内的所有的质数全部放到子集 T 中，再添加一个非 1 并且没有出现过的数字，必定存在两个数的 gcd 不为 1 。

所以素数筛这个区间的素数，最后的结果加 2 即可。

注意 -1 的情况。

代码

#include <cmath>
#include <cstdio>
#include <string>
#include <iostream>
#include <algorithm>

using namespace std;
#define endl "\n" 
typedef long long ll;
typedef unsigned long long ull;
const ll mod = 1e9+7;
const int PSZ = 1e7+7;
int isprimes[PSZ];
int primes[PSZ/10];
int Prime(int n) {
    int cnt = 0;
    isprimes[0] = isprimes[1] = 0;
    for (int i = 2; i <= n; ++ i) {
        isprimes[i] = 1;
    }
    for (int i = 2; i <= n; ++ i) {
        if (isprimes[i]) {
            primes[++cnt] = i;
        }
        for (int j = 1; j <= cnt && 1ll*primes[j]*i <= n; ++ j) {
            isprimes[i*primes[j]] = 0;
            if (i % primes[j] == 0) break;
        }
    }
    for (int i = 1; i <= n; ++ i) {
        isprimes[i] += isprimes[i-1];
    }
    return cnt;
}

int main() {
    int n;
    cin >> n;
    int x = Prime(n);
    x += 2;
    if (x > n) {
        cout << -1 << endl;
    }
    else cout << x << endl;
    return 0;
}

E. 牛牛的加法

题解

模拟吧，注意坑点，

删除前导 0
如果所有的结果都为 0 的话，记得输出一个 0

代码

#include <cmath>
#include <cstdio>
#include <string>
#include <iostream>
#include <algorithm>

using namespace std;
#define endl "\n" 
typedef long long ll;
typedef unsigned long long ull;
const ll mod = 1e9+7;

const int maxn = 2e5+555;
string s, t;
int a[maxn] = {0};
int main() {
    cin >> s >> t;
    int n = max(s.length(), t.length());
    for (int i = 0; i < s.length(); ++ i) {
        a[i] += s[s.length()-1-i] - '0';
    }
    for (int i = 0; i < t.length(); ++ i) {
        a[i] += t[t.length()-1-i] - '0';
    }
    for (int i = 0; i < n; ++ i) {
        a[i] %= 10;
    }
    bool flag = false;
    for (int i = n-1; i >= 0; -- i) {
        if (a[i]) flag = true;
        if (flag) cout << a[i];
    }
    if (!flag) cout << 0;
    cout << endl;
    return 0;
}

F. 石子合并

题解

首先考虑一下，最后的结果为 2n-2 个石子的和并且每个石子至少没选择的了一次。

根据这个原则的话，优先选择，除了最大的石子以外的石子都被选择一次，其余的 n-1 个石子都为最大的才是最优的结果。

代码

#include <cmath>
#include <cstdio>
#include <iostream>
#include <algorithm>

using namespace std;
#define endl "\n" 
typedef long long ll;
typedef unsigned long long ull;
const ll mod = 1e9+7;

int main() {
    int n;
    cin >> n;
    ll maxx = 0, ans = 0, x;
    for (int i = 0; i < n; ++ i) {
        cin >> x;
        maxx = max(maxx, x);
        ans += x;
    }
    ans -= maxx;
    ans += maxx*(n-1);
    cout << ans << endl;
    return 0;
}

G. 滑板比赛

题解

双指针算法，模拟+贪心

对于牛妹的每个值优先选择大于它的第一个值就好了。

代码

#include <queue>
#include <cmath>
#include <cstdio>
#include <string>
#include <iostream>
#include <algorithm>

using namespace std;
#define endl "\n" 
typedef long long ll;
typedef unsigned long long ull;
const ll mod = 1e9+7;
const int maxn = 2e5+55;
int a[maxn], b[maxn];
int main() {
    int n, m;
    cin >> n >> m;
    for (int i = 0; i < n; ++ i) {
        cin >> a[i];
    }
    for (int i = 0; i < m; ++ i) {
        cin >> b[i];
    }
    sort(a, a+n);
    sort(b, b+m);
    int pos = 0;
    for (int i = 0; i < n; ++ i) {
        if (pos < m && a[i] > b[pos]) {
            pos ++;
        }
    }
    cout << pos << endl;
    return 0;
}

H. 第 k 小

题解

优先队列+模拟？

代码

#include <queue>
#include <cmath>
#include <cstdio>
#include <string>
#include <iostream>
#include <algorithm>

using namespace std;
#define endl "\n" 
typedef long long ll;
typedef unsigned long long ull;
const ll mod = 1e9+7;
priority_queue<int> que;
const int maxn = 2e5+55;
int a[maxn];
int main() {
    int n, m, k;
    cin >> n >> m >> k;
    for (int i = 0; i < n; ++ i) {
        cin >> a[i];
    }
    sort(a, a+n);
    for (int i = 0; i < n && i < k; ++ i) {
        que.push(a[i]);
    }
    int cnt = min(n, k);

    while (m --) {
        int x, y;
        cin >> x;
        if (x == 1) {
            cin >> y;
            cnt ++;
            que.push(y);
            if (cnt > k) {
                cnt --;
                que.pop();
            }
        }
        else {
            if (cnt == k) {
                cout << que.top() << endl;
            }
            else cout << -1 << endl;
        }
    }
    return 0;
}

I. 区间异或

题解

枚举区间 [l,r] 之间的所有异或值。

for (int i = 0; i < n; ++ i) {
    int ans = 0;
    for (int j = i; j < n; ++ j) {
        ans ^= a[j];
        dp[i][j] = ans;
    }
}
for (int i = 0; i < n; ++ i) {
    for (int j = i; j < n; ++ j) {
        v.push_back({dp[i][j], j-i+1});
    }
}

排序+单调栈处理一下，目的是为了求出，区间 [l,r] 的最大值是多少。

sort(v.begin(), v.end(), [](node x, node y) {
    return x.val < y.val;
});
for (int i = 0; i < (int)v.size(); ++ i) {
    while (u.size() && v[i].site <= u[u.size()-1].site) {
        u.pop_back();
    }
    u.push_back(v[i]);
}
for (int i = 0; i < (int)u.size(); ++ i) {
    xt[i] = u[i].val;
}

m 次查询，二分查找一下。

while (m --) {
    int x; 
    cin >> x;
    if (x > u[u.size()-1].val) {
        cout << -1 << endl;
    } 
    else {
        int pos = lower_bound(xt, xt+u.size(), x) - xt;
        cout << u[pos].site << endl;
    }
}

代码

#include <queue>
#include <cmath>
#include <vector>
#include <cstdio>
#include <string>
#include <iostream>
#include <algorithm>

using namespace std;
#define endl "\n" 
typedef long long ll;
typedef unsigned long long ull;
const ll mod = 1e9+7;
const int maxn = 3000+55;
int dp[maxn][maxn];
int a[maxn];
struct node {
    int val, site;
};
vector<node> v, u;
int xt[maxn];
int main() {
    int n, m;
    cin >> n >> m;
    for (int i = 0; i < n; ++ i) {
        cin >> a[i];
    }
    for (int i = 0; i < n; ++ i) {
        int ans = 0;
        for (int j = i; j < n; ++ j) {
            ans ^= a[j];
            dp[i][j] = ans;
        }
    }

    for (int i = 0; i < n; ++ i) {
        for (int j = i; j < n; ++ j) {
            v.push_back({dp[i][j], j-i+1});
        }
    }
    sort(v.begin(), v.end(), [](node x, node y) {
        return x.val < y.val;
    });
    for (int i = 0; i < (int)v.size(); ++ i) {
        while (u.size() && v[i].site <= u[u.size()-1].site) {
            u.pop_back();
        }
        u.push_back(v[i]);
    }
    for (int i = 0; i < (int)u.size(); ++ i) {
        xt[i] = u[i].val;
    }
    while (m --) {
        int x; 
        cin >> x;
        if (x > u[u.size()-1].val) {
            cout << -1 << endl;
        } 
        else {
            int pos = lower_bound(xt, xt+u.size(), x) - xt;
            cout << u[pos].site << endl;
        }
    }
    return 0;
}

J. 小游戏

题解

dp 瞎搞。。。。。

代码

#include <queue>
#include <cmath>
#include <cstdio>
#include <string>
#include <iostream>
#include <algorithm>

using namespace std;
#define endl "\n" 
typedef long long ll;
typedef unsigned long long ull;
const ll mod = 1e9+7;
const int maxn = 2e5+555;
ll dp[maxn] = {0};
int main() {
    int n, x;
    cin >> n;
    for (int i = 0; i < n; ++ i) {
        cin >> x;
        dp[x] += x;
    }
    ll maxx = 0;
    for (int i = 0; i <= 200000; ++ i) {
        if (i == 2) dp[i] += dp[i-2];
        if (i >= 3) dp[i] += max(dp[i-2], dp[i-3]);
    }
    for (int i = 0; i <= 200050; ++ i) {
//        cout << dp[i] << " ";
        maxx = max(maxx, dp[i]);
    }
//    cout << endl;
    cout << maxx << endl;
    return 0;
}