Codeforces Round #756 (Div. 3)

Codeforces 1611

A

末尾是偶数0

第一位是偶数1

如果有偶数是2

否则是-1

#include 
using namespace std;

std::mt19937 rng(std::random_device{}());
typedef long double ld;
typedef long long ll;
typedef unsigned long long ull;
typedef const int& cint;
typedef const ll& cll;
typedef pair<int, int> pii;
typedef pair<int, ll> pil;

#define ls (loc<<1)
#define rs ((loc<<1)|1)

const int mod1 = 1e9+7;
const int mod2 = 998244353;
const int inf_int = 0x7fffffff;
const int hf_int = 0x3f3f3f3f;
const ll inf_ll = 0x7fffffffffffffff;
const double ept = 1e-9;

int t;
char s[100100];

int main() {
    //freopen("1.in", "r", stdin);
    //cout.flags(ios::fixed); cout.precision(8);
    ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    int T_=1;
    std::cin >> T_;
    for(int _T=1; _T<=T_; _T++) {
        cin >> s;
        int le = strlen(s);
        if(!((s[le-1]-'0')%2)) cout << 0 << endl;
        else if(!((s[0]-'0')%2)) cout << 1 << endl;
        else {
            bool flag = 0;
            for(int i=0; i<le; i++)
                if(!((s[i]-'0')%2)) flag = 1;
            if(flag) cout << 2 << endl;
            else cout << -1 << endl;
        }
    }
    return 0;
}

B

假定 $a<b$

如果 $a\ast 3<b$ , 仅有 $a$ 能取完

否则全部都能取完

#include 
using namespace std;

std::mt19937 rng(std::random_device{}());
typedef long double ld;
typedef long long ll;
typedef unsigned long long ull;
typedef const int& cint;
typedef const ll& cll;
typedef pair<int, int> pii;
typedef pair<int, ll> pil;

#define ls (loc<<1)
#define rs ((loc<<1)|1)

const int mod1 = 1e9+7;
const int mod2 = 998244353;
const int inf_int = 0x7fffffff;
const int hf_int = 0x3f3f3f3f;
const ll inf_ll = 0x7fffffffffffffff;
const double ept = 1e-9;

ll a, b;

int main() {
    //freopen("1.in", "r", stdin);
    //cout.flags(ios::fixed); cout.precision(8);
    ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    int T_=1;
    std::cin >> T_;
    for(int _T=1; _T<=T_; _T++) {
        cin >> a >> b;
        if(a > b) swap(a, b);
        if(a*3 < b) cout << a << endl;
        else cout << (a+b)/4 << endl;
    }
    return 0;
}

C

如果最大值在左边或右边一定成立,否则一定不成立

构造答案考虑将最大值放在边上,其余值倒序一下

#include 
using namespace std;

std::mt19937 rng(std::random_device{}());
typedef long double ld;
typedef long long ll;
typedef unsigned long long ull;
typedef const int& cint;
typedef const ll& cll;
typedef pair<int, int> pii;
typedef pair<int, ll> pil;

#define ls (loc<<1)
#define rs ((loc<<1)|1)

const int mod1 = 1e9+7;
const int mod2 = 998244353;
const int inf_int = 0x7fffffff;
const int hf_int = 0x3f3f3f3f;
const ll inf_ll = 0x7fffffffffffffff;
const double ept = 1e-9;

int n;
int a[200200];
int to[200200];

int main() {
    //freopen("1.in", "r", stdin);
    //cout.flags(ios::fixed); cout.precision(8);
    ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    int T_=1;
    std::cin >> T_;
    for(int _T=1; _T<=T_; _T++) {
        cin >> n;
        for(int i=1; i<=n; i++) cin >> a[i];
        for(int i=1; i<=n; i++) to[a[i]] = i;
        vector<int> ans;
        if(to[n] != 1 && to[n] != n) cout << -1;
        else {
            if(to[n] == n) {
                for(int i=n-1; i>=1; i--) ans.push_back(a[i]);
                ans.push_back(n);
            }
            else {
                ans.push_back(n);
                for(int i=n; i>1; i--) ans.push_back(a[i]);
            }
        }
        for(int i: ans) cout << i << ' ';
        cout << endl;
    }
    return 0;
}

D

按顺序枚举给定排列,如果轮到一个点时他的父亲还没有被访问过,那么不成立,否则成立

构造答案考虑到根距离等于上一个枚举的点到根的距离+1,然后计算路径长度

#include 
using namespace std;

std::mt19937 rng(std::random_device{}());
typedef long double ld;
typedef long long ll;
typedef unsigned long long ull;
typedef const int& cint;
typedef const ll& cll;
typedef pair<int, int> pii;
typedef pair<int, ll> pil;

#define ls (loc<<1)
#define rs ((loc<<1)|1)

const int mod1 = 1e9+7;
const int mod2 = 998244353;
const int inf_int = 0x7fffffff;
const int hf_int = 0x3f3f3f3f;
const ll inf_ll = 0x7fffffffffffffff;
const double ept = 1e-9;

int n;
int b[200200], p[200200];
int dis[200200];
int ans[200200];

int main() {
    //freopen("1.in", "r", stdin);
    //cout.flags(ios::fixed); cout.precision(8);
    ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    int T_=1;
    std::cin >> T_;
    for(int _T=1; _T<=T_; _T++) {
        cin >> n;
        for(int i=1; i<=n; i++) cin >> b[i];
        for(int i=1; i<=n; i++) cin >> p[i];
        bool flag = 0;
        for(int i=1; i<=n; i++) dis[i] = -1;
        if(b[p[1]] != p[1]) flag = 1;
        else { dis[p[1]] = 0; ans[p[1]] = 0; }
        for(int i=2; i<=n; i++) {
            if(flag) break;
            int fa = b[p[i]];
            if(dis[fa] < 0) flag = 1;
            else {
                dis[p[i]] = dis[p[i-1]] + 1;
                ans[p[i]] = dis[p[i]] - dis[fa];
            }
        }
        if(flag) cout << -1 << endl;
        else {
            for(int i=1; i<=n; i++)
                cout << ans[i] << ' ';
            cout << endl;
        }
    }
    return 0;
}

E

e1:因为是树,朋友的最优行动肯定是贪心往上走,记录每一个点被朋友所占据的时间数,然后从1开始dfs,如果到某个节点的时间大于等于朋友占据的时间,那么肯定走不到,返回,能到一个叶子说明自己能赢

e2:在e1的基础上,如果不能赢的话,记录dfs中碰到的朋友数,就是答案

#include 
using namespace std;

std::mt19937 rng(std::random_device{}());
typedef long double ld;
typedef long long ll;
typedef unsigned long long ull;
typedef const int& cint;
typedef const ll& cll;
typedef pair<int, int> pii;
typedef pair<int, ll> pil;

#define ls (loc<<1)
#define rs ((loc<<1)|1)

const int mod1 = 1e9+7;
const int mod2 = 998244353;
const int inf_int = 0x7fffffff;
const int hf_int = 0x3f3f3f3f;
const ll inf_ll = 0x7fffffffffffffff;
const double ept = 1e-9;

int n, k;
bool pe[200200];
vector<int> to[200200];
int tim[200200];
int ans;

void dfs(cint u, cint fa) {
    for(int v: to[u])
        if(v != fa) dfs(v, u);
    if(pe[u]) tim[u] = 0;
    else tim[u] = hf_int;
    for(int v: to[u])
        if(v != fa) tim[u] = min(tim[u], tim[v]+1);
    // cout << u << ' ' << fa << ' ' << tim[u] << endl;
}

bool check(cint u, cint fa, cint step) {
    if(tim[u] <= step) { ++ans; return 0; }
    if(to[u].size() == 1 && u != 1) return 1;
    bool flag = 0;
    for(int v: to[u])
        if(v != fa) flag |= check(v, u, step+1);
    return flag;
}

int main() {
    // freopen("1.in", "r", stdin);
    //cout.flags(ios::fixed); cout.precision(8);
    ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    int T_=1;
    std::cin >> T_;
    for(int _T=1; _T<=T_; _T++) {
        cin >> n >> k;
        ans = 0;
        for(int i=1; i<=n; i++)
            pe[i] = 0;
        for(int i=1; i<=k; i++) {
            int tmp;
            cin >> tmp;
            pe[tmp] = 1;
        }
        int u, v;
        for(int i=1; i<=n; i++)
            to[i].clear();
        for(int i=1; i<n; i++) {
            cin >> u >> v;
            to[u].push_back(v);
            to[v].push_back(u);
        }
        dfs(1, 1);
        if(check(1, 1, 0)) cout << -1 << endl;
        else cout << ans << endl;
    }
    return 0;
}

F

可以二分+st表

也可以双指针,下面说双指针

假设从第一个人服务,确定了一个r,那么如果要让r更大,就需要前面扣得钱更少,这是一个递增的过程,所以在移动l的过程中不会有中间无解的情况出现

#include 
using namespace std;

std::mt19937 rng(std::random_device{}());
typedef long double ld;
typedef long long ll;
typedef unsigned long long ull;
typedef const int& cint;
typedef const ll& cll;
typedef pair<int, int> pii;
typedef pair<int, ll> pil;

#define ls (loc<<1)
#define rs ((loc<<1)|1)

const int mod1 = 1e9+7;
const int mod2 = 998244353;
const int inf_int = 0x7fffffff;
const int hf_int = 0x3f3f3f3f;
const ll inf_ll = 0x7fffffffffffffff;
const double ept = 1e-9;

int n;
ll s;
ll a[200200];

int main() {
    // freopen("1.in", "r", stdin);
    //cout.flags(ios::fixed); cout.precision(8);
    ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    int T_=1;
    std::cin >> T_;
    for(int _T=1; _T<=T_; _T++) {
        cin >> n >> s;
        for(int i=1; i<=n; i++) cin >> a[i];
        for(int i=1; i<=n; i++) a[i] += a[i-1];
        int r = 1, ans = -1, lst = -inf_int;
        int al, ar;
        for(int i=1; i<=n; i++) {
            if(r < i) { r = i; lst = -inf_int; }
            if(-a[i-1] >= lst && a[i]-a[i-1]+s >= 0) {
                while(r <= n && a[r]-a[i-1]+s >= 0) ++r;
                if(r-i+1 > ans) {
                    ans = r-i+1;
                    al = i;
                    ar = r-1;
                }
                lst = -a[i-1];
            }
        }
        if(ans <= 0) cout << -1 << endl;
        else cout << al << ' ' << ar << endl;
    }
    return 0;
}

G

不会