Codeforces Round #592 (Div. 2)

C题全员fst，不过幸好过了D、E，上了62分。现在Rating：1815，上紫指日可待。

先贴个代码，心情好的时候回来补题解QwQ。

Codeforces1244A. Pens and Pencils（水题）

#include <bits/stdc++.h>
#define fast ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
#define N 100005
#define M 100005
#define INF 0x3f3f3f3f
#define mk(x) (1<<x) // be conscious if mask x exceeds int
#define sz(x) ((int)x.size())
#define mp(a,b) make_pair(a, b)
#define endl '
'
#define lowbit(x) (x&-x)
 
using namespace std;
typedef long long ll;
typedef double db;
 
/** fast read **/
template <typename T>
inline void read(T &x) {
    x = 0; T fg = 1; char ch = getchar();
    while (!isdigit(ch)) {
        if (ch == '-') fg = -1;
        ch = getchar();
    }
    while (isdigit(ch)) x = x*10+ch-'0', ch = getchar();
    x = fg * x;
}
template <typename T, typename... Args>
inline void read(T &x, Args &... args) { read(x), read(args...); }
 
int main()
{
    int t;
    cin >> t;
    while (t--){
        int a, b, c, d, k;
        read(a, b, c, d, k);
        int ans1 = a/c + (a%c>0);
        int ans2 = b/d + (b%d>0);
        if (ans1 + ans2 <= k) {
            cout << ans1 << ' ' << ans2 << endl;
        }
        else {
            cout << -1 << endl;
        }
    }
    return 0;
}

Codeforces1244B. Rooms and Staircases（两种情况枚举一下，水题）

#include <bits/stdc++.h>
#define fast ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
#define N 100005
#define M 100005
#define INF 0x3f3f3f3f
#define mk(x) (1<<x) // be conscious if mask x exceeds int
#define sz(x) ((int)x.size())
#define mp(a,b) make_pair(a, b)
#define endl '
'
#define lowbit(x) (x&-x)
 
using namespace std;
typedef long long ll;
typedef double db;
 
/** fast read **/
template <typename T>
inline void read(T &x) {
    x = 0; T fg = 1; char ch = getchar();
    while (!isdigit(ch)) {
        if (ch == '-') fg = -1;
        ch = getchar();
    }
    while (isdigit(ch)) x = x*10+ch-'0', ch = getchar();
    x = fg * x;
}
template <typename T, typename... Args>
inline void read(T &x, Args &... args) { read(x), read(args...); }
 
 
int main()
{
    int t;
    cin >> t;
    while (t--) {
        int n; read(n);
        string s; cin >> s;
        int ans = n;
        int pre = -1;
        for (int i = 0; i < n; i++) {
            if (s[i] == '1') {
                pre = i;
                break;
            }
        }
        if (pre != -1) {
            ans = max(ans, 2*(n - pre));
        }
        pre = -1;
        for (int i = n-1; i >= 0; i--) {
            if (s[i] == '1') {
                pre = i;
                break;
            }
        }
        if (pre != -1) {
            ans = max(ans, 2*(pre+1));
        }
        cout << ans << endl;
    }
    return 0;
}

Codeforces1244C. The Football Season（暴力，可以证明循环节长度不超过max(w,d)，全民fst）

UPD：

很多人WA on test 62，是因为算出了负数的答案。TLE on test 66的话可能是忘记break了。

#include <bits/stdc++.h>
#define fast ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
#define N 100005
#define M 100005
#define INF 0x3f3f3f3f
#define mk(x) (1<<x) // be conscious if mask x exceeds int
#define sz(x) ((int)x.size())
#define mp(a,b) make_pair(a, b)
#define endl '
'
#define lowbit(x) (x&-x)
 
using namespace std;
typedef long long ll;
typedef double db;
 
/** fast read **/
template <typename T>
inline void read(T &x) {
    x = 0; T fg = 1; char ch = getchar();
    while (!isdigit(ch)) {
        if (ch == '-') fg = -1;
        ch = getchar();
    }
    while (isdigit(ch)) x = x*10+ch-'0', ch = getchar();
    x = fg * x;
}
template <typename T, typename... Args>
inline void read(T &x, Args &... args) { read(x), read(args...); }
 
int main()
{
    ll n, p, w, d;
    read(n, p, w, d);
    ll cntw = 0, cntd = 0, cntl = 0;
    if (d < w) {
        ll sum = 0;
        for (; cntd <= w; cntd++) {
            if ((p-sum) % w == 0) {
                break;
            }
            sum += d;
        }
        cntw = (p-sum) / w;
        cntl = n-cntw-cntd;
        if ((p-sum)%w != 0 || cntw+cntd > n) {
            puts("-1");
        }
        else {
            if (cntw < 0 || cntd < 0 || cntl < 0) {
                return 0 * puts("-1");
            }
            printf("%I64d %I64d %I64d
", cntw, cntd, cntl);
        }
    }
    else if (d >= w) {
        ll sum = 0;
        for (; cntw <= d; cntw++) {
            if ((p-sum) % d == 0) {
                break;
            }
            sum += w;
        }
        cntw = (p-sum) / d;
        cntl = n-cntw-cntd;
        if ((p-sum)%d != 0 || cntw+cntd > n) {
            puts("-1");
        }
        else {
            if (cntw < 0 || cntd < 0 || cntl < 0) {
                return 0 * puts("-1");
            }
            printf("%I64d %I64d %I64d
", cntw, cntd, cntl);
        }
    }
    return 0;
}

Codeforces1244D. Paint the Tree

UPD：

可以发现给出的树不是链的情况都是不可能染色的。对于链的情况，枚举前三个节点的颜色，暴力跑一边取最小值就好了。

#include <bits/stdc++.h>
#define fast ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
#define N 100005
#define M 100005
#define INF 0x3f3f3f3f
#define mk(x) (1<<x) // be conscious if mask x exceeds int
#define sz(x) ((int)x.size())
#define mp(a,b) make_pair(a, b)
#define endl '
'
#define lowbit(x) (x&-x)
 
using namespace std;
typedef long long ll;
typedef double db;
 
/** fast read **/
template <typename T>
inline void read(T &x) {
    x = 0; T fg = 1; char ch = getchar();
    while (!isdigit(ch)) {
        if (ch == '-') fg = -1;
        ch = getchar();
    }
    while (isdigit(ch)) x = x*10+ch-'0', ch = getchar();
    x = fg * x;
}
template <typename T, typename... Args>
inline void read(T &x, Args &... args) { read(x), read(args...); }
 
/** graph **/
int tot = 0;
int head[N], nxt[M<<1], ver[M<<1];
int deg[N];
void addEdge(int u, int v) {
    nxt[++tot] = head[u], ver[tot] = v, head[u] = tot;
    deg[u]++;
}
 
int n;
ll c[N][4];
 
void input() {
    read(n);
    memset(head, -1, sizeof head);
    for (int i = 1; i <= 3; i++) {
        for (int j = 1; j <= n; j++) {
            read(c[j][i]);
        }
    }
    for (int i = 1; i <= n-1; i++) {
        int u, v;
        read(u, v);
        addEdge(u, v);
        addEdge(v, u);
    }
}
 
bool vis[N];
vector <int> nodes;
void dfs(int u) {
    vis[u] = true;
    nodes.push_back(u);
    for (int i = head[u]; i != -1; i = nxt[i]) {
        int v = ver[i];
        if (vis[v])
            continue;
        dfs(v);
    }
}
ll f[N][4][4];//i-th node choose j-th color's and previous node is k-th color tot cost
int pre[N][4][4];
int ans[N];
ll dp() {
    ll res = 1e18;
    vector <int> vs;
    for (int i = 1; i <= 3; i++)
        vs.push_back(i);
    do {
        ll tmpres = 0;
        for (int i = 0; i < sz(nodes); i++) {
            int u = nodes[i];
            tmpres += c[u][vs[i%3]];
        }
        if (res > tmpres) {
            res = tmpres;
            for (int i = 0; i < sz(nodes); i++) {
                int u = nodes[i];
                ans[u] = vs[i%3];
            }
        }
    }while (next_permutation(vs.begin(), vs.end()));
    return res;
}
int main()
{
    input();
    bool ok = true;
    int st = 1;
    for (int i = 1; i <= n; i++) {
        if (deg[i] == 1) {
            st = i;
        }
        if (deg[i] >= 3) {
            ok = false;
            break;
        }
    }
    if (!ok) {
        puts("-1");
    }
    else {
        memset(vis, false, sizeof vis);
        dfs(st);
        ll res = dp();
        cout << res << endl;
        for (int i = 1; i <= n; i++) {
            printf("%d%c", ans[i], i == n ? '
' : ' ');
        }
    }
 
    return 0;
}

Codeforces1244E. Minimizing Difference（一个很经典的模拟题）

UPD：

我的做法是把所有数值的数量统计好，并把所有数值放在一个vector里面排序去重。

用l和r指针分别指向最小的数和最大的数。

每次考虑移动数量比较小的一边，并考虑是否能向l+1（或r-1）合并。能合并的话，就把两个数量加起来，并最小值变成l+1（或最大值变成r-1）；不能合并说明k要用完了，看最多能改变多少最小值（或最大值），修改之后就直接出答案了。

#include <bits/stdc++.h>
#define fast ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
#define N 100005
#define M 100005
#define INF 0x3f3f3f3f
#define mk(x) (1<<x) // be conscious if mask x exceeds int
#define sz(x) ((int)x.size())
#define mp(a,b) make_pair(a, b)
#define endl '
'
#define lowbit(x) (x&-x)
 
using namespace std;
typedef long long ll;
typedef double db;
 
/** fast read **/
template <typename T>
inline void read(T &x) {
    x = 0; T fg = 1; char ch = getchar();
    while (!isdigit(ch)) {
        if (ch == '-') fg = -1;
        ch = getchar();
    }
    while (isdigit(ch)) x = x*10+ch-'0', ch = getchar();
    x = fg * x;
}
template <typename T, typename... Args>
inline void read(T &x, Args &... args) { read(x), read(args...); }
 
ll a[N];
unordered_map <ll, ll> mcnt;
vector <ll> v;
int main()
{
    int n; ll k;
    read(n, k);
    for (int i = 1; i <= n; i++) {
        read(a[i]);
        v.push_back(a[i]);
        mcnt[a[i]]++;
    }
    sort(v.begin(), v.end());
    v.erase(unique(v.begin(), v.end()), v.end());
    int pl = 0, pr = sz(v)-1;
    ll vall = v[pl], valr = v[pr];
    ll cntl = mcnt[vall], cntr = mcnt[valr];
 
    ll _min = vall, _max = valr;
    while (pl < pr && k > 0) {
        vall = v[pl], valr = v[pr];
        if (cntl < cntr) {
            // pl+
            ll vallr = v[pl+1];
            ll cutk = (vallr - vall) * cntl;
            if (k >= cutk) {
                k -= cutk;
                pl++;
                cntl += mcnt[vallr];
                _min = vallr;
            }
            else {
                ll add = k / cntl;
                k = 0;
                _min += add;
            }
        }
        else if (cntl >= cntr) {
            ll valrl = v[pr-1];
            ll cutk = (valr - valrl) * cntr;
            if (k >= cutk) {
                k -= cutk;
                pr--;
                cntr += mcnt[valrl];
                _max = valrl;
            }
            else {
                ll add = k / cntr;
                k = 0;
                _max -= add;
            }
        }
    }
 
    ll ans = _max - _min;
    cout << ans << endl;
 
    return 0;
}

Codeforces1244F. Chips（模拟）

UPD：

补完这题发现这场真的是手速场，一堆模拟题。错失上分的大好机会QAQ。

用手模拟一下样例，就可以发现两个结论：

1、连续的W或者B不论执行多少次操作，都不会改变。

2、连续的形如WBWB的子串，每次操作会翻转一次。

形如WWBWBWW，中间的BWB在执行一次操作之后就会翻转成WBW，两侧的WBW会合并到两侧的W中。

此时我们就可以模拟了。将所有形如WBWB的子串的两端（l，r）记录下来，每次操作将两段的点修改为和相邻的连续的W或者B相同，再l++，r--就好了。

特别地，如果没有连续的W或者B，每次操作都会使整个字符串翻转，如果翻转偶数次，则不变，奇数次每个W改成B，B改成W输出就好了。

#include <bits/stdc++.h>
#define fast ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
#define N 200005
#define M 100005
#define INF 0x3f3f3f3f
#define mk(x) (1<<x) // be conscious if mask x exceeds int
#define sz(x) ((int)x.size())
#define upperdiv(a,b) (a/b + (a%b>0))
#define mp(a,b) make_pair(a, b)
#define endl '
'
#define lowbit(x) (x&-x)

using namespace std;
typedef long long ll;
typedef double db;

/** fast read **/
template <typename T>
inline void read(T &x) {
    x = 0; T fg = 1; char ch = getchar();
    while (!isdigit(ch)) {
        if (ch == '-') fg = -1;
        ch = getchar();
    }
    while (isdigit(ch)) x = x*10+ch-'0', ch = getchar();
    x = fg * x;
}
template <typename T, typename... Args>
inline void read(T &x, Args &... args) { read(x), read(args...); }

int n, k;
bool vis[N];
int posl[N], posr[N];
string s;
void input() {
    read(n, k);
    cin >> s;
    for (int i = 0; i < n; i++) {
        if (i == 0) posl[i] = n-1;
        else posl[i] = i-1;

        if (i == n-1) posr[i] = 0;
        else posr[i] = i+1;

        vis[i] = false;
    }
}
struct SNode{
    int p;
    int type; // l1, r2;
    bool operator < (const SNode& x) const {
        return p < x.p;
    }
};
int main()
{
    input();
    queue <SNode> ps, tmpps;
    for (int l = 0; l < n; l++) {
        int r = l;
        while (r+1 < n && s[r+1] == s[r]) {
            ++r;
        }
        if (r-l+1 >= 2) {
            for (int i = l; i <= r; i++)
                vis[i] = true;
            ps.push(SNode{l, 1});
            ps.push(SNode{r, 2});
            l = r;
        }
    }
    if (!vis[0] && !vis[n-1] && s[0] == s[n-1]) {
        ps.push(SNode{n-1, 1});
        ps.push(SNode{0, 2});
        vis[0] = vis[n-1] = true;
    }

    /** 000000pr0000pl000 **/
    bool can_reverse = false;
    for (int i = 1; i <= k; i++, can_reverse = !can_reverse) {
        if (ps.empty())
            break;
        while (!ps.empty()) {
            SNode tmpr = ps.front(); ps.pop();
            if (tmpr.type == 1) {
                ps.push(tmpr);
                continue;
            }
            SNode tmpl = ps.front(); ps.pop();
            int pl = tmpl.p, pr = tmpr.p;
            if (posl[pl] != pr && posl[posl[pl]] != pr) {
                int prr = posr[pr], pll = posl[pl];
                s[prr] = s[pr]; vis[prr] = true; tmpps.push(SNode{prr, 2});
                s[pll] = s[pl]; vis[pll] = true; tmpps.push(SNode{pll, 1});
            }
            else if (posl[posl[pl]] == pr) {
                int p = posl[pl]; vis[p] = true;
                if (s[pl] == s[pr])
                    s[p] = s[pl];
                else if (can_reverse) {
                    if (s[p] == 'B') s[p] = 'W';
                    else if (s[p] == 'W') s[p] = 'B';
                }
            }
        }

        while (!tmpps.empty()) {
            ps.push(tmpps.front()); tmpps.pop();
        }
    }
    if (can_reverse) {
        for (int i = 0; i < n; i++) {
            if (!vis[i]) {
                if (s[i] == 'B') s[i] = 'W';
                else if (s[i] == 'W') s[i] = 'B';
            }
        }
    }
    cout << s << endl;

    return 0;
}
/*
6 1
BWBBWW

*/

Codeforces1244G. Running in Pairs（构造，大水题！！！！）

UPD:

显然最小的和是$sum_{i=1}^{n}i$= n*（n+1）/2。如果k小于这个最小值，那显然是不行的。

否则必定是可行的。不妨先将p和q都设1，2，……，n。sum = n*（n+1）/2。

从n到1枚举i，考虑是否能把q中的元素i放到q中第一个没被修改的点，从而让sum接近但不超过k。如果超过k就跳过当前的i。最多能接近$sum_{i=1}^{n/2}(n-i+1)-i$，这样的操作最后会把步长收敛到1，所以是最优的。

#include <bits/stdc++.h>
#define fast ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
#define N 1000005
#define M 100005
#define INF 0x3f3f3f3f
#define mk(x) (1<<x) // be conscious if mask x exceeds int
#define sz(x) ((int)x.size())
#define mp(a,b) make_pair(a, b)
#define endl '
'
#define lowbit(x) (x&-x)
 
using namespace std;
typedef long long ll;
typedef double db;
 
/** fast read **/
template <typename T>
inline void read(T &x) {
    x = 0; T fg = 1; char ch = getchar();
    while (!isdigit(ch)) {
        if (ch == '-') fg = -1;
        ch = getchar();
    }
    while (isdigit(ch)) x = x*10+ch-'0', ch = getchar();
    x = fg * x;
}
template <typename T, typename... Args>
inline void read(T &x, Args &... args) { read(x), read(args...); }
 
int ans1[N], ans2[N];
int main()
{
    ll n, k;
    read(n, k);
    ll minres = n*(n+1) / 2;
    ll maxres = 0;
    for (int i = 1; i <= n; i++) {
        ans1[i] = i;
        ans2[i] = n-i+1;
        maxres += max(ans1[i], ans2[i]);
    }
    if (k < minres)
        return 0 * puts("-1");
    else if (k >= maxres) {
        k = maxres;
    }
    else {
        int l = 1, r = n;
        ll sum = minres;
        for (int i = 1; i <= n; i++) {
            if (r >= ans1[i] && sum + r - ans1[i] <= k) {
                sum += r - ans1[i];
                ans2[i] = r--;
            }
            else {
                ans2[i] = l++;
            }
        }
    }
    cout << k << endl;
    for (int i = 1; i <= n; i++)
        printf("%d%c", ans1[i], " 
"[i==n]);
    for (int i = 1; i <= n; i++)
        printf("%d%c", ans2[i], " 
"[i==n]);
 
    return 0;
}
/*
5 20
*/