Codeforces Round #780 (Div. 3)

By xiaruize

My Performance

水调歌头
苏轼

丙辰中秋，欢饮达旦，大醉，作此篇，兼怀子由。
明月几时有？把酒问青天。
不知天上宫阙，今夕是何年？
我欲乘风归去，又恐琼楼玉宇，高处不胜寒。
起舞弄清影，何似在人间？

转朱阁，低绮户，照无眠。
不应有恨，何事长向别时圆？
人有悲欢离合，月有阴晴圆缺，此事古难全。
但愿人长久，千里共婵娟。

Codeforces rating涨了 qwq

A. Vasya and Coins

Method

分两种情况：

情况一

他有≥1\geq1个1-buble的硬币

ans=a+2bans=a+2b

情况二

他没有1-buble的硬币

则他一定付不出1-buble这个金额

∴ans=1\therefore ans=1

Code

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// Problem: A. Vasya and Coins
// Contest: Codeforces Round #780 (Div. 3)
// URL: https://codeforces.com/contest/1660/problem/A
// Memory Limit: 256 MB
// Time Limit: 1000 ms
//
// Powered by CP Editor (https://cpeditor.org)

/*
    Name:
    Author: xiaruize
    Date:
*/
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define ull unsigned long long
#define ALL(a) (a).begin(), (a).end()
#define pb push_back
#define mk make_pair
#define pii pair<int, int>
#define pis pair<int, string>
#define sec second
#define fir first
#define sz(a) int((a).size())
#define rep(i, x, y) for (int i = x; i <= y; i++)
#define repp(i, x, y) for (int i = x; i >= y; i--)
#define Yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define No cout << "No" << endl
#define NO cout << "NO" << endl
#define double long double
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
const int N = 1e5 + 10;

// bool st;
int t;
// bool en;

void solve()
{
    int a, b;
    cin >> a >> b;
    if (a == 0)
        cout << "1" << endl;
    else
        cout << a + 2 * b + 1 << endl;
}

signed main()
{
    // freopen(".in","r",stdin);
    // freopen(".out","w",stdout);
    // cerr<<(&en-&st)/1024.0/1024.0<<endl;
    cin >> t;
    while (t--)
        solve();
    return 0;
}

B. Vlad and Candies

Method

每次操作，我们优先降低最大值，只要最大值和次大值的差≤1\leq1就一定可以完成

Code

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// Problem: B. Vlad and Candies
// Contest: Codeforces Round #780 (Div. 3)
// URL: https://codeforces.com/contest/1660/problem/B
// Memory Limit: 256 MB
// Time Limit: 1000 ms
//
// Powered by CP Editor (https://cpeditor.org)

/*
    Name:
    Author: xiaruize
    Date:
*/
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define ull unsigned long long
#define ALL(a) (a).begin(), (a).end()
#define pb push_back
#define mk make_pair
#define pii pair<int, int>
#define pis pair<int, string>
#define sec second
#define fir first
#define sz(a) int((a).size())
#define rep(i, x, y) for (int i = x; i <= y; i++)
#define repp(i, x, y) for (int i = x; i >= y; i--)
#define Yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define No cout << "No" << endl
#define NO cout << "NO" << endl
#define double long double
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
const int N = 2e5 + 10;

// bool st;
int t;
int n;
int a[N];
int mx, smx;
// bool en;

void solve()
{
    mx = smx = 0;
    cin >> n;
    for (int i = 1; i <= n; i++)
        cin >> a[i];
    for (int i = 1; i <= n; i++)
    {
        if (a[i] > mx)
        {
            smx = mx;
            mx = a[i];
        }
        else if (a[i] > smx)
            smx = a[i];
    }
    if (mx - smx <= 1)
        YES;
    else
        NO;
}

signed main()
{
    // freopen(".in","r",stdin);
    // freopen(".vout","w",stdout);
    // cerr<<(&en-&st)/1024.0/1024.0<<endl;
    cin >> t;
    while (t--)
        solve();
    return 0;
}

C. Get an Even String

Feeling

赛时在这题上卡了很久。。。

Method

贪心策略：

按顺序遍历，如果存在配对则一定优先选取。

Code

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// Problem: C. Get an Even String
// Contest: Codeforces Round #780 (Div. 3)
// URL: https://codeforces.com/contest/1660/problem/C
// Memory Limit: 256 MB
// Time Limit: 1000 ms
//
// Powered by CP Editor (https://cpeditor.org)

/*
    Name:
    Author: xiaruize
    Date:
*/
#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
const int N = 2e5 + 10;

// bool st;
int t;
string s;
int la[30];
int now = 0;
int ans = 0;
// bool en;

void solve()
{
    ans = 0;
    now = 0;
    s = "";
    memset(la, 0, sizeof(la));
    cin >> s;
    s = ' ' + s;
    for (int i = 1; i < s.size(); i++)
    {
        if (la[s[i] - 'a'] && now < la[s[i] - 'a'])
        {
            for (int j = max(1, now); j < i; j++)
                la[s[j] - 'a'] = 0;
            ans += i - now - 2;
            now = i;
            la[s[i] - 'a'] = 0;
        }
        else
        {
            la[s[i] - 'a'] = i;
        }
        // cerr << i << ' ' << ans << endl;
    }
    for (int i = 0; i < 26; i++)
        if (la[i])
            ans++;
    cout << ans << endl;
    return;
}

signed main()
{
    // freopen(".in","r",stdin);
    // freopen(".out","w",stdout);
    // cerr<<(&en-&st)/1024.0/1024.0<<endl;
    cin >> t;
    while (t--)
        solve();
    return 0;
}

D. Maximum Product Strikes Back

Method

并不是很难，思路很简单

可以用每个00为界线，把数组分成几段，下面我们讨论对每一段求最大值：

因为全部删去时的值为11,所以任意一段的极值一定为正。

可以用前缀和维护负数的个数。(设到第ii位为sumisum_i)

对于每一段，可以O(n)O(n)找到包含区间最大的一对(i,j)(i,j)使得sumi≡sumj(mod2)sum_i\equiv sum_j(mod 2)

可以证明区间[i+1,j]为当前段中的最优答案

2100大于longlong的范围，统计一定不要统计值qwq2^{100}大于long long的范围，统计一定不要统计值qwq

Code

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// Problem: D. Maximum Product Strikes Back
// Contest: Codeforces Round #780 (Div. 3)
// URL: https://codeforces.com/contest/1660/problem/D
// Memory Limit: 256 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)

/*
    Name:
    Author: xiaruize
    Date:
*/
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define ull unsigned long long
#define ALL(a) (a).begin(), (a).end()
#define pb push_back
#define mk make_pair
#define pii pair<int, int>
#define pis pair<int, string>
#define sec second
#define fir first
#define sz(a) int((a).size())
#define rep(i, x, y) for (int i = x; i <= y; i++)
#define repp(i, x, y) for (int i = x; i >= y; i--)
#define Yes cout << "Yes" << endl
#define YES cout << "YES" << endl
#define No cout << "No" << endl
#define NO cout << "NO" << endl
#define double long double
const int INF = 0x3f3f3f3f;
const int MOD = 1000000007;
const int N = 2e5 + 10;

// bool st;
int t;
int n;
int a[N];
int cnt[N];
int sum[N];
int fo, fe;
int lo, le;
// bool en;

void solve()
{
    memset(sum, 0, sizeof(sum));
    le = lo = fe = fo = 0;
    cin >> n;
    for (int i = 1; i <= n; i++)
    {
        cin >> a[i];
        sum[i] = sum[i - 1] + (a[i] == 2 || a[i] == -2);
    }
    int mx = 0;
    pii ans;
    for (int i = 1; i <= n; i++)
    {
        if (a[i] != 0)
        {
            cnt[i] = cnt[i - 1] + (a[i] < 0);
            if (cnt[i] & 1 && !fo)
                fo = i;
            if (cnt[i] & 1)
                lo = i;
            else
                le = i;
        }
        if (a[i] == 0)
        {
            if (fo && (sum[lo] - sum[fo]) > mx)
            {
                mx = (sum[lo] - sum[fo]);
                ans = {fo, lo};
            }
            if ((sum[le] - sum[fe]) > mx)
            {
                mx = (sum[le] - sum[fe]);
                ans = {fe, le};
            }
            fe = le = i;
            fo = lo = 0;
            cnt[i] = 0;
        }
    }
    if (fo && (sum[lo] - sum[fo]) > mx)
    {
        mx = (sum[lo] - sum[fo]);
        ans = {fo, lo};
    }
    if ((sum[le] - sum[fe]) > mx)
    {
        mx = (sum[le] - sum[fe]);
        ans = {fe, le};
    }
    cout << ans.fir << ' ' << n - ans.sec << endl;
}

signed main()
{
    // freopen(".in","r",stdin);
    // freopen(".out","w",stdout);
    // cerr<<(&en-&st)/1024.0/1024.0<<endl;
    cin >> t;
    while (t--)
        solve();
    return 0;
}

Happy-April-Fool’s-Day!