2025-11-19

CF

Problem - 1418C - Codeforces（dp+贪心好题！）（1500）

dp操作，要分开判断先手和后手

#include <bits/stdc++.h>
using namespace std;
#define LL long long
const LL mod = 998244353;
const int N=2e5+10;
int a[N],dp[N][2];
//dp[i][0] 最后一次是后手,dp[i][1] 最后一次是先手
//dp[][]记录先手取 1的最小数量
int inf = 1e9;

void solve()
{
    int n;
    cin >> n;
    for (int i = 1; i <= n;i++){
        cin >> a[i];
    }
    for (int i = 0; i <= n;i++){
        dp[i][0] = inf;
        dp[i][1] = inf;
    }
    dp[0][0] = 0;
    dp[1][1] = a[1];
    for (int i = 2; i <= n;i++){
        dp[i][1] = min(dp[i - 1][0] + a[i], dp[i - 2][0] + a[i] + a[i - 1]);
        dp[i][0] = min(dp[i - 1][1], dp[i - 2][1]);
    }
    cout << min(dp[n][0], dp[n][1]) << endl;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int T;
    cin >> T;
    while (T--)
    {
        solve();
    }
}

一个很妙的贪心解法
找全为1的长度段
分成3个1，自己拿2个，朋友skip一个，这满足最小
0的话留给朋友就行了

#include <bits/stdc++.h>
using namespace std;
#define LL long long
const LL mod = 998244353;
const int N=2e5+10;
int a[N];

void solve()
{
    int n;
    int cnt = 0,ans=0;
    cin >> n;
    cin >> a[1];
    for (int i = 2; i <= n;i++){
        cin >> a[i];
        if(a[i]==1){
            cnt++;
        }else{
            ans+=cnt/3;
            cnt = 0;
        }
    }
    ans += cnt / 3;
    cout << ans + a[1] << endl;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int T;
    cin >> T;
    while (T--)
    {
        solve();
    }
}

Problem - 1753A2 - Codeforces(贪心)（双指针）

#include <bits/stdc++.h>
using namespace std;
#define LL long long
const LL mod = 998244353;
const int N=2e5+10;
int a[N];

void solve()
{
    int n;
    cin >> n;
    for (int i = 1; i <= n;i++)
        cin >> a[i];
    int sum = 0;
    for (int i = 1; i <= n;i++){
        sum += a[i];
    }
    if(sum%2){
        cout << -1 << endl;
        return;
    }
    int l = 1, r;
    vector<pair<int, int>> ans;
    while(l<=n){
        if(a[l]==0){
            ans.push_back({l, l});
            l++;
            continue;
        }
        r = l + 1;
        while(a[r]==0)
            r++;
        if(a[l]==a[r]){
            if((r-l+1)%2==0)
                ans.push_back({l, r});
            else{
                ans.push_back({l, l});
                ans.push_back({l + 1, r});
            }
        }else{
            if((r-l+1)%2==1){
                ans.push_back({l, r});
            }else{
                ans.push_back({l, l});
                ans.push_back({l + 1, r});
            }
        }
        l = r + 1;
    }
    cout << ans.size() << endl;
    for(auto x:ans){
        cout << x.first << " " << x.second << endl;
    }
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int T;
    cin >> T;
    while (T--)
    {
        solve();
    }
}

Problem - 1307C - Codeforces（贪心）

一道考验观察能力的题
需要发现最大可能要不1个字符，要不2个字符
所以只要贪心就行啦

#include <bits/stdc++.h>
using namespace std;
#define LL long long
const LL mod = 998244353;
const int N=2e5+10;
LL c, cnt;

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    string s;
    cin >> s;
    LL maxx = 0;
    for (int t = 0; t < 26; t++)
    {
        for (int j = 0; j < 26; j++)
        {
            cnt = 0, c = 0;
            for (int i = 0; i < s.size(); i++)
            {
                if (s[i] - 'a' == j)
                    cnt += c;
                if (s[i] - 'a' == t)
                    c++;
            }
            maxx = max(maxx, cnt);
        }
        maxx = max(maxx, c);
    }
    cout << maxx << endl;
}

碎碎念

今天考完人工智能导论了，很忙的期末周马上开始了