2025-11-16

Problem - 1924A - Codeforces（构造）

要判断s字符串是否满足是所有前k个字符的子数组
则需要把s分段，每一段都包含前k个字符
如果段数>=n长度，即满足
否则，找最后一段不满足的字符
构造一个不满足的字符串

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

void solve()
{
    set<char> s;
    int n, k, m;
    int cnt = 0;
    cin >> n >> k >> m;
    string str,ans;
    cin >> str;
    for(auto x:str){
        s.insert(x);
        if(s.size()==k){
            s.clear();
            cnt++;
            ans += x;
        }
    }
    if(ans.size()<n){
        cout << "NO\n";
        char ch;
        for (char i = 'a'; i <= 'z';i++){
            if(!s.count(i)){
                ch = i;
                break;
            }
        }
        while(ans.size()<n){
            ans += ch;
        }
        cout << ans << endl;
    }else{
        cout << "YES\n";
    }
}

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

Problem - 1547E - Codeforces（dp）(1500)

从前往后推一遍
从后往前推一遍

#include <bits/stdc++.h>
using namespace std;
#define LL long long
const LL mod = 998244353;
const int N=3e5+10;
LL a[N], t[N],dp[N];
LL inf = 1e18;

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

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

Problem - 1976C - Codeforces（贪心）（1600）

贪心，求出n,m+1和n+1,m
然后遍历减去

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

void solve()
{
    int n, m;
    cin >> n >> m;
    for (int i = 0; i < n + m + 1;i++)
        cin >> a[i];
    for (int i = 0; i < n + m + 1;i++)
        cin >> b[i];
    LL ca1 = n, cb1 = m + 1, ca2 = n + 1, cb2 = m;
    LL ans1 = 0, ans2 = 0;
    for (int i = 0; i < n + m + 1;i++){
        if(a[i]>b[i]){
            if(ca1>0){
                ans1 += a[i];
                ca1--;
            }else{
                ans1 += b[i];
                cb1--;
            }

            if(ca2>0){
                ans2 += a[i];
                ca2--;
            }else{
                ans2 += b[i];
                cb2--;
            }
        }
        else
        {
            if (cb1 > 0)
            {
                ans1 += b[i];
                cb1--;
            }
            else
            {
                ans1 += a[i];
                ca1--;
            }

            if (cb2 > 0)
            {
                ans2 += b[i];
                cb2--;
            }
            else
            {
                ans2 += a[i];
                ca2--;
            }
        }
    }
    int ca = n, cb = m;
    for (int i = 0; i < n + m + 1;i++){
        if(ca>0&&cb>0){
            if(a[i]>b[i]){
                cout << ans2 - a[i] << " ";
                ca--;
            }else{  
                cout << ans1 - b[i] << " ";
                cb--;
            }
        }else{
            if(ca==0){
                cout << ans1 - b[i] << " ";
            }else
                cout << ans2 - a[i] << " ";
        }
    }
    cout << endl;
}

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

Problem - 1811E - Codeforces（妙）（思维题）（1500）

初看以为是复杂的数位dp+二分
看了题解发现只要把十进制转化成9进制

#include <bits/stdc++.h>
using namespace std;
#define LL long long
const LL mod = 998244353;
int x,a[50];
int m[12] = {0, 1, 2, 3, 5, 6, 7, 8, 9};

void solve()
{
    LL n;
    x = 0;
    cin >> n;
    while(n){
        a[++x] = m[n % 9];
        n /= 9;
    }
    for (int i = x; i >= 1;i--){
        cout << a[i];
    }
    cout << endl;
}

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