2025-11-21

CF

Problem - 1234C - Codeforces（贪心）

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

void solve()
{
    int n;
    cin >> n;
    cin >> s[0] >> s[1];
    int x = 0;
    for (int i = 0; i < n;i++){
        if(s[x][i]>'2'){
            x ^= 1;
            if(s[x][i]<='2'){
                cout << "NO\n";
                return;
            }
        }
    }
    if(x==0){
        cout << "NO\n";
    }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 - 1689C - Codeforces（树形dp）（1500）

一道很板的树形dp还看了好久

#include <bits/stdc++.h>
using namespace std;
#define LL long long
const LL mod = 998244353;
const int N=3e5+10;
vector<int> e[N];
int cnt[N],dp[N];
//dp[i]:表示 i 被感染,最多能保留多少结点

void dfs(int u,int fa){
    dp[u] = 0, cnt[u] = 1;
    int sum = 0;
    for (int i = 0; i < e[u].size();i++){
        int v = e[u][i];
        if(v==fa)
            continue;
        dfs(v, u);
        sum += dp[v];//sum存u的两个子结点都被感染，最多保存的结点数
        cnt[u] += cnt[v];
    }
    for (int i = 0; i < e[u].size();i++){
        int v = e[u][i];
        if(v==fa)
            continue;
        dp[u] = max(dp[u], sum - dp[v] + cnt[v] - 1);
        //dp[v1]+cnt[v2]-1,相当于保留 v2结点,加上v1被感染后最多能保留的结点数
    }
}

void solve()
{
    int n;
    cin >> n;
    for (int i = 0; i <= n;i++){
        e[i].clear();
    }
    for (int i = 1; i < n; i++)
    {
        int u, v;
        cin >> u >> v;
        e[u].push_back(v);
        e[v].push_back(u);
    }
    dfs(1, 0);
    cout << dp[1] << endl;
}

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

Problem - 44H - Codeforces（dp）（1700虚高）

要注意看数组有没有越界

#include <bits/stdc++.h>
using namespace std;
#define LL long long
const LL mod = 998244353;
const int N=2e5+10;
LL dp[55][10];//记得开LL！！！
int a[55];

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    string s;
    cin >> s;
    for (int i = 0; i < s.size();i++)
        a[i] = s[i] - '0';
    for (int i = 0; i <= 9;i++){
        dp[0][i] = 1;
    }
    for (int i = 0; i < s.size()-1;i++){
        for (int j = 0; j <= 9;j++){
            if((j+a[i+1])%2==0){
                dp[i + 1][(j + a[i + 1]) / 2] += dp[i][j];
            }
            else{
                dp[i + 1][(j + a[i + 1]) / 2] += dp[i][j];
                dp[i + 1][(j + a[i + 1]) / 2 + 1] += dp[i][j];
            }
        }
    }
    LL ans = 0;
    for (int i = 0; i<= 9;i++){
        ans += dp[s.size() - 1][i];
    }
    int flag = 1;
    for (int i = 0; i < s.size()-1;i++){
        if(abs(a[i]-a[i+1])>1)
            flag = 0;
    }
    ans -= flag;
    cout << ans << endl;
}

dfs解法！

要分清局部变量和全局变量
如果是递归的话，要用局部变量！！！

#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N=2e5+10;
int a[N];
int cnt,res;
int f[N][15];
int n;

int dfs(int x,int lst){
    if(x==n)
        return 1;
    if(~f[x][lst])
        return f[x][lst];
    int cnt = a[x + 1] + lst;//注意这里
    f[x][lst] = dfs(x + 1, cnt / 2);
    if(cnt&1)
        f[x][lst] += dfs(x + 1, cnt / 2 + 1);
    return f[x][lst];
}

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    string s;
    cin >> s;
    n = s.size();
    s = " " + s;
    for (int i = 1; i <= n;i++){
        a[i] = s[i] - '0';
    }
    memset(f, -1, sizeof f);
    for (int i = 0; i <= 9;i++){
        res += dfs(1, i);
    }
    int flag = 1;
    for (int i = 1; i < n;i++){
        if(abs(a[i]-a[i+1])>1){
            flag = 0;
            break;
        }
    }
    res -= flag;
    cout << res << endl;
}