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;
}