終于開始補提了
重點 : C， E的倒著算， F的染色，G的相鄰的轉換；
B - Hard Calculation

#include <iostream> #include <algorithm> #include <cstring> #include <vector> #include <cmath> #include <stack> #include <map> #define mid (l+r>>1) #define lowbit(x) (x&-x) using namespace std; typedef long long LL; typedef pair<int,int> PII; const int N = 4e5+10, mod = 1e9+7; void mull(int &a, LL b){a = a*b%mod;return ;} void add(int &a, LL b){a = (a+b)%mod;return ;}int main() {LL a, b;scanf("%lld%lld", &a, &b);while(a||b){if(a%10 + b%10 >= 10)return puts("Hard"), 0;a /= 10; b /= 10;}puts("Easy");return 0; }

C - Cheese
思路 : 倒著算，直接減，和上海熱身賽B都好神奇。

#include <iostream> #include <algorithm> #include <cstring> #include <vector> #include <cmath> #include <stack> #include <map> #define mid (l+r>>1) #define lowbit(x) (x&-x) using namespace std; typedef long long LL; typedef pair<int,int> PII; const int N = 3e5+10, mod = 998244353; void mull(int &a, LL b){a = a*b%mod;return ;} void add(int &a, LL b){a = (a+b)%mod;return ;}int n, w; PII a[N]; int main() {scanf("%d%d", &n, &w);for(int i = 1;i <= n;i ++){int x, y;scanf("%d%d", &x, &y);a[i] = {-x, y};}sort(a+1, a+n+1);LL ans = 0;for(int i = 1;i <= n;i ++){int mi = min(a[i].second, w);ans += (LL)mi*a[i].first;w -= mi;}cout<<-ans<<endl;return 0; }

D - Longest X
思路 : 雙指針或二分，練一下雙指針，哎。

#include <iostream> #include <algorithm> #include <cstring> #include <vector> #include <cmath> #include <stack> #include <map> #define mid (l+r>>1) #define lowbit(x) (x&-x) using namespace std; typedef long long LL; typedef pair<int,int> PII; const int N = 2e5+10, mod = 1e9+7; void mull(int &a, LL b){a = a*b%mod;return ;} void add(int &a, LL b){a = (a+b)%mod;return ;}char c[N]; int main() {int k;scanf("%s%d", c, &k);int num = 0, l = 0, r = 0, ans = 0;while(c[r]){if(c[r] == 'X') r++;else if(num == k) num -= c[l++]=='.';else if(num < k) num ++, r++;ans = max(ans, r-l);}cout<<ans<<endl;return 0; }

E - Graph Destruction
思路 ：從后往前算，倒著并查集合并。

#include <iostream> #include <algorithm> #include <cstring> #include <vector> #include <cmath> #include <stack> #include <queue> #define mid (l+r>>1) #define lowbit(x) (x&-x) using namespace std; typedef long long LL; typedef pair<int,int> PII; const int N = 2e5+10, mod = 1e9+7;vector<int>G[N]; int fa[N], num = 0; int find(int u){return fa[u] == u ? fa[u]:fa[u] = find(fa[u]);}void dfs(int u) {if(!u)return ;int t = num;num++;int y = find(u);for(auto x : G[u]){int a = find(x);if(a != y) num--, fa[a] = y;}dfs(u-1);cout<<t<<endl;return ; } int main() {int n, m;scanf("%d%d", &n, &m);for(int i = 1;i <= n;i ++) fa[i] = i;while(m --){int a, b;scanf("%d%d", &a, &b);if(a > b) swap(a, b);G[a].push_back(b);}dfs(n);return 0; }

F - Make Bipartite
思路 ： dp，給每個點染色之后就可以清晰的得到要刪哪條邊了，但是這是個封閉圖形，題解就確定了起點的顏色，然后算到第n個點，這時候先不考慮n到1的邊 。 然后再枚舉n和1的四種顏色組合，枚舉的時候要注意n到0的邊已經刪過了，因為這個wa了兩次。

#include <iostream> #include <algorithm> #include <cstring> #include <vector> #include <cmath> #include <stack> #include <map> #define mid (l+r>>1) #define lowbit(x) (x&-x) using namespace std; typedef long long LL; typedef pair<int,int> PII; const int N = 3e5+10, mod = 998244353; void mull(int &a, LL b){a = a*b%mod;return ;} void add(int &a, LL b){a = (a+b)%mod;return ;}LL dp[N][2][2]; int a[N], b[N];int main() {int n;scanf("%d", &n);for(int i = 1;i <= n;i ++)scanf("%d", a+i);for(int i = 1;i <= n;i ++)scanf("%d", b+i);memset(dp, 0x7f, sizeof dp);dp[1][0][0] = a[1];dp[1][1][1] = 0;for(int i = 2;i <= n;i ++){dp[i][1][0] = min(dp[i-1][1][0]+b[i-1], dp[i-1][0][0]);dp[i][1][1] = min(dp[i-1][1][1]+b[i-1], dp[i-1][0][1]);dp[i][0][0] = min(dp[i-1][1][0], dp[i-1][0][0]+b[i-1])+a[i];dp[i][0][1] = min(dp[i-1][1][1], dp[i-1][0][1]+b[i-1])+a[i];}cout<<min(min(dp[n][0][0], dp[n][1][1]) + b[n], min(dp[n][1][0], dp[n][0][1]));return 0; }

G - Longest Y

#include <iostream> #include <algorithm> #include <cstring> #include <vector> #include <cmath> #include <stack> #include <map> #define mid (l+r>>1) #define lowbit(x) (x&-x) using namespace std; typedef long long LL; typedef pair<int,int> PII; const int N = 3e5+10, mod = 998244353; void mull(int &a, LL b){a = a*b%mod;return ;} void add(int &a, LL b){a = (a+b)%mod;return ;}int b[N], idx; // 看哪個題解的，放到一起連續就是坐標減去指數相等。 string c; LL k, a[N]; // a數組是前綴和，方便計算一段區間放到的最小步數； bool ch(int len) {for(int r = len;r <= idx;r ++){int l = r-len+1, lle = mid-l+1;LL sum = (LL)b[mid]*lle - (a[mid]-a[l-1]) + a[r]-a[mid] - (LL)b[mid]*(len-lle);if(sum<=k)return 1; }return 0; }int main() {cin>>c>>k;for(int i = 0;i < c.size();i ++)if(c[i] == 'Y')b[++idx] = i-idx;sort(b+1, b+idx+1);for(int i = 1;i <= idx;i ++)a[i] = a[i-1]+b[i];int l = 0, r = idx+1;while(l < r) // 二分都是找到連續區間的右邊的左端點； {if(ch(mid))l = mid+1;else r = mid; }cout<<--l<<endl;return 0; }

總結

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