C - Multiple Gift

題解

首項是X，每次乘個2，暴力統(tǒng)計

代碼

#include <bits/stdc++.h> #define fi first #define se second #define pii pair<int,int> #define pdi pair<db,int> #define mp make_pair #define pb push_back #define enter putchar('\n') #define space putchar(' ') #define eps 1e-8 #define mo 974711 #define MAXN 2005 //#define ivorysi using namespace std; typedef long long int64; typedef double db; template<class T> void read(T &res) {res = 0;char c = getchar();T f = 1;while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}while(c >= '0' && c <= '9') {res = res * 10 + c - '0';c = getchar();}res *= f; } template<class T> void out(T x) {if(x < 0) {x = -x;putchar('-');}if(x >= 10) {out(x / 10);}putchar('0' + x % 10); } int64 X,Y;int main() { #ifdef ivorysifreopen("f1.in","r",stdin); #endifread(X);read(Y);int l = 1;while(X <= Y / 2) {++l;X *= 2;}out(l);enter; }

D - Wide Flip

題解

從大到小枚舉K（也可以二分）
如果\(2K <= N\)，此時\(K\)一定合法
否則看前\(K\)個和后\(K\)個交出來的中間一段，這段的前面和后面1和0是可以單個位置隨便改的（用一次K + 1的區(qū)間再用一次K的區(qū)間）
只要看中間一段是不是全1或者全0即可

代碼

#include <bits/stdc++.h> #define fi first #define se second #define pii pair<int,int> #define pdi pair<db,int> #define mp make_pair #define pb push_back #define enter putchar('\n') #define space putchar(' ') #define eps 1e-8 #define mo 974711 #define MAXN 2005 //#define ivorysi using namespace std; typedef long long int64; typedef double db; template<class T> void read(T &res) {res = 0;char c = getchar();T f = 1;while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}while(c >= '0' && c <= '9') {res = res * 10 + c - '0';c = getchar();}res *= f; } template<class T> void out(T x) {if(x < 0) {x = -x;putchar('-');}if(x >= 10) {out(x / 10);}putchar('0' + x % 10); } char s[1000005]; int sum[100005],N; int main() { #ifdef ivorysifreopen("f1.in","r",stdin); #endifscanf("%s",s + 1);N = strlen(s + 1);for(int i = 1 ; i <= N ; ++i) {sum[i] = sum[i - 1];if(s[i] == '1') sum[i]++;}for(int K = N ; K >= 1 ; --K) {int r = K;int l = N - K + 1;if(l > r) {out(K);enter;return 0;}if(sum[r] - sum[l - 1] == 0 || sum[r] - sum[l - 1] == r - l + 1) {out(K);enter;return 0;}} }

E - Papple Sort

題解

只要每次把靠邊的一個字符的匹配點找出來同樣后移然后一起刪除即可
就是討論一下
...A...A...B...B..
A要和這個一對B交換兩次
B如果要過去也要換兩次，所以都一樣
A..B..A..B
A要和B換一次，如果B到外面也是和A換一次，所以都一樣
樹狀數(shù)組維護(hù)一下前綴有多少數(shù)即可

代碼

#include <bits/stdc++.h> #define fi first #define se second #define pii pair<int,int> #define pdi pair<db,int> #define mp make_pair #define pb push_back #define enter putchar('\n') #define space putchar(' ') #define eps 1e-8 #define mo 974711 #define MAXN 200005 //#define ivorysi using namespace std; typedef long long int64; typedef double db; template<class T> void read(T &res) {res = 0;char c = getchar();T f = 1;while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}while(c >= '0' && c <= '9') {res = res * 10 + c - '0';c = getchar();}res *= f; } template<class T> void out(T x) {if(x < 0) {x = -x;putchar('-');}if(x >= 10) {out(x / 10);}putchar('0' + x % 10); } char s[MAXN]; int N; vector<int> c[30]; bool vis[MAXN]; int tr[MAXN]; void Init() {scanf("%s",s + 1);N = strlen(s + 1); } int lowbit(int x) {return x & -x;} void Insert(int x,int v) {while(x <= N) {tr[x] += v;x += lowbit(x);} } int Query(int x) {int res = 0;while(x > 0) {res += tr[x];x -= lowbit(x);}return res; } void Solve() {int cnt = 0;for(int i = N ; i >= 1 ; --i) {c[s[i] - 'a'].pb(i);}for(int i = 0 ; i < 26 ; ++i) {if(c[i].size() & 1) ++cnt;}if(cnt >= 2) {puts("-1");return;}for(int i = 1 ; i <= N ; ++i) Insert(i,1);int64 ans = 0;int k = 0;for(int i = N ; i >= 1 ; --i) {if(vis[i]) continue;int t = c[s[i] - 'a'].back();c[s[i] - 'a'].pop_back();if(t == i){ ++k;continue;}ans += Query(t - 1) + k;vis[t] = 1;Insert(t,-1);}out(ans);enter; } int main() { #ifdef ivorysifreopen("f1.in","r",stdin); #endifInit();Solve(); }

F - Christmas Tree

題解

啊這個只要dp一下第一個值再二分。。。二分判的方法是……
woc這不NOIPD1T3啊怎么一模一樣啊

代碼

#include <bits/stdc++.h> #define fi first #define se second #define pii pair<int,int> #define pdi pair<db,int> #define mp make_pair #define pb push_back #define enter putchar('\n') #define space putchar(' ') #define eps 1e-8 #define mo 974711 #define MAXN 100005 //#define ivorysi using namespace std; typedef long long int64; typedef double db; template<class T> void read(T &res) {res = 0;char c = getchar();T f = 1;while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}while(c >= '0' && c <= '9') {res = res * 10 + c - '0';c = getchar();}res *= f; } template<class T> void out(T x) {if(x < 0) {x = -x;putchar('-');}if(x >= 10) {out(x / 10);}putchar('0' + x % 10); } int N; struct node {int to,next; }E[MAXN * 2]; int head[MAXN],sumE; int dp[MAXN],A,lim,res; int val[MAXN],tot; bool vis[MAXN]; void add(int u,int v) {E[++sumE].to = v;E[sumE].next = head[u];head[u] = sumE; } void Init() {read(N);int a,b;for(int i = 1 ; i < N ; ++i) {read(a);read(b);add(a,b);add(b,a);}} void dfs(int u,int fa) {int son = 0;for(int i = head[u] ; i ; i = E[i].next) {int v = E[i].to;if(v != fa) {dfs(v,u);++son;dp[u] += dp[v];}}dp[u] -= son / 2;if(son % 2 == 0 && fa) dp[u]++; } struct BIT {int tr[MAXN],s;void clear() {for(int i = 1 ; i <= s ; ++i) tr[i] = 0;}int lowbit(int x) {return x & -x;}int query(int x) {int res = 0;while(x > 0) {res += tr[x];x -= lowbit(x);}return res;}void insert(int x,int v) {while(x <= s) {tr[x] += v;x += lowbit(x);}}int find_max() {int L = 1,R = s;int x = query(s);if(x == 0) return 0;while(L < R) {int m = (L + R + 1) >> 1;if(query(m - 1) != x) L = m;else R = m - 1;}return L;} }BST; void calc(int u,int fa) {for(int i = head[u] ; i ; i = E[i].next) {int v = E[i].to;if(v != fa) {calc(v,u);}}tot = 0;for(int i = head[u] ; i ; i = E[i].next) {int v = E[i].to;if(v != fa) {val[++tot] = dp[v];}}if(!tot) {dp[u] = 1;++res;return;}BST.clear();sort(val + 1,val + tot + 1);for(int i = 1 ; i <= tot ; ++i) vis[i] = 1;BST.s = tot;int p = 0;for(int i = tot ; i >= 1 ; --i) {if(!vis[i]) continue;while(p < i && val[p + 1] + val[i] <= lim) {++p;BST.insert(p,1);}if(p >= i) BST.insert(i,-1);int t = BST.find_max();if(t) {vis[t] = 0;vis[i] = 0;BST.insert(t,-1);--res;}}if(fa) {dp[u] = 1;++res;for(int i = 1 ; i <= tot ; ++i) {if(vis[i]) {if(val[i] + 1 > lim) break;dp[u] += val[i];--res;break;}}} } bool check(int m) {lim = m;res = 0;calc(1,0);return res <= A; } void Solve() {dfs(1,0);out(dp[1]);space;A = dp[1];int L = 1,R = N - 1;while(L < R) {int mid = (L + R) >> 1;if(check(mid)) R = mid;else L = mid + 1;}out(R);enter; } int main() { #ifdef ivorysifreopen("f1.in","r",stdin); #endifInit();Solve(); }

轉(zhuǎn)載于:https://www.cnblogs.com/ivorysi/p/10170335.html

總結(jié)

以上是生活随笔為你收集整理的【AtCoder】ARC088的全部內(nèi)容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網(wǎng)站內(nèi)容還不錯，歡迎將生活随笔推薦給好友。