C - Splitting Pile
枚举从哪里开始分的即可
#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define MAXN 200005
#define eps 1e-12
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {res = 0;T f = 1;char c = getchar();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;
int64 s[MAXN];
void Solve() {read(N);for(int i = 1 ; i <= N ; ++i) read(s[i]);for(int i = 1 ; i <= N ; ++i) s[i] += s[i - 1];int64 ans = abs(s[N] - 2 * s[1]);for(int i = 1 ; i < N ; ++i) {ans = min(ans,abs(s[N] - 2 * s[i]));}out(ans);enter;
}
int main() {
#ifdef ivorysifreopen("f1.in","r",stdin);
#endifSolve();
}D - Fennec VS. Snuke
看树上这段链从Fennec开始数第K / 2和K / 2+1的边断开之后，分成的两个子树哪个结点多
Fennec只有当节点数大于Snuke才会胜利
#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define MAXN 100005
#define eps 1e-12
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {res = 0;T f = 1;char c = getchar();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);
}
struct node {int to,next;
}E[MAXN * 2];
int head[MAXN],sumE,N;
int dep[MAXN],fa[MAXN],siz[MAXN];
void add(int u,int v) {E[++sumE].to = v;E[sumE].next = head[u];head[u] = sumE;
}
void dfs(int u) {dep[u] = dep[fa[u]] + 1;siz[u] = 1;for(int i = head[u] ; i ; i = E[i].next) {int v = E[i].to;if(v != fa[u]) {fa[v] = u;dfs(v);siz[u] += siz[v];}}
}
void Solve() {read(N);int u,v;for(int i = 1 ; i < N ; ++i) {read(u);read(v);add(u,v);add(v,u);}dfs(1);int t = dep[N] / 2 - 1;u = N;while(t--) {u = fa[u];}if(N - siz[u] > siz[u]) puts("Fennec");else puts("Snuke");
}
int main() {
#ifdef ivorysifreopen("f1.in","r",stdin);
#endifSolve();
}E - Awkward Response
如果不为1后面接的只有0的形式，那么问出第一个\(10^k\)为N则证明数字有k位，然后可以通过二分，判断中间值是否小于当前值可以把mid扩大10倍，这样可以知道中间值的字典序是否大于还是小于n，因为长度相等字典序顺序就是大小顺序，所以可行
如果是1后面接的只有0，那么问出第一个合法的k个9，这个数就是\(10^{k - 1}\)
#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define MAXN 100005
#define eps 1e-12
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {res = 0;T f = 1;char c = getchar();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);
}
bool Query(int64 t) {putchar('?');space;out(t);enter;fflush(stdout);char s[5];scanf("%s",s);return s[0] == 'Y';
}
void Solve() {int64 v = 1;bool f = 0;for(int i = 1 ; i <= 10 ; ++i) {if(!Query(v)) {f = 1;v /= 10;break;}v *= 10;}if(!f) {v = 1;for(int i = 1 ; i <= 10 ; ++i) {if(Query(v * 10 - 1)) {putchar('!');space;out(v);enter;return;}v *= 10;}}int64 L = v,R = min(v * 10 - 1,(int64)1e9);while(L < R) {int64 mid = (L + R + 1) >> 1;if(!Query(mid * 10)) L = mid;else R = mid - 1;}putchar('!');space;out(L + 1);enter;
}
int main() {
#ifdef ivorysifreopen("f1.in","r",stdin);
#endifSolve();
}F - Mole and Abandoned Mine
为啥算完2s跑出来不到0.1s？？？
atc扩充我的想象力系列？？？
这个就是把这唯一一条路径挑出来，肯定是希望这条路径和路径上每个点上挂的一个联通块价值最大，然后用总路径价值减掉
然后dp[i][S]表示当前走到第i个点，已经扩充的点集是S，就是每次路径往下走一个点，或者扩充一个包括i其余的点不在S中的点集即可，复杂度\(O(N\times 3^{N})\)
然后我过于智障写错了好几遍，cao
#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define MAXN 10005
#define eps 1e-12
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {res = 0;T f = 1;char c = getchar();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,M;
struct node {int to,next,val;
}E[1005];
int sumE,head[25],pos[(1 << 15) + 5];
int f[16][(1 << 15) + 5],h[16][(1 << 15) + 5],sum[(1 << 15) + 5];
void add(int u,int v,int c) {E[++sumE].to = v;E[sumE].next = head[u];E[sumE].val = c;head[u] = sumE;
}
inline int lowbit(int x) {return x & (-x);
}
void Init() {read(N);read(M);int u,v,c;for(int i = 1 ; i <= M ; ++i) {read(u);read(v);read(c);add(u,v,c);add(v,u,c);h[u][1 << v - 1] = c;h[v][1 << u - 1] = c;}for(int i = 1 ; i <= N ; ++i) {for(int j = 1 ; j < (1 << N) ; ++j) {if(lowbit(j) == j) continue;h[i][j] = h[i][j - lowbit(j)] + h[i][lowbit(j)];}}for(int i = 1 ; i <= N ; ++i) pos[(1 << i - 1)] = i;for(int i = 1 ; i < (1 << N) ; ++i) {sum[i] = sum[i - lowbit(i)] + h[pos[lowbit(i)]][i - lowbit(i)];}
}
void Solve() {for(int i = 1 ; i <= N ; ++i) {for(int j = 0 ; j < (1 << N) ; ++j) {f[i][j] = -1e9;}}f[1][1] = 0;for(int j = 0 ; j < (1 << N - 1) ; ++j) {for(int i = 1 ; i <= N ; ++i) {int S = j << 1 | 1;if(!(S & (1 << i - 1))) continue;int L = S ^ (1 << i - 1);for(int T = L; T ; T = (T - 1) & L) {if(f[i][S ^ T] >= 0)f[i][S] = max(f[i][S],f[i][S ^ T] + sum[T ^ (1 << i - 1)]);}if(f[i][S] >= 0) {for(int k = head[i] ; k ; k = E[k].next) {int v = E[k].to;if(!(S & (1 << v - 1))) {f[v][S ^ (1 << v - 1)] = max(f[v][S ^ (1 << v - 1)],f[i][S] + E[k].val);}}}}}out(sum[(1 << N) - 1] - f[N][(1 << N) - 1]);enter;
}
int main() {
#ifdef ivorysifreopen("f1.in","r",stdin);
#endifInit();Solve();
}
)
 Backpropagation and Neural Network)
