【AtCoder】ARC078

C - Splitting Pile

枚舉從哪裏開始分的便可

#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 ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}

D - Fennec VS. Snuke

看樹上這段鏈從Fennec開始數第K / 2和K / 2+1的邊斷開以後，分紅的兩個子樹哪一個結點多
Fennec只有當節點數大於Snuke纔會勝利

#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 ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}

E - Awkward Response

若是不爲1後面接的只有0的形式，那麼問出第一個\(10^k\)爲N則證實數字有k位，而後能夠經過二分，判斷中間值是否小於當前值能夠把mid擴大10倍，這樣能夠知道中間值的字典序是否大於仍是小於n，由於長度相等字典序順序就是大小順序，因此可行
若是是1後面接的只有0，那麼問出第一個合法的k個9，這個數就是\(10^{k - 1}\)

#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 ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}

F - Mole and Abandoned Mine

爲啥算完2s跑出來不到0.1s？？？
atc擴充個人想象力系列？？？
這個就是把這惟一一條路徑挑出來，確定是但願這條路徑和路徑上每一個點上掛的一個聯通塊價值最大，而後用總路徑價值減掉
而後dp[i][S]表示當前走到第i個點，已經擴充的點集是S，就是每次路徑往下走一個點，或者擴充一個包括i其他的點不在S中的點集便可，複雜度\(O(N\times 3^{N})\)
而後我過於智障寫錯了好幾遍，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 ivorysi
    freopen("f1.in","r",stdin);
#endif
    Init();
    Solve();
}