AtCoder Regular Contest 078

C

    #include <cstdio>    #include <cstring>    #include <map>    #include <algorithm>    #include <iostream>    #include <vector>         using namespace std;    #define LL long long    #define pb push_back    #define mk make_pair    #define pill pair<int, int>    #define mst(a, b)memset(a, b, sizeof a)    #define REP(i, x, n)for(int i = x; i <= n; ++i)    const int MOD = 1e9 + 7;    const int qq = 2e5 + 10;    LL sum[qq];         int main(){    int n;scanf("%d", &n);    for(int i = 1; i <= n; ++i){    scanf("%lld", sum + i);    sum[i] += sum[i - 1];    }    LL minx = 1e18;    for(int i = 1; i < n; ++i){    minx = min(minx, abs(sum[i] - (sum[n] - sum[i])));    }    printf("%lld\n", minx);    return 0;    }

D

题意：n个节点n-1条边组成的一棵树，首先1是黑色，n是白色，黑色可以染黑色的相邻节点，白色同理，问最后谁染的节点多，黑色先手。

思路：1到n节点实际上只有一条路径，那么抢占这一条路径上的节点成了决策关键性，我的方法是以1节点为根节点建树，然后模拟1和n染色过程，n染色的最后一个节点，以这个节点为根节点的所有节点即为白色，否则是黑色

    #include <cstdio>    #include <cstring>    #include <map>    #include <algorithm>    #include <iostream>    #include <vector>         using namespace std;    #define LL long long    #define pb push_back    #define mk make_pair    #define pill pair<int, int>    #define mst(a, b)memset(a, b, sizeof a)    #define REP(i, x, n)for(int i = x; i <= n; ++i)    const int MOD = 1e9 + 7;    const int qq = 2e5 + 10;    vector<int> vt[qq];    int n, a, b;    int dp[qq], num[qq];    int pre[qq];    bool vis[qq];    void Dfs(int v, int fa, int dis){    dp[v] = dis;    pre[v] = fa;    for(int i = 0; i < (int)vt[v].size(); ++i){    int u = vt[v][i];    if(u == fa)continue;    Dfs(u, v, dis + 1);    num[v] += num[u];    }    num[v] = num[v] + 1;    }    int node[qq];         int main(){    scanf("%d", &n);    REP(i, 1, n - 1){    scanf("%d%d", &a, &b);    vt[a].pb(b), vt[b].pb(a);    }    Dfs(1, -1, 0);    int cnt = 0;    int x = n;    while(pre[x] != -1){    node[++cnt] = x;    x = pre[x];    }    node[++cnt] = 1;    int l = 1, r = cnt;    while(l < r){    if(l + 1 >= r - 1)break;    l++, r--;    }    int F, S;    F = S = 0;    S = num[node[l]];    F = n - S;    if(F > S)puts("Fennec");    elseputs("Snuke");    return 0;    }