Codeforces Round 665 (div2)

2020.8.22

装修完了我的博客，喜欢这个造型，挂上友链就更好了

昨天cf就是一个彻头彻尾的悲剧，本来能上蓝，结果因为在A题耽误时间过多导致掉了30分，不过没关系，这算是一个小波动吧，影响不了什么，现在稳定到每场3题打底了，下场打回来就行了。、

A题. Distance and Axis

这道题我刚跟朋友聊完天，乍一看有些懵，想了半个小时最后才出来。谢谢木下巨佬给的tips，就是首先观察到当OB 变动1，那么差距变动2，所以我们一定是需要一个偶数的差距才能不移动，如果n比m大，那么挪过去就行了，其实很简单，但是当时怎么也想象不出这个动作

#include <bits/stdc++.h>
using namespace std;
#define limit (10000000 + 5)//防止溢出
#define INF 0x3f3f3f3f
#define inf 0x3f3f3f3f3f
#define lowbit(i) i&(-i)//一步两步
#define EPS 1e-6
#define FASTIO  ios::sync_with_stdio(false);cin.tie(0);
#define ff(a) printf("%d\n",a );
#define pi(a,b) pair<a,b>
#define rep(i, a, b) for(ll i = a; i <= b ; ++i)
#define per(i, a, b) for(ll i = b ; i >= a ; --i)
#define MOD 998244353
#define traverse(u) for(int i = head[u]; ~i ; i = edge[i].next)
#define FOPEN freopen("C:\\Users\\tiany\\CLionProjects\\acm_01\\data.txt", "rt", stdin)
#define FOUT freopen("C:\\Users\\tiany\\CLionProjects\\acm_01\\dabiao.txt", "wt", stdout)
#define debug(x) cout<<x<<endl
typedef long long ll;
typedef unsigned long long ull;
inline ll read(){
    ll sign = 1, x = 0;char s = getchar();
    while(s > '9' || s < '0' ){if(s == '-')sign = -1;s = getchar();}
    while(s >= '0' && s <= '9'){x = (x << 3) + (x << 1) + s - '0';s = getchar();}
    return x * sign;
}//快读
void write(ll x){
    if(x < 0) putchar('-'),x = -x;
    if(x / 10) write(x / 10);
    putchar(x % 10 + '0');
}
int n,kase,m;
int a[limit];
int main() {
#ifdef LOCAL
    FOPEN;
#endif
    cin>>kase;
    while (kase--){
        cin>>n>>m;
        if(n < m){
            cout<<abs(n - m)<<endl;
        }else if((n - m) & 1){
            cout<<1<<endl;
        }else{
            cout<<0<<endl;
        }

    }
    return 0;
}

AC Code

B题. Ternary Sequence

这道题就是问怎么安排序列使得序列和最大，给出来了规则，那么就看啊，首先肯定是2越多越好，而且只有2，因为无论是1还是0乘积都只可能有0，好的。然后用剩下的1互相抵消，因为避免对方的2和己方1结合扣分，然后用1减0，最后安排不了的才和2结合从答案中减去，就不用管了，反正其余对答案的贡献铁定为0，这个逻辑好绕啊，我昨天不知道是怎么硬着头皮写出来的，代码太丑陋了，但还是过了。听说有人写完79行if else直接关电脑的，

#include <bits/stdc++.h>
using namespace std;
#define limit (10000000 + 5)//防止溢出
#define INF 0x3f3f3f3f
#define inf 0x3f3f3f3f3f
#define lowbit(i) i&(-i)//一步两步
#define EPS 1e-6
#define FASTIO  ios::sync_with_stdio(false);cin.tie(0);
#define ff(a) printf("%d\n",a );
#define pi(a,b) pair<a,b>
#define rep(i, a, b) for(ll i = a; i <= b ; ++i)
#define per(i, a, b) for(ll i = b ; i >= a ; --i)
#define MOD 998244353
#define traverse(u) for(int i = head[u]; ~i ; i = edge[i].next)
#define FOPEN freopen("C:\\Users\\tiany\\CLionProjects\\acm_01\\data.txt", "rt", stdin)
#define FOUT freopen("C:\\Users\\tiany\\CLionProjects\\acm_01\\dabiao.txt", "wt", stdout)
#define debug(x) cout<<x<<endl
typedef long long ll;
typedef unsigned long long ull;
inline ll read(){
    ll sign = 1, x = 0;char s = getchar();
    while(s > '9' || s < '0' ){if(s == '-')sign = -1;s = getchar();}
    while(s >= '0' && s <= '9'){x = (x << 3) + (x << 1) + s - '0';s = getchar();}
    return x * sign;
}//快读
void write(ll x){
    if(x < 0) putchar('-'),x = -x;
    if(x / 10) write(x / 10);
    putchar(x % 10 + '0');
}
int n,kase,m;
int a,b,c;
int main() {
#ifdef LOCAL
    FOPEN;
#endif
    cin>>kase;
    while (kase--){
        int x,y,z;
        cin>>x>>y>>z;
        cin>>a>>b>>c;
        ll ans = 0;
        int minn2 = min(z,b);
        ans += minn2 * 2;
        b -= minn2, z -= minn2;
        int minn = min(y,a);
        a-=minn,y-=minn;
        minn2 = min(z,c);
        z -= minn2, c -= minn2;
        minn = min(y,b);
        y -= minn, b -= minn;
        minn = min(a,x);
        a -= minn, x -= minn;
        ans -= min(y,c) * 2;

        cout<<ans<<endl;

    }
    return 0;
}

AC Code

C题 Mere Array

这题问的是如果序列里任意两个元素的gcd等于最小值就能交换，问能不能变成非严格上升序列。首先说升序，那么就想到sort一遍拷贝之后的数组，然后比较，那么一个元素能换到其本身应该所处位置的充要条件是这个数字能够整除最小值，所以做法如下，往常的话我今天已经铁蓝了，但A耽误时间太长了，还wa了两次，到C的排名才3k9，今天必须做到d才能不掉分

#include <bits/stdc++.h>
using namespace std;
#define limit (10000000 + 5)//防止溢出
#define INF 0x3f3f3f3f
#define inf 0x3f3f3f3f3f
#define lowbit(i) i&(-i)//一步两步
#define EPS 1e-6
#define FASTIO  ios::sync_with_stdio(false);cin.tie(0);
#define ff(a) printf("%d\n",a );
#define pi(a,b) pair<a,b>
#define rep(i, a, b) for(ll i = a; i <= b ; ++i)
#define per(i, a, b) for(ll i = b ; i >= a ; --i)
#define MOD 998244353
#define traverse(u) for(int i = head[u]; ~i ; i = edge[i].next)
#define FOPEN freopen("C:\\Users\\tiany\\CLionProjects\\acm_01\\data.txt", "rt", stdin)
#define FOUT freopen("C:\\Users\\tiany\\CLionProjects\\acm_01\\dabiao.txt", "wt", stdout)
#define debug(x) cout<<x<<endl
typedef long long ll;
typedef unsigned long long ull;
inline ll read(){
    ll sign = 1, x = 0;char s = getchar();
    while(s > '9' || s < '0' ){if(s == '-')sign = -1;s = getchar();}
    while(s >= '0' && s <= '9'){x = (x << 3) + (x << 1) + s - '0';s = getchar();}
    return x * sign;
}//快读
void write(ll x){
    if(x < 0) putchar('-'),x = -x;
    if(x / 10) write(x / 10);
    putchar(x % 10 + '0');
}
int n,kase,m;
int a[limit];
int f[limit];
int main() {
#ifdef LOCAL
    FOPEN;
#endif
    cin>>kase;
    while (kase--){
        cin>>n;
        int minn = INF;
        int flag = 1;
        rep(i,1,n){
            cin>>a[i];

            minn = min(minn, a[i]);
            f[i] = a[i];
        }
        sort(f + 1, f + 1 + n);
        flag = 0;
        rep(i ,1,n){
            if(a[i] != f[i] && a[i] % minn != 0){
                flag = 1;
                cout<<"NO"<<endl;
                break;
            }
        }
        if(!flag){
            cout<<"YES"<<endl;
        }
    }
    return 0;
}

AC Code

D题. Maximum Distributed Tree

D题是给出一棵树，然后给出若干个乘积为k的质因子，问怎么才能让∑i=1|n-1∑j=i+1|n f(i,j)最大，其中f(u,v)为从u到v的链上的权值，这道题刚开始我觉得可能是树链剖分，直接优先往重链上放prime factor，但是想了下这样做似乎不妥，1e5您打算怎么统计呢？之后队友说是不是可以统计边的使用次数，一语点醒梦中人，使用次数最多的优先安排大prime factor不就好了？那么这个使用次数，通过手动模拟法发现，对于任意一点u，u以上的节点想要走到任意u为根的子树节点，那么必须要经过这条边，次数也就好固定了，是子节点size * （n - 子节点size）。我给每条边打上一个号，从1-n-1，然后开个数组记录，用堆优先选取次数达的，然后考虑n -1小于m的情况，将后面的贡献全给第一个就行了，难得d的思路能这么清晰，然后大草就发生了，数组开小了tle，最后没交上，赛后一发过了？？？？这事儿怎么老发生在我和weeping demon身上？？？？？然后掉了36，下次再打回来就是了

#include <bits/stdc++.h>
using namespace std;
#define limit (500000 + 5)//防止溢出
#define INF 0x3f3f3f3f
#define inf 0x3f3f3f3f3f
#define lowbit(i) i&(-i)//一步两步
#define EPS 1e-6
#define FASTIO  ios::sync_with_stdio(false);cin.tie(0);
#define ff(a) printf("%d\n",a );
#define pi(a,b) pair<a,b>
#define rep(i, a, b) for(ll i = a; i <= b ; ++i)
#define per(i, a, b) for(ll i = b ; i >= a ; --i)
#define MOD 998244353
#define traverse(u) for(int i = head[u]; ~i ; i = edge[i].next)
#define FOPEN freopen("C:\\Users\\tiany\\CLionProjects\\acm_01\\data.txt", "rt", stdin)
#define FOUT freopen("C:\\Users\\tiany\\CLionProjects\\acm_01\\dabiao.txt", "wt", stdout)
#define debug(x) cout<<x<<endl
typedef long long ll;
typedef unsigned long long ull;
inline ll read(){
    ll sign = 1, x = 0;char s = getchar();
    while(s > '9' || s < '0' ){if(s == '-')sign = -1;s = getchar();}
    while(s >= '0' && s <= '9'){x = (x << 3) + (x << 1) + s - '0';s = getchar();}
    return x * sign;
}//快读
void write(ll x){
    if(x < 0) putchar('-'),x = -x;
    if(x / 10) write(x / 10);
    putchar(x % 10 + '0');
}
int  sizes[limit], head[limit<<1] ,cnt;

struct node{
    int to, next, num;
}edge[limit<<1];
void init(){
    memset(head, -1, sizeof(head));
    cnt = 0;
}
void add(int u ,int v,int num){
    edge[cnt].to = v;
    edge[cnt].next = head[u];
    edge[cnt].num = num;
    head[u] = cnt++;
}
ll dp[limit];
int n;
void dfs(int u, int pre){
    sizes[u] = 1;
    for(int i = head[u]; ~i; i = edge[i].next){
        int v = edge[i].to;
        if(v != pre) {
            dfs(v, u);
            sizes[u] += sizes[v];
            dp[edge[i].num] = (sizes[v]) * (n - sizes[v]);
        }
    }
}
int kase,m;
ll fact[limit];
int main() {
#ifdef LOCAL
    FOPEN;
#endif
    FASTIO
    cin>>kase;
    const ll mod = 1e9 + 7;
    while (kase--){
        cin>>n;
        init();
        rep(i,1,n-1){
            int x,y;
            cin>>x>>y;
            add(x,y,i);
            add(y,x,i);
        }
        cin>>m;
        rep(i,1,m){
            cin>>fact[i];
        }
        sort(fact + 1, fact + 1 + m, greater<>());
        dfs(1,1);
        priority_queue<ll>q;
        rep(i,1,n-1){
            q.push(dp[i]);
        }
        if(n -1 < m){//如果小于
            rep(i,n,m){
                ((fact[1] %= mod) *= (fact[i] %= mod)) %= mod;
            }
            m = n - 1;
        }
        ll ans = 0;
        rep(i ,1,m){
            ans += (q.top() * fact[i]) % mod;
            ans %= mod;
            q.pop();
        }
        while (q.size()){
            ans += q.top();
            ans %= mod;
            q.pop();
        }
        cout<<(ans + mod) % mod<<endl;
    }
    return 0;
}

AC Code