cf div2 round 688 题解

爆零了，自闭了

小张做项目入职字节

小李ak wf入职ms

我比赛爆零月薪3k

我们都有光明的前途

好吧，这场感觉有一点难了，昨天差点卡死在B上，要不受O爷出手相救我就boom zero了

第一题，看上去很唬人，我觉得还得记录变量什么的，1分钟后发现只要查出两个集合的交集大小就行了，连变量增量都不用，拿一血，此时rk 1k6，我人没了

#include <bits/stdc++.h>
using namespace std;
#define limit (315000 + 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 kase;
int n,m,k;
int a[limit];
void solve(){
    cin>>n>>m;
    set<int>s;
    rep(i,1,n){
        int x;
        cin>>x;
        s.insert(x);
    }
    int tot = 0;
    rep(i,1,m){
        int x;
        cin>>x;
        if(s.find(x) != s.end())++tot;
    }
    cout<<tot<<endl;
}
int main() {
#ifdef LOCAL
    FOPEN;
#endif
    cin>>kase;
    while (kase--){
        solve();
    }
    return 0;
}

B题看上去么得啥思路，后来O神说是记录两边的插值，然后每次记录当前这全都设置成某一位的最小值就行，dbq我现在也没怎么搞懂这个怎么搞，于是滚去看C

#include <bits/stdc++.h>
using namespace std;
#define limit (3150000 + 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 kase;
int n,m,k;
ll a[limit];
ll dp2[limit],dp[limit];
void solve(){
    cin>>n;
    ll tot = 0,ans = 0x3f3f3f3f3f3f3f;
    rep(i,1,n){
        cin>>a[i];
        dp[i] = 0;
        dp2[i] = 0;
    }
    dp2[1] = max(dp2[1],llabs(a[2] - a[1]));
    rep(i,2,n){
        ll tmp = llabs(a[i] - a[i - 1]);
        tot += tmp;
        dp[i] = tmp;
    }
    rep(i,1,n){
        if(i == 1){
            ans = min(ans , tot - dp[2]);
        }else if(i == n){
            ans = min(ans,tot - dp[n]);
        }
        else if((a[i] <= a[i + 1] && a[i] <= a[i - 1])){
            ans = min(ans,tot + abs(a[i - 1] - a[i + 1]) - dp[i + 1 ] - dp[i]);
        }else if(a[i] >= a[i + 1] && a[i] >= a[i - 1]){
            ans = min(ans,tot + abs(a[i + 1] - a[i - 1]) - dp[i + 1] - dp[i]);
        }
    }
    cout<<ans<<endl;
}
int main() {
#ifdef LOCAL
    FOPEN;
#endif
    cin>>kase;
    while (kase--){
        solve();
    }
    return 0;
}

C题一看就很阅读理解，先是花半个小时看题，之后大概清楚了什么意思，大概是一个0-9构成的棋盘，每次找最大的又i（i = 0 - 9）三角形构成的平行四边形面积，其中一边必须平行于坐标轴，每次可以在棋盘上改变不超过一个不影响后续的元素，求以0-9各个为顶点的最大面积

首先显然想到，三角形面积 = 底 X 高 / 2，不用/2了，好耶。大概思路就是预处理一下左右，上下两个维度的第一个元素每一行或者每一列最靠边缘的元素位置，然后枚举看每一行，对于0-9，存在以下几种对答案产生贡献的情况

1. 这行有至少两个数字i，以最边上两个元素为底寻找一个与其垂直的维度的点，或者1，n这两个端点，构成的三角形比原先大

2. 这行存在一个数字，但用次行/列第一个或者最后一个元素改变为顶点数字作为底，构成的三角形比原先大

3. 这行的情况与1相同，不添加任何数字，不做改动，构成的三角形就比原先大

然后我赛场写了100+行大模拟，好在过了，写完一身冷汗，代码如下：

#include <bits/stdc++.h>
using namespace std;
#define limit (3150000 + 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 kase;
int n,m,k;
ll a[limit];
ll dp2[limit],dp[limit];
void solve(){
    cin>>n;
    ll tot = 0,ans = 0x3f3f3f3f3f3f3f;
    rep(i,1,n){
        cin>>a[i];
        dp[i] = 0;
        dp2[i] = 0;
    }
    dp2[1] = max(dp2[1],llabs(a[2] - a[1]));
    rep(i,2,n){
        ll tmp = llabs(a[i] - a[i - 1]);
        tot += tmp;
        dp[i] = tmp;
    }
    rep(i,1,n){
        if(i == 1){
            ans = min(ans , tot - dp[2]);
        }else if(i == n){
            ans = min(ans,tot - dp[n]);
        }
        else if((a[i] <= a[i + 1] && a[i] <= a[i - 1])){
            ans = min(ans,tot + abs(a[i - 1] - a[i + 1]) - dp[i + 1 ] - dp[i]);
        }else if(a[i] >= a[i + 1] && a[i] >= a[i - 1]){
            ans = min(ans,tot + abs(a[i + 1] - a[i - 1]) - dp[i + 1] - dp[i]);
        }
    }
    cout<<ans<<endl;
}
int main() {
#ifdef LOCAL
    FOPEN;
#endif
    cin>>kase;
    while (kase--){
        solve();
    }
    return 0;
}

D题没时间搞了，但后来和Dxtst老哥讨论了下似乎是当前位上0或1，如果是1对于答案的贡献为2^k, 其他结论今晚再搞

奥里给！