cf div2 round 688 题解
- 2020 年 12 月 5 日
- 笔记
- ACM/ICPC, Codeforces
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!