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简单分析一下，对于x<y，求a[x]>=y 同时a[y]>=x 再简化一下，求1-a[y]区间内大于>=y的个数。。。主席树牛逼 #include<iostream> #include<stdio.h> #include<string.h> #include<algorithm> #include<vector> #define LL long long using namespace std; const int maxx = 2e5+6; struct node{ int l,r; int cnt; }tree[maxx*40]; int a[maxx]; int b[maxx]; int root[maxx]; int cnt; vector<int>v; int getval(int x){ return lower_bound(v.begin(),v.end(),x)-v.begin()+1; } void inserts(int l,int r,int pre,int &now,int pos){ now=++cnt; tree[now]=tree[pre]; tree[now].cnt++; if(l==r){ return ; } int mid=(l+r)>>1; if (pos<=mid){ inserts(l,mid,tree[pre].l,tree[now]

Solution A. There Are Two Types Of Burgers 题意： 做一个 \(A\) 需要两个 \(b\) 和两个 \(p\) ，能卖 \(h\) 元；做一个 \(B\) 需要两个 \(b\) 和两个 \(f\) ，能卖 \(c\) 元。给出 \(b,p,f\) 的数量，求最多卖多少元。 思路： 比赛的时候用循环搞了，判断 \(h,c\) 的大小关系，决定先做什么。也可以 \(b=b/2\) ，然后判断大小，每次取最小值，利润直接算，然后更新 \(b\) ，再进行一次。 //#define DEBUG #include<bits/stdc++.h> using namespace std; #define lson (rt<<1) #define rson (rt<<1|1) const int N=100010; const int inf=0X3f3f3f3f; const long long INF = 0x3f3f3f3f3f3f3f3f; const double eps = 1e-6; const double pi = acos(-1.0); const int mod = 1000000007; typedef long long ll; int main() { #ifdef DEBUG freopen("in.txt","r"

A. Best Subsegment 题意：给你一组数字，让你找到最长且平均数最大且没有小数的长度 解：平均数最大，直接找最大值，然后找最大值的长度（注意处理最后一个） #include <bits/stdc++.h> using namespace std; const int maxn=1e5+10; typedef long long ll; int a[maxn]; int main(){ int n; cin>>n; int maxx=-1; for(int i=1;i<=n;i++){ cin>>a[i]; maxx=max(maxx,a[i]); } int ans=1; int cnt=0; for(int i=1;i<=n;i++){ if(a[i]==maxx&&a[i+1]==maxx&&i!=n){ cnt++; } else{ if(a[i]==a[i-1])cnt++; ans=max(cnt,ans); cnt=0; } //cout<<cnt<<endl; } cout<<ans<<endl; return 0; } View Code B. Emotes 题意：n个数，让你找出m个数，使得和最大，每个数可以重复选择，最多选择k次； 解：排序，选k个最大+1个次大搭配； #include <bits/stdc++.h> using namespace

A. There Are Two Types Of Burgers 思路：签到题，价格高的先做就行。 AC代码： 1 #include<bits/stdc++.h> 2 using namespace std; 3 int main() 4 { 5 std::ios::sync_with_stdio(false); 6 int t; 7 cin >> t; 8 int b,p,f; 9 int h, c; 10 while(t--) 11 { 12 cin >> b >> p >> f; 13 cin >> h >> c; 14 b /= 2; 15 int ans = 0; 16 if(h > c) 17 { 18 while(p && b) 19 { 20 ans += h; 21 p--; 22 b--; 23 } 24 while(f && b) 25 { 26 ans += c; 27 f--; 28 b--; 29 } 30 } 31 else{ 32 while(f && b) 33 { 34 ans += c; 35 f--; 36 b--; 37 } 38 while(p && b) 39 { 40 ans += h; 41 p--; 42 b--; 43 } 44 } 45 cout << ans << endl; 46 } 47 return 0; 48 }

A. Digits Sequence Dividing 题意：给你一个数字串，只包含1-9，让你至少分成两段，使得每一段的数字大于前一段的数字； 解：对n特判，如果n为2，那么比较一下两个数字大小，如果n>2，那么就可以直接分成两部分，第一部分1个数字，剩下都为第二部分； #include <bits/stdc++.h> using namespace std; typedef long long ll; const int maxn=1e3+10; const int mod=998244353; ll ans[maxn]; int main(){ int T; cin>>T; while(T--){ int n; string s; cin>>n>>s; if(n==2){ if(s[0]<s[1]){ cout<<"YES"<<endl; cout<<"2"<<endl; cout<<s[0]<<' '<<s[1]<<endl; } else cout<<"NO"<<endl; } else { cout<<"YES"<<endl; cout<<"2"<<endl; cout<<s[0]<<' '; for(int i=1;i<n;i++)cout<<s[i]; cout<<endl; } } return 0; } View Code B. Digital root 题意

原文链接： https://www.cnblogs.com/xwl3109377858/p/11405773.html Educational Codeforces Round 71 (Rated for Div. 2) D - Number Of Permutations You are given a sequence of n pairs of integers: (a1,b1),(a2,b2),…,(an,bn). This sequence is called bad if it is sorted in non-descending order by first elements or if it is sorted in non-descending order by second elements. Otherwise the sequence is good. There are examples of good and bad sequences: s=[(1,2),(3,2),(3,1)] is bad because the sequence of first elements is sorted: [1,3,3]; s=[(1,2),(3,2),(1,2)] is bad because the sequence of second elements is

原文链接： https://www.cnblogs.com/xwl3109377858/p/11404321.html Educational Codeforces Round 71 (Rated for Div. 2) C - Gas Pipeline You are responsible for installing a gas pipeline along a road. Let's consider the road (for simplicity) as a segment [0,n] on OX axis. The road can have several crossroads, but for simplicity, we'll denote each crossroad as an interval (x,x+1) with integer x. So we can represent the road as a binary string consisting of n characters, where character 0 means that current interval doesn't contain a crossroad, and 1 means that there is a crossroad. Usually, we can

原文链接： https://www.cnblogs.com/xwl3109377858/p/11404261.html Educational Codeforces Round 71 (Rated for Div. 2) B - Square Filling You are given two matrices A and B. Each matrix contains exactly n rows and m columns. Each element of A is either 0 or 1; each element of B is initially 0. You may perform some operations with matrix B. During each operation, you choose any submatrix of B having size 2×2, and replace every element in the chosen submatrix with 1. In other words, you choose two integers x and y such that 1≤x<n and 1≤y<m, and then set Bx,y, Bx,y+1, Bx+1,y and Bx+1,y+1 to 1. Your goal

本文链接： https://www.cnblogs.com/xwl3109377858/p/11404050.html Educational Codeforces Round 71 (Rated for Div. 2) A - There Are Two Types Of Burgers There are two types of burgers in your restaurant — hamburgers and chicken burgers! To assemble a hamburger you need two buns and a beef patty. To assemble a chicken burger you need two buns and a chicken cutlet. You have b buns, p beef patties and f chicken cutlets in your restaurant. You can sell one hamburger for h dollars and one chicken burger for c dollars. Calculate the maximum profit you can achieve. You have to answer t independent queries.

A There Are Two Types Of Burgers 传送门 简单的分类讨论而已 #include <iostream> #include <cstdio> #include <algorithm> #include <cstring> #define ll long long using namespace std; const int maxn=1e6+5; int main(){ int t; cin>>t; while(t--){ int b,p,f; cin>>b>>p>>f; int h,c; cin>>h>>c; int con=b/2; if(con<=min(p,f)){//面包少的话 if(h>=c){//牛肉贵，买牛肉 printf("%d\n",con*h); }else{ printf("%d\n",con*c); } }else{//面包够的话 if(h>=c){//牛肉贵，先买牛肉 if(con>=p){//大于牛肉的数,牛肉全卖加鸡肉 if(con>=(p+f)){ printf("%d\n",h*p+f*c); }else{ printf("%d\n",h*p+(con-p)*c); //printf("aa\n"); } }else{//全卖牛肉 printf("%d\n",con*h); //printf("aa\n"); }