Codeforces Beta Round #72 (Div. 2 Only)

A、总感觉在哪里见过，n个人和n个人对打，一回合每个人至多干掉一个人，两方轮流行动

问你三回合之后最多死多少个人，保证n是偶数

猜个结论，第一回合干掉另一边的1/2，第二、第三回合同理

代码：

#include <bits/stdc++.h>
#define int long long
#define sc(a) scanf("%lld",&a)
#define scc(a,b) scanf("%lld %lld",&a,&b)
#define sccc(a,b,c) scanf("%lld %lld %lld",&a,&b,&c)
#define schar(a) scanf("%c",&a)
#define pr(a) printf("%lld",a)
#define fo(i,a,b) for(int i=a;i<b;++i)
#define re(i,a,b) for(int i=a;i<=b;++i)
#define rfo(i,a,b) for(int i=a;i>b;--i)
#define rre(i,a,b) for(int i=a;i>=b;--i)
#define prn() printf("\n")
#define prs() printf(" ")
#define mkp make_pair
#define pii pair<int,int>
#define pub(a) push_back(a)
#define pob() pop_back()
#define puf(a) push_front(a)
#define pof() pop_front()
#define fst first
#define snd second
#define frt front()
#define bak back()
#define mem0(a) memset(a,0,sizeof(a))
#define memmx(a) memset(a,0x3f3f,sizeof(a))
#define memmn(a) memset(a,-0x3f3f,sizeof(a))
#define debug
#define db double
#define yyes cout<<"YES"<<endl;
#define nno cout<<"NO"<<endl;
using namespace std;
typedef vector<int> vei;
typedef vector<pii> vep;
typedef map<int,int> mpii;
typedef map<char,int> mpci;
typedef map<string,int> mpsi;
typedef deque<int> deqi;
typedef deque<char> deqc;
typedef priority_queue<int> mxpq;
typedef priority_queue<int,vector<int>,greater<int> > mnpq;
typedef priority_queue<pii> mxpqii;
typedef priority_queue<pii,vector<pii>,greater<pii> > mnpqii;
const int maxn=500005;
const int inf=0x3f3f3f3f3f3f3f3f;
const int MOD=100000007;
const db eps=1e-10;
int qpow(int a,int b){int tmp=a%MOD,ans=1;while(b){if(b&1){ans*=tmp,ans%=MOD;}tmp*=tmp,tmp%=MOD,b>>=1;}return ans;}
int lowbit(int x){return x&-x;}
int max(int a,int b){return a>b?a:b;}
int min(int a,int b){return a<b?a:b;}
int mmax(int a,int b,int c){return max(a,max(b,c));}
int mmin(int a,int b,int c){return min(a,min(b,c));}
void mod(int &a){a+=MOD;a%=MOD;}
bool chk(int now){}
int half(int l,int r){while(l<=r){int m=(l+r)/2;if(chk(m))r=m-1;else l=m+1;}return l;}
int ll(int p){return p<<1;}
int rr(int p){return p<<1|1;}
int mm(int l,int r){return (l+r)/2;}
int lg(int x){if(x==0) return 1;return (int)log2(x)+1;}
bool smleql(db a,db b){if(a<b||fabs(a-b)<=eps)return true;return false;}
db len(db a,db b,db c,db d){return sqrt((a-c)*(a-c)+(b-d)*(b-d));}
bool isp(int x){if(x==1)return false;if(x==2)return true;for(int i=2;i*i<=x;++i)if(x%i==0)return false;return true;}
inline int read()
{char ch=getchar();int s=0,w=1;while(ch<48||ch>57){if(ch=='-')w=-1;ch=getchar();}while(ch>=48&&ch<=57){s=(s<<1)+(s<<3)+ch-48;ch=getchar();}return s*w;
}
inline void write(int x)
{if(x<0)putchar('-'),x=-x;if(x>9)write(x/10);putchar(x%10+48);
}int n,m;signed main(){ios_base::sync_with_stdio(0);cin.tie(0),cout.tie(0);cin>>n;cout<<n+n/2;return 0;
}

B、给你一个数列，所有的子区间[l,r]满足[l,r]区间里的最大值等于最小值的区间称为好区间

问一个数列里一共有多少个好区间

预处理一下连续区间的个数，对答案的贡献就是(区间长度+1)*区间长度/2

代码：

#include <bits/stdc++.h>
#define int long long
#define sc(a) scanf("%lld",&a)
#define scc(a,b) scanf("%lld %lld",&a,&b)
#define sccc(a,b,c) scanf("%lld %lld %lld",&a,&b,&c)
#define schar(a) scanf("%c",&a)
#define pr(a) printf("%lld",a)
#define fo(i,a,b) for(int i=a;i<b;++i)
#define re(i,a,b) for(int i=a;i<=b;++i)
#define rfo(i,a,b) for(int i=a;i>b;--i)
#define rre(i,a,b) for(int i=a;i>=b;--i)
#define prn() printf("\n")
#define prs() printf(" ")
#define mkp make_pair
#define pii pair<int,int>
#define pub(a) push_back(a)
#define pob() pop_back()
#define puf(a) push_front(a)
#define pof() pop_front()
#define fst first
#define snd second
#define frt front()
#define bak back()
#define mem0(a) memset(a,0,sizeof(a))
#define memmx(a) memset(a,0x3f3f,sizeof(a))
#define memmn(a) memset(a,-0x3f3f,sizeof(a))
#define debug
#define db double
#define yyes cout<<"YES"<<endl;
#define nno cout<<"NO"<<endl;
using namespace std;
typedef vector<int> vei;
typedef vector<pii> vep;
typedef map<int,int> mpii;
typedef map<char,int> mpci;
typedef map<string,int> mpsi;
typedef deque<int> deqi;
typedef deque<char> deqc;
typedef priority_queue<int> mxpq;
typedef priority_queue<int,vector<int>,greater<int> > mnpq;
typedef priority_queue<pii> mxpqii;
typedef priority_queue<pii,vector<pii>,greater<pii> > mnpqii;
const int maxn=500005;
const int inf=0x3f3f3f3f3f3f3f3f;
const int MOD=100000007;
const db eps=1e-10;
int qpow(int a,int b){int tmp=a%MOD,ans=1;while(b){if(b&1){ans*=tmp,ans%=MOD;}tmp*=tmp,tmp%=MOD,b>>=1;}return ans;}
int lowbit(int x){return x&-x;}
int max(int a,int b){return a>b?a:b;}
int min(int a,int b){return a<b?a:b;}
int mmax(int a,int b,int c){return max(a,max(b,c));}
int mmin(int a,int b,int c){return min(a,min(b,c));}
void mod(int &a){a+=MOD;a%=MOD;}
bool chk(int now){}
int half(int l,int r){while(l<=r){int m=(l+r)/2;if(chk(m))r=m-1;else l=m+1;}return l;}
int ll(int p){return p<<1;}
int rr(int p){return p<<1|1;}
int mm(int l,int r){return (l+r)/2;}
int lg(int x){if(x==0) return 1;return (int)log2(x)+1;}
bool smleql(db a,db b){if(a<b||fabs(a-b)<=eps)return true;return false;}
db len(db a,db b,db c,db d){return sqrt((a-c)*(a-c)+(b-d)*(b-d));}
bool isp(int x){if(x==1)return false;if(x==2)return true;for(int i=2;i*i<=x;++i)if(x%i==0)return false;return true;}
inline int read()
{char ch=getchar();int s=0,w=1;while(ch<48||ch>57){if(ch=='-')w=-1;ch=getchar();}while(ch>=48&&ch<=57){s=(s<<1)+(s<<3)+ch-48;ch=getchar();}return s*w;
}
inline void write(int x)
{if(x<0)putchar('-'),x=-x;if(x>9)write(x/10);putchar(x%10+48);
}int n,a[maxn];
int ans=0;
vei now;signed main(){ios_base::sync_with_stdio(0);cin.tie(0),cout.tie(0);cin>>n;re(i,1,n) cin>>a[i];int cnt=1;a[n+1]=inf;re(i,1,n){if(a[i]==a[i+1]) cnt++;else now.pub(cnt),cnt=1;}
//    fo(i,0,now.size()) cout<<now[i]<<' ';cout<<endl;fo(i,0,now.size()){ans+=now[i]*(now[i]+1)/2;}cout<<ans;return 0;
}

C、题意是给你一堆圆心在坐标轴上的圆和一堆点，问你每一个圆第一次被哪个点“击中”

“击中”的定义是点在圆内或者在圆上
输入数据保证所有的圆没有内含或相交关系，但是可以相切

考虑对每个圆建立结构体，重载运算符使得其按照圆心的横坐标由小到大排序
对每个点，二分找到圆心横坐标最接近点的横坐标的圆，如果点到该圆圆心的距离小于等于半径，并且该圆尚未被标记，则标记该圆
但是这样还不够，还要检查与这个圆最接近的两个圆，因为可能存在相切的情况。

代码：

#include <bits/stdc++.h>
#define int long long
#define sc(a) scanf("%lld",&a)
#define scc(a,b) scanf("%lld %lld",&a,&b)
#define sccc(a,b,c) scanf("%lld %lld %lld",&a,&b,&c)
#define schar(a) scanf("%c",&a)
#define pr(a) printf("%lld",a)
#define fo(i,a,b) for(int i=a;i<b;++i)
#define re(i,a,b) for(int i=a;i<=b;++i)
#define rfo(i,a,b) for(int i=a;i>b;--i)
#define rre(i,a,b) for(int i=a;i>=b;--i)
#define prn() printf("\n")
#define prs() printf(" ")
#define mkp make_pair
#define pii pair<int,int>
#define pub(a) push_back(a)
#define pob() pop_back()
#define puf(a) push_front(a)
#define pof() pop_front()
#define fst first
#define snd second
#define frt front()
#define bak back()
#define mem0(a) memset(a,0,sizeof(a))
#define memmx(a) memset(a,0x3f3f,sizeof(a))
#define memmn(a) memset(a,-0x3f3f,sizeof(a))
#define debug
#define db double
#define yyes cout<<"YES"<<endl;
#define nno cout<<"NO"<<endl;
using namespace std;
typedef vector<int> vei;
typedef vector<pii> vep;
typedef map<int,int> mpii;
typedef map<char,int> mpci;
typedef map<string,int> mpsi;
typedef deque<int> deqi;
typedef deque<char> deqc;
typedef priority_queue<int> mxpq;
typedef priority_queue<int,vector<int>,greater<int> > mnpq;
typedef priority_queue<pii> mxpqii;
typedef priority_queue<pii,vector<pii>,greater<pii> > mnpqii;
const int maxn=500005;
const int inf=0x3f3f3f3f3f3f3f3f;
const int MOD=100000007;
const db eps=1e-10;
int qpow(int a,int b){int tmp=a%MOD,ans=1;while(b){if(b&1){ans*=tmp,ans%=MOD;}tmp*=tmp,tmp%=MOD,b>>=1;}return ans;}
int lowbit(int x){return x&-x;}
int max(int a,int b){return a>b?a:b;}
int min(int a,int b){return a<b?a:b;}
int mmax(int a,int b,int c){return max(a,max(b,c));}
int mmin(int a,int b,int c){return min(a,min(b,c));}
void mod(int &a){a+=MOD;a%=MOD;}
bool chk(int now){}
int half(int l,int r){while(l<=r){int m=(l+r)/2;if(chk(m))r=m-1;else l=m+1;}return l;}
int ll(int p){return p<<1;}
int rr(int p){return p<<1|1;}
int mm(int l,int r){return (l+r)/2;}
int lg(int x){if(x==0) return 1;return (int)log2(x)+1;}
bool smleql(db a,db b){if(a<b||fabs(a-b)<=eps)return true;return false;}
db len(db a,db b,db c,db d){return sqrt((a-c)*(a-c)+(b-d)*(b-d));}
bool isp(int x){if(x==1)return false;if(x==2)return true;for(int i=2;i*i<=x;++i)if(x%i==0)return false;return true;}
inline int read()
{char ch=getchar();int s=0,w=1;while(ch<48||ch>57){if(ch=='-')w=-1;ch=getchar();}while(ch>=48&&ch<=57){s=(s<<1)+(s<<3)+ch-48;ch=getchar();}return s*w;
}
inline void write(int x)
{if(x<0)putchar('-'),x=-x;if(x>9)write(x/10);putchar(x%10+48);
}int n,m;
struct node{int x,r,id;node(){}node(int X,int R):x(X),r(R){}bool operator < (const node& o)const{return x<o.x;}
}o[maxn];
int a,b,ans[maxn],tot=0;bool ok(int a,int b,int p){int x=o[p].x;int r=o[p].r;if(smleql(len((db)a,(db)b,(db)x,0.0),(db)r)) return 1;return 0;
}signed main(){ios_base::sync_with_stdio(0);cin.tie(0),cout.tie(0);cin>>n;re(i,1,n) cin>>o[i].x>>o[i].r,o[i].id=i;sort(o+1,o+1+n);re(i,1,n) ans[i]=-1;cin>>m;re(i,1,m){cin>>a>>b;int p=lower_bound(o+1,o+1+n,node(a,0))-o;if(p<=n&&ans[o[p].id]==-1&&ok(a,b,p)) ans[o[p].id]=i,tot++;p++;if(p<=n&&ans[o[p].id]==-1&&ok(a,b,p)) ans[o[p].id]=i,tot++;p-=2;if(p<=n&&ans[o[p].id]==-1&&ok(a,b,p)) ans[o[p].id]=i,tot++;}cout<<tot<<endl;re(i,1,n) cout<<ans[i]<<' ';return 0;
}

D、题意是有一堆动物去看病，编号从1到n，每只动物看病的次数是有限的，看完病了就会直接回家，否则会排到队伍的末尾去

医生今天打算看k次病，问你看完k次之后队伍的排列是怎样的

k是1e14级别的，直接模拟不现实，我一开始胡了一个线段树的做法，每次在不更改队列顺序的情况下对序列中每个数字减去一个最小值，但是T在第39个点

实际上应该把a[i]看成一个楼梯，二分找医生最多能看“几轮”病人，看一轮病人指队列中从头到尾的病人全部看过一遍

每一轮相当于在楼梯上了一个台阶

显然看一轮病人是不改变原序列的顺序的

之后减去最多看的轮数之后上deque大模拟就行，时间复杂度是O(n)级别的

代码：

#include <bits/stdc++.h>
#define int long long
#define sc(a) scanf("%lld",&a)
#define scc(a,b) scanf("%lld %lld",&a,&b)
#define sccc(a,b,c) scanf("%lld %lld %lld",&a,&b,&c)
#define schar(a) scanf("%c",&a)
#define pr(a) printf("%lld",a)
#define fo(i,a,b) for(int i=a;i<b;++i)
#define re(i,a,b) for(int i=a;i<=b;++i)
#define rfo(i,a,b) for(int i=a;i>b;--i)
#define rre(i,a,b) for(int i=a;i>=b;--i)
#define prn() printf("\n")
#define prs() printf(" ")
#define mkp make_pair
#define pii pair<int,int>
#define pub(a) push_back(a)
#define pob() pop_back()
#define puf(a) push_front(a)
#define pof() pop_front()
#define fst first
#define snd second
#define frt front()
#define bak back()
#define mem0(a) memset(a,0,sizeof(a))
#define memmx(a) memset(a,0x3f3f,sizeof(a))
#define memmn(a) memset(a,-0x3f3f,sizeof(a))
#define debug
#define db double
#define yyes cout<<"YES"<<endl;
#define nno cout<<"NO"<<endl;
using namespace std;
typedef vector<int> vei;
typedef vector<pii> vep;
typedef map<int,int> mpii;
typedef map<char,int> mpci;
typedef map<string,int> mpsi;
typedef deque<int> deqi;
typedef deque<char> deqc;
typedef priority_queue<int> mxpq;
typedef priority_queue<int,vector<int>,greater<int> > mnpq;
typedef priority_queue<pii> mxpqii;
typedef priority_queue<pii,vector<pii>,greater<pii> > mnpqii;
const int maxn=500005;
const int inf=0x3f3f3f3f3f3f3f3f;
const int MOD=100000007;
const db eps=1e-10;
int qpow(int a,int b){int tmp=a%MOD,ans=1;while(b){if(b&1){ans*=tmp,ans%=MOD;}tmp*=tmp,tmp%=MOD,b>>=1;}return ans;}
int lowbit(int x){return x&-x;}
int max(int a,int b){return a>b?a:b;}
int min(int a,int b){return a<b?a:b;}
int mmax(int a,int b,int c){return max(a,max(b,c));}
int mmin(int a,int b,int c){return min(a,min(b,c));}
void mod(int &a){a+=MOD;a%=MOD;}
int ll(int p){return p<<1;}
int rr(int p){return p<<1|1;}
int mm(int l,int r){return (l+r)/2;}
int lg(int x){if(x==0) return 1;return (int)log2(x)+1;}
bool smleql(db a,db b){if(a<b||fabs(a-b)<=eps)return true;return false;}
db len(db a,db b,db c,db d){return sqrt((a-c)*(a-c)+(b-d)*(b-d));}
bool isp(int x){if(x==1)return false;if(x==2)return true;for(int i=2;i*i<=x;++i)if(x%i==0)return false;return true;}
inline int read()
{char ch=getchar();int s=0,w=1;while(ch<48||ch>57){if(ch=='-')w=-1;ch=getchar();}while(ch>=48&&ch<=57){s=(s<<1)+(s<<3)+ch-48;ch=getchar();}return s*w;
}
inline void write(int x)
{if(x<0)putchar('-'),x=-x;if(x>9)write(x/10);putchar(x%10+48);
}int n,k,a[maxn],sum=0;
deque<int> q;int chk(int now){int res=0;re(i,1,n) res+=min(now,a[i]);return res;
}int half(int l,int r){while(l<=r){int m=(l+r)/2;if(chk(m)<=k)l=m+1;else r=m-1;}return r;}signed main(){ios_base::sync_with_stdio(0);cin.tie(0),cout.tie(0);cin>>n>>k;int mx=-inf;re(i,1,n) cin>>a[i],mx=max(mx,a[i]),sum+=a[i];if(sum<k) cout<<-1;else if(sum==k);else{int t=half(1,mx);k-=chk(t);re(i,1,n) {a[i]-=t;if(a[i]>0) q.pub(i);}while(k--){int f=q.frt;q.pof();a[f]--;if(a[f]>0) q.pub(f);}fo(i,0,q.size()) cout<<q[i]<<' ';}return 0;
}

转载于:https://www.cnblogs.com/oneman233/p/11494216.html

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