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P2801 教主的魔法(分块)

时间:2019-10-10 01:22:40      阅读:40      评论:0      收藏:0      [点我收藏+]

标签:getch   {}   air   double   else   二分   

分块儿练习,维护每个区间有序,二分找大于它的即可

注意的是数组别开小,并且要建立两个数组用来处理边角问题

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 scs(a) scanf("%s",a) 
#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;
#define all(i,a) for(auto i=a.begin();i!=a.end();++i)
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=1000005;
const int inf=0x3f3f3f3f3f3f3f3f;
const int MOD=100000007;
const db eps=1e-10;
const db pi=3.1415926535;
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;}
bool bigeql(db a,db b){if(a>b||fabs(a-b)<=eps)return true;return false;}
bool eql(db a,db b){if(fabs(a-b)<eps) return 1;return 0;}
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 gcd(int a, int b){
	if(a==0) return b;
	if(b==0) return a;
	if(!(a&1)&&!(b&1)) return gcd(a>>1,b>>1)<<1;
	else if(!(b&1)) return gcd(a,b>>1);
	else if(!(a&1)) return gcd(a>>1,b);
	else return gcd(abs(a-b),min(a,b));
}
int lcm(int x,int y){return x*y/gcd(x,y);}

int n,q,a[maxn],b[maxn],l,r,x;
char op[5];
int tot,id[maxn],from[maxn],to[maxn];
int add[maxn];

void pre(){
	re(i,1,n) a[i]=b[i];
	tot=sqrt(n);
	re(i,1,tot)
		from[i]=(i-1)*tot+1,
		to[i]=i*tot;
	if(to[tot]<n) tot++,from[tot]=to[tot-1]+1,to[tot]=n;
	for(int i=1;i<=tot;++i)
		for(int j=from[i];j<=to[i];++j)
			id[j]=i;
	for(int i=1;i<=tot;++i)
		sort(a+from[i],a+to[i]+1);
}

void modi(int l,int r,int x){
	if(id[l]==id[r]){
		re(i,l,r) b[i]+=x;
		re(i,from[id[l]],to[id[l]]) a[i]=b[i];
		sort(a+from[id[l]],a+to[id[l]]+1);
		return;
	}
	re(i,l,to[id[l]]) b[i]+=x;
	re(i,from[id[l]],to[id[l]]) a[i]=b[i];
	sort(a+from[id[l]],a+to[id[l]]+1);
	re(i,from[id[r]],r) b[i]+=x;
	re(i,from[id[r]],to[id[r]]) a[i]=b[i];
	sort(a+from[id[r]],a+to[id[r]]+1);
	re(i,id[l]+1,id[r]-1)
		add[i]+=x;
}

int ask(int l,int r,int x){
	int ret=0;
	if(id[l]==id[r]){
		re(i,l,r) if(b[i]+add[id[i]]>=x) ret++;
		return ret;
	}
	re(i,l,to[id[l]]) if(b[i]+add[id[i]]>=x) ret++;
	re(i,from[id[r]],r) if(b[i]+add[id[i]]>=x) ret++;
	re(i,id[l]+1,id[r]-1){
		int L=from[i],R=to[i];
		int p=lower_bound(a+L,a+R+1,x-add[i])-a;
//		cout<<"??? "<<L<<‘ ‘<<p<<endl;
		ret+=R-p+1;
	}
	return ret;
}

signed main(){
    ios_base::sync_with_stdio(0);
    cin.tie(0),cout.tie(0);
    n=read(),q=read();
    re(i,1,n) b[i]=read();
    pre();
    while(q--){
		scanf("%s",op);
		l=read(),r=read(),x=read();
		if(op[0]==‘A‘) write(ask(l,r,x)),prn();
		else if(op[0]==‘M‘) modi(l,r,x);
	}
    return 0;
}
/*
5 3
1 2 3 4 5
A 1 5 4
M 3 5 1
A 1 5 4
*/

  

P2801 教主的魔法(分块)

标签:getch   {}   air   double   else   二分   

原文:https://www.cnblogs.com/oneman233/p/11645171.html

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