luogu P3353 [the stars shining in your window]

This problem can prefix and maintenance, mainly talk about prefixes and

- and the prefix

provided $ pre_i $ represents $ \ sum ^ {i} _ {1} a_ {i} $

there is
$ pre_i pre_ = {I}. 1-a_i $ +

- and query prefix

With pre $ $ array if we claim $ [l, r] $ and has

$ pre_. 1} = {L-A_2 A_1 + + + ... + A_3 A_. 1-L {} $

$ R & lt pre_} = { a_1 + a_2 ... a_ {l- 1} + a_ {l} + ... a_r $

be a difference, found twenty-two offset, we get

$$ \ sum ^ r_ {i = l} a_i = pre_r- {L}. 1-pre_ $$

(pen can not understand the math)
- Code

code can achieve such

#include<iostream>
#include<cstdio>
using namespace std;
inline int read(){
	register int x=0,v=1,ch=getchar();
	while(!isdigit(ch)){if(ch=='-')v=-1;ch=getchar();}
	while(isdigit(ch)){x=(x<<3)+(x<<1)+(ch^'0');ch=getchar();}
	return x*v;
}
const int MAX=100005;
int n,k,L,R,l,r;
int x,ans,res[MAX],pre[MAX];
int main(){
	n=read(),k=read();
	for(register int i=1;i<=n;++i){
		x=read();res[x]+=read();
		L=min(L,x),R=max(R,x);//查询最小 位置和最大位置
	}
	for(register int i=L;i<=R;++i){//前缀和
		pre[i]=pre[i-1]+res[i];
	}
	for(register int i=L;i+k-1<=R;++i){
		l=i,r=i+k-1;
		ans = max (ans, pre [ r] -pre [l-1]); // determines whether the maximum and 
	} 
	the printf ( "% D \ n-", ANS); // output answer 
	return 0; 
}