UVA 1451 Average

A DNA sequence consists of four letters, A, C, G, and T. The GC-ratio of a DNA sequence is the
number of Cs and Gs of the sequence divided by the length of the sequence. GC-ratio is important
in gene nding because DNA sequences with relatively high GC-ratios might be good candidates for
the starting parts of genes. Given a very long DNA sequence, researchers are usually interested in
locating a subsequence whose GC-ratio is maximum over all subsequences of the sequence. Since short
subsequences with high GC-ratios are sometimes meaningless in gene nding, a length lower bound is
given to ensure that a long subsequence with high GC-ratio could be found. If, in a DNA sequence,
a 0 is assigned to every A and T and a 1 to every C and G, the DNA sequence is transformed into a
binary sequence of the same length. GC-ratios in the DNA sequence are now equivalent to averages in
the binary sequence.

题目大意:给出一个01序列，求长度至少为L的子序列，使得平均值最大
解题报告:
比较简单，但是花了许久时间，还是太渣.
开始以为是简单贪心，长度固定为L即可，发现这个单调性只有排序后才有QWQ,所以拍WA后改写斜率优化DP:我们要求的是((sum[i]-sum[j])/(i-j+1))最大值，明显对应平面上的斜率，所以直接做就好，注意要把0加进去，调了很久

#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <cmath>
#define RG register
#define il inline
#define iter iterator
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
using namespace std;
const int N=1e5+5;const double eps=1e-6;
int a[N],n,m,sum[N],q[N];char s[N];
int fy(int i,int j){
	return sum[i]-sum[j];
}
int fx(int i,int j){
	return i-j;
}
void work()
{
	scanf("%d%d",&n,&m);
 	scanf("%s",s+1);
	for(int i=1;i<=n;i++){
		a[i]=s[i]-'0',sum[i]=sum[i-1]+a[i];
		if(m==1 && a[i]){
			printf("%d %d
",i,i);
			return ;
		}
	}
	if(m==1){puts("1 1");return ;}
	int l=1,r=0,j,k,L=0,R=m;int tot;
	for(int i=m;i<=n;i++){
		while(r-l>=1){
			j=q[r];k=q[r-1];
			if(fy(i-m,j)*fx(i-m,k)<=fy(i-m,k)*fx(i-m,j))r--;
			else break;
		}
		q[++r]=i-m;
		while(r-l>=1){
			j=q[l+1];k=q[l];
			if(fy(i,j)*fx(i,k)>=fy(i,k)*fx(i,j))l++;
			else break;
		}
		tot=fy(i,q[l])*fx(R,L)-fy(R,L)*fx(i,q[l]);
		if(tot>0 || (tot==0 && i-q[l]<R-L)){
			L=q[l];R=i;
		}
	}
	printf("%d %d
",L+1,R);
}

int main()
{
	int T;cin>>T;
	while(T--)work();
	return 0;
}