Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

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 finding 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 finding, 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.

For the binary sequence above, if the length lower bound is 7, the maximum average is 6/8 which happens in the subsequence [7,14]. Its length is 8, which is greater than the length lower bound 7. If the length lower bound is 5, then the subsequence [7,11] gives the maximum average 4/5. The length is 5 which is equal to the length lower bound. For the subsequence [7,11], 7 is its starting index and 11 is its ending index.

Given a binary sequence and a length lower bound L, write a program to find a subsequence of the binary sequence whose length is at least L and whose average is maximum over all subsequences of the binary sequence. If two or more subsequences have the maximum average, then find the shortest one; and if two or more shortest subsequences with the maximum average exist, then find the one with the smallest starting index.

Input 

Your program is to read from standard input. The input consists of T test cases. The number of test cases T is given in the first line of the input. Each test case starts with a line containing two integers n (1n100, 000) and L (1L1, 000) which are the length of a binary sequence and a length lower bound, respectively. In the next line, a string, binary sequence, of length n is given.

Output 

Your program is to write to standard output. Print the starting and ending index of the subsequence.

The following shows sample input and output for two test cases.

Sample Input 

2 
17 5 
00101011011011010 
20 4 
11100111100111110000

Sample Output 

7 11 
6 9

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轉換成 01 字串，問題等價求最大平均值問題，

最大平均值問題，使用單調&斜率優化的方式去解，
證明方式轉換成幾何問題。

效率 O(n)

證明請網路搜一下這篇論文 算法合集之《浅谈数形结合思想在信息学竞赛中的应用》

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
double get_avg(int l, int r, int sum[]) {
    return (double)(sum[r]-sum[l])/(r-l);
}
int q[100005], sum[100005];
double maxAvgProblem(int n, int L, int sum[]) {
    double ans = (double)sum[n]/n;
    int front, rear, i, j;
    int retL = 0, retR = n;
    front = 0, rear = -1;
    for(i = L; i <= n; i++) {
        int tmp = i-L;
        while(front < rear && get_avg(q[rear], tmp, sum) <= get_avg(q[rear-1], q[rear], sum))
            rear--;
        q[++rear] = tmp;
        while(front < rear && get_avg(q[front], i, sum) <= get_avg(q[front+1], i, sum))
            front++;
        double tans = get_avg(q[front], i, sum);
        if(tans > ans)
            ans = tans, retL = q[front], retR = i;
        if(tans == ans && i - q[front] < retR - retL)
			retL = q[front], retR = i;
    }
    printf("%d %d\n", retL + 1, retR);
    return ans;
}
int main() {
	int testcase, n, L;
	int i, j, k;
	char s[100005];
	scanf("%d", &testcase);
	while(testcase--) {
		scanf("%d %d", &n, &L);
		scanf("%s", s+1);
		for(i = 1, sum[0] = 0; i <= n; i++)
			sum[i] = sum[i-1] + (s[i] == '1');
		maxAvgProblem(n, L, sum);
	}
	return 0;
}