一道ACM题，求O(n) 算法
大意，给你一01串，你能够把其中的k个0改成一,求改变后连续1的最大数目
Description

You are given a 01 sequence whose size is n. you can change at most k 0 into 1 in the sequence.
Now I want to know how many consecutive 1 in the sequce at most after you do the change.

Input

The first of input is an integer t which stands for the number of test cases.
for each test case, the first line contains two number n, k ( 1 <= n <= 100000, 0 <= k <= n ).
the next contains n element of the sequence, whose value is 0 or 1. Output Output the answer in one line for each test case.

Simple Input

2
4 2
1 0 0 1
4 1
1 0 1 0

Simple Output
4
3


[解决办法]
刚AC的，热乎乎的。要做到O(n)最主要的办法就是下标不要回溯。
代码大意为2个下标l,h，h一直向前走，遇到1，则结果加1，遇到0则看看是否还有机会(t)补1.
如果有，t--。
如果没有，l++且如果a[l]为0的话，机会+1，下一次循环。

C/C++ code

void max_01(int* a,int n,int t)    {        int l = 0,h = 0,res = 0,tmp = 0;        while(h < n)        {            if(a[h] == 0)            {                if(t == 0)                {                    --tmp;                    if(a[l++] == 0) ++t;                    continue;                }                --t;            }            ++h;            ++tmp;            if(res < tmp) res = tmp;        }        printf("%d\n",res);    }    int main()    {        int t,n,k,a[100001];        scanf("%d",&t);        while(t--)        {            scanf("%d %d",&n,&k);            for(int i = 0 ; i < n ; ++ i)                scanf("%d",&a[i]);            max_01(a,n,k);        }            return 0;    }