现在已经会用分块卡了。大概就是在所有偶数位上数的位置+1，在所有奇数位数的位置-1，然后判定所有的前缀和是否为0或1.

然而只有80pts

inline void add(int k, int tp)
{
    int value = tp ? 1 : -1;
    //printf("%d\n", value);
    int startBlock = (k - 1) / blksize + 1;
    mn[startBlock] = 0x3f3f3f3f;
    mx[startBlock] = -0x3f3f3f3f;
    for (RI i = (startBlock - 1) * blksize + 1; i <= startBlock * blksize; ++i)
    {
        if (i >= k)
            val[i] += value;
        mn[startBlock] = min(val[i] + tag[startBlock], mn[startBlock]);
        mx[startBlock] = max(val[i] + tag[startBlock], mx[startBlock]);
    }
    for (RI i(startBlock + 1), e(blknum); i <= e; ++i)
    {
        tag[i] += value;
        mn[i] += value;
        mx[i] += value;
    }
}

inline bool judge()
{
    for (RI i = 1; i <= blknum; ++i)
    {
        if (mn[i] < 0)
            return false;
        if (mx[i] > 1)
            return false;
    }
    return true;
}

int main()
{
    freopen("sort.in", "r", stdin);
    freopen("sort.out", "w", stdout);
    F_IO.read(n);
    F_IO.read(m);
    blksize = 1005;
    blknum = (MAXN - 1) / blksize + 1;
    for (RI i(1), e(blknum); i <= e; ++i)
    {
        tag[i] = 0;
    }
    for (RI i(1), e(n); i <= e; ++i)
    {
        F_IO.read(a[i]);
        add(a[i], i & 1);
    }
    RI x, y;
    while (m--)
    {
        F_IO.read(x);
        F_IO.read(y);
        add(a[x], (x & 1) ^ 1);
        a[x] = y;
        add(a[x], (x & 1));
        puts(judge() ? "Yes" : "No");
    }
    F_IO.buf_flush();
    return 0;
}

求调

数据规模不给怎么选算法（

看起来直接排序然后模拟会 T……

不知道诶……我想想