2018 Multi-University Training Contest 3 Solution

A - Problem A. Ascending Rating

题意：给出n个数，给出区间长度m。对于每个区间，初始值的max为0，cnt为0.遇到一个a[i] > ans, 更新ans并且cnt++。计算

$A = \sum_{i = 1}^{i = n - m +1} (max \oplus i)$

$B = \sum_{i = 1}^{i = n - m +1} (cnt \oplus i)$

思路：单调队列，倒着扫一遍，对于每个区间的cnt就是队列的长度，扫一遍即可。

 1 #include<bits/stdc++.h>
 2 
 3 using namespace std;
 4 
 5 const int maxn = 1e7 + 10;
 6 typedef long long ll;
 7 
 8 int t;
 9 int n,m,k;
10 ll p, q, r, mod;
11 ll arr[maxn];
12 ll brr[maxn];
13 
14 int main()
15 {
16     scanf("%d", &t);
17     while(t--)
18     {
19         scanf("%d %d %d %lld %lld %lld %lld", &n, &m, &k, &p, &q, &r, &mod);
20         for(int i = 1; i <= n; ++i)
21         {
22             if(i <= k) scanf("%lld", arr + i);
23             else arr[i] = (p * arr[i - 1] + q * i + r) % mod;
24         }
25         ll A = 0, B = 0;
26         int head = 0, tail = -1;
27         for(int i = n; i > (n - m + 1); --i)
28         {
29             while(head <= tail && arr[i] >= arr[brr[tail]]) tail--;
30             brr[++tail] = i;
31         }
32         for(int i = n - m + 1; i >= 1; --i)
33         {
34             while(head <= tail && arr[i] >= arr[brr[tail]]) tail--;
35             brr[++tail] = i;
36             while(brr[head] - i >= m) head++;
37             A += arr[brr[head]] ^ i;
38             B += (tail - head + 1) ^ i;
39         }
40         printf("%lld %lld\n", A, B);
41     }
42     return 0;
43 }

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