Problem A. Ascending Rating
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 533 Accepted Submission(s): 122
Problem Description
Before the start of contest, there are n ICPC contestants waiting in a long queue. They are labeled by 1 to n from left to right. It can be easily found that the i -th contestant's QodeForces rating is ai .
Little Q, the coach of Quailty Normal University, is bored to just watch them waiting in the queue. He starts to compare the rating of the contestants. He will pick a continous interval with length m , say [l,l+m−1] , and then inspect each contestant from left to right. Initially, he will write down two numbers maxrating=0 and count=0 . Everytime he meets a contestant k with strictly higher rating than maxrating , he will change maxrating to ak and count to count+1 .
Little T is also a coach waiting for the contest. He knows Little Q is not good at counting, so he is wondering what are the correct final value of maxrating and count . Please write a program to figure out the answer.
Input
The first line of the input contains an integer T(1≤T≤2000) , denoting the number of test cases.
In each test case, there are 7 integers n,m,k,p,q,r,MOD(1≤m,k≤n≤107,5≤p,q,r,MOD≤109) in the first line, denoting the number of contestants, the length of interval, and the parameters k,p,q,r,MOD .
In the next line, there are k integers a1,a2,...,ak(0≤ai≤109) , denoting the rating of the first k contestants.
To reduce the large input, we will use the following generator. The numbers p,q,r and MOD are given initially. The values ai(k<i≤n) are then produced as follows :
ai=(p×ai−1+q×i+r)modMOD
It is guaranteed that ∑n≤7×107 and ∑k≤2×106 .
Output
Since the output file may be very large, let's denote maxratingi and counti as the result of interval [i,i+m−1] .
For each test case, you need to print a single line containing two integers A and B , where :
AB==∑i=1n−m+1(maxratingi⊕i)∑i=1n−m+1(counti⊕i)
Note that ``⊕ '' denotes binary XOR operation.
Sample Input
1 10 6 10 5 5 5 5 3 2 2 1 5 7 6 8 2 9
Sample Output
46 11
【小结】
看了标程，才知道我有多菜。
【题意】
输入那一堆东西不想说了，反正就是最后输出俩数。而且卡时间，必须线性复杂度O(n)
解法1：赛时一种很烂的思路，建议直接看解法2.
单调栈预处理出每个位置的下一个首次大于自己的位置。
dp思想预处理出每个位置最多能往后升高几次。
最后跑一遍，单调栈维护区间最值（当栈空时，说明当前区间所有数比之前区间存储的数都要小）
解法2：题解思路，太强了%%%%
从后往前扫一遍，单调栈（我更愿意说是单调队列）维护滑动区间的递减值（就像在维护一个台阶）
【代码1】
/****
***author winter2121
****/
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mod=1e9+7;
const int MAX=1e7+5;
const int INF=0x3f3f3f3f;
int n,m,k,p,q,r,MOD,T;
int a[MAX];
int st[MAX]; //栈
int fir[MAX]; //i右边第一个大于a[i]的位置
int dp[MAX]; //以i起始的最长上坡
int main()
{
scanf("%d",&T);
while(T--)
{
scanf("%d%d%d%d%d%d%d",&n,&m,&k,&p,&q,&r,&MOD);
for(int i=1;i<=k;i++)scanf("%d",&a[i]);
for(int i=k+1;i<=n;i++)a[i]=(1ll*p*a[i-1]+1ll*q*i+r)%MOD;
int top=0;
for(int i=1;i<=n;i++)
{
while(top&&a[ st[top] ]<a[i])
fir[ st[top--] ]=i;
st[++top]=i;
}
for(int i=n;i>=1;i--)dp[i]=dp[fir[i]]+1;
top=0;
unsigned long long ans1=0,ans2=0;
for(int i=1,j=1;i<=n-m+1;i++)
{
while(top&&st[top]<i)top--;
if(top==0)j=i; //若空，则从新开始
while(j<=i+m-1)
{
if(a[j]>a[ st[top] ])st[++top]=j;
j++;
}
int t=st[top];
int Count=dp[i]-dp[t]+1;
int maxrating=a[t];
ans1+=maxrating^i;
ans2+=Count^i;
}
printf("%llu %llu\n",ans1,ans2);
}
}
【代码2】
#include<bits/stdc++.h>
using namespace std;
const int MAX=1e7+5;
int T,n,m,k,p,q,r,MOD,a[MAX],st[MAX];
int main()
{
scanf("%d",&T);
while(T--)
{
scanf("%d%d%d%d%d%d%d",&n,&m,&k,&p,&q,&r,&MOD);
for(int i=1;i<=k;i++)scanf("%d",&a[i]);
for(int i=k+1;i<=n;i++)a[i]=(1ll*p*a[i-1]+1ll*q*i+r)%MOD;
long long A=0,B=0;
for(int i=n,pre=0,tail=1;i>=1;i--)
{
while(pre>=tail&&st[tail]>=i+m)tail++; //超出区间的部分去掉
while(pre>=tail&&a[ st[pre] ]<=a[i])pre--; //遇到的新的更大值，之前的不需要了
st[++pre]=i;
if(i>n-m+1)continue;
A+=a[ st[tail] ]^i;
B+=(pre-tail+1)^i;
}
printf("%lld %lld\n",A,B);
}
}