北邮OJ 884. 16校赛-Average Modulo

时间限制 5000 ms 内存限制 65536 KB

题目描述

We define function g on an array as:

g([a0,a1,⋯,an−1])=(Σn−1l=0al) mod pn

g([a_0,a_1,\cdots,a_{n-1}]) = \frac{(\Sigma_{l=0}^{n-1} a_l) ~ mod ~ p}{n}

Here, p is a given constant, and n is the length of the array.
You are given p and an array A and a length k.
You need to find the maximum value of g over all contiguous subarrays of A that are of length ≥ k.

输入格式

The first line of input contains T, the number of test cases. Each test case contains two lines. The first line of each test case contains three integers: N, p and k, respectively.
The second line of each test case contains N integers contained in array A.

1≤T≤101 \le T \le 10
1≤N≤5∗1041 \le N \le 5*10^4
1≤K≤min(300,N),1≤P≤1091 \le K \le min(300,N) ,1 \le P \le 10^9
1≤Ai≤1091 \le A_i \le 10^9

输出格式

For each test case, output one line containing two space separated integers: p and q. If the answer is an integer x, then output x and 1 separated by a space.
Otherwise, output p and q if pq is the answer is in its simplest form.

Explanation of example
In the first test case, when length ≥2, the maximum g value is obtained when you take the entire array, so g is 53\frac{5}{3}.
In the second case, all f values are even. Hence, when we take modulo 2, the answer is 0 for any array.

In the third case, p is 1, and the value is 0 for any array.

输入样例

3
3 100 2
2 1 2
5 2 1
2 10 4 6 8
5 1 2
100 213142 3123 123 321

输出样例

5 3
0 1
0 1

题意为，找出给定序列中的一个满足长度≥k的连续子序列，使得这个子序列的序列和模上p再除以n尽可能的大。
比如a1,a2,a3,a4,a5,a6,a7,a8,a9,a10，找出其中长度≥2的一段连续子序列，使得他们相加之后mod p = tmp，tmp / n= tmp2，使得tmp2尽可能的大。
乍看之下又是一道枚举题，但是n的范围最大是5∗1045*10^4，如果从length=k枚举到length=n，O(n2n^2)的复杂度明显就超时了，这时候有两种思路，第一种思路是自己打表出数据，用O(n2n^2)的枚举法测一下答案的连续子序列长度最大是多少，

打表代码如下

#include<bits/stdc++.h>
#define K 5
using namespace std;
int main(){srand(time(NULL));freopen("tsx.txt","w",stdout);int n=50000,k=200,ts=1000,p=1000000000;printf("%d\n",ts);while(ts--){printf("%d %d %d\n",n,p,k);for(int i=0;i<n;i++)printf("%d%c",rand()%p+1,i==n-1?'\n':' ');}return 0;
}

然后用裸的O(n2n^2)的枚举测了几千组自己生成的样例，答案长度最大小于k的1.5倍，比赛时稳妥起见开了k的3倍，时间5s绰绰有余。

第二种思路呢，是用数学思想证明，首先(a+b)%p =(a%p + b%p)%p

令a=Σ2ki=0ai\Sigma_{i=0}^{2k} a_i，b=Σ2k+xi=2k+1ai\Sigma_{i=2k+1}^{2k+x} a_i

(a+b)%p2k+x\frac{(a+b) \%p}{2k+x}=(a%p+b%p)%p2k+x\frac{(a\%p+b\%p) \%p}{2k+x}
≤ a%p2k+x\frac{a\%p}{2k+x}+ b%p2k+x\frac{b\%p}{2k+x}
≤max(a%p2k\frac{a\%p}{2k} , b%px\frac{b\%p}{x} )

即当枚举到长度≥2k但≤4k的时候（更长的情况可以递推），设长度为2k+x，一定可以划分为一个2k和x长度的两个连续子序列，它们的值一定可以代替（大于等于）原始2k+x串的值。

解决代码如下

#include<cstdio>
#include<cmath>
#include<algorithm>
#define MA 50050
using namespace std;
int a[MA],sum[MA],n,k,p;
int gcd(int a,int b){return b==0?a:gcd(b,a%b);}
int main(){int i,j,k,len,t;for(scanf("%d",&t);t--;){scanf("%d%d%d",&n,&p,&k);for(i=1;i<=n;i++)scanf("%d",&a[i]);for(i=1;i<=n;i++)sum[i]=(sum[i-1]+a[i])%p;int up=0,down=1;int lim=k*3;//根据数学验证，取2k即可for(len=k;len<=lim;len++){for(i=0;i+len<=n;i++){int tmpa=(p+sum[i+len]-sum[i])%p;int tmpb=len;double ta=1.0*up/down;double tb=1.0*tmpa/tmpb;if(ta<tb){up=tmpa;down=tmpb;}}}int gcd_=gcd(up,down);printf("%d %d\n",up/gcd_,down/gcd_);}return 0;
}