题解:一开始做这个题的时候想过尺取法,但是因为没有逆元的知识,不知道该如何不断删除左端元素。其实这题并不难想,设l,r为两端开始都置为1,当长度小于k的时候不断乘右端元素并取余,当长度等于k时删除左端元素并且乘上右端端元素。注意:若右端元素为0时,就将两个端点都移到下一位从新开始。
#pragma GCC optimize(3,"Ofast","inline")
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <math.h>
#include <string>
#include <list>
#include <set>
#include <map>
#include <stack>
#include <algorithm>
#define maxn 110
#define MaxN 0x3f3f3f
#define MinN 0xc0c0c0
typedef long long ll;
using namespace std;
const int mod=998244353;
ll a[300010];ll quickpow(ll a,ll b){ll ans=1;while(b){if(b%2==1)ans=ans*a%mod;a=a*a%mod;b=b/2;}return ans;
}
int main()
{int n,k;cin>>n>>k;for(int i=1;i<=n;i++) scanf("%lld",&a[i]);ll l=1,r=1;ll imax=0;ll sum=1;while(r<=n){if(r-l<k-1&&a[r]!=0){sum=(sum*a[r])%mod;r++;}else if(r-l==k-1&&a[r]!=0){sum=(sum*a[r])%mod;imax=max(imax,sum);r++;}else if(r-l==k&&a[r]!=0){sum=(sum%mod*a[r]%mod*quickpow(a[l],mod-2))%mod; //费马小定理不会的小伙伴可以去百度一下l++,r++;imax=max(imax,sum);}else if(a[r]==0){r++;l=r;sum=1;}}cout<<imax<<endl;return 0;
}
后续。。。
刚学会线段树就拿来写了一下这个题目
以上是逆元+尺取法,又现学了一下线段树用线段树莽一波也是不错的解法,本题线段树只涉及了插入和计算“和”的部分
#pragma GCC optimize(3,"Ofast","inline")
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <math.h>
#include <string>
#include <list>
#include <set>
#include <map>
#include <stack>
#include <algorithm>
#define maxn 1000000
#define MaxN 0x3f3f3f
#define MinN 0xc0c0c0
typedef long long ll;
using namespace std;
const int mod=998244353;
ll arr[maxn];
ll tree[maxn]={1};
void build_tree(ll node,ll start,ll ends){if(start==ends){tree[node]=arr[start];}else{ll mid=(start+ends)/2;ll left_node=2*node+1;ll right_node=2*node+2;build_tree(left_node,start,mid);build_tree(right_node,mid+1,ends);tree[node]=(tree[left_node]*tree[right_node])%mod;}return ;
}ll query_tree(ll node,ll start,ll ends,ll l,ll r,ll sum){if(r<start||l>ends) return 1;else if(l<=start&&r>=ends){return (tree[node]*sum)%mod;}else{ll mid=(start+ends)/2;ll left_node=2*node+1;ll right_node=2*node+2;ll sum_left=query_tree(left_node,start,mid,l,r,sum);ll sum_right=query_tree(right_node,mid+1,ends,l,r,sum);return ((sum_left*sum_right)*sum%mod)%mod;}
}
int main()
{int n,m,k;cin>>n>>k;for(int i=0;i<n;i++) scanf("%lld",&arr[i]);build_tree(0,0,n-1);//for(int i=0;i<100;i++) cout<<tree[i]<<" ";ll imax=0;for(int i=0;i+k-1<n;i++){ll ans;ans=query_tree(0,0,n-1,i,i+k-1,1);imax=max(imax,ans);}cout<<imax<<endl;return 0;}