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题意:
- 给你n个数组成的序列,让你求从k+1位置开始,序列以p个数为周期。最小的k+p是多少。
做法:
- kmp 算法的一个应用,即最小循环节是i-next[i] (吃了没文化的亏_(:з」∠)_
详细讲解:KMP算法 —— next 数组的应用 --- 前缀中最小循环节,最大重复次数
AC代码:
#include<bits/stdc++.h>
#define IO ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0)
#define pb(x) push_back(x)
#define sz(x) (int)(x).size()
#define sc(x) scanf("%d",&x)
#define abs(x) ((x)<0 ? -(x) : x)
#define all(x) x.begin(),x.end()
#define mk(x,y) make_pair(x,y
#define fin freopen("in.txt","r",stdin)
#define fout freopen("out.txt","w",stdout)
using namespace std;
typedef long long ll;
typedef pair<int,int> PII;
const int mod = 1e9+7;
const double PI = 4*atan(1.0);
const int maxm = 1e5+5;
const int maxn = 1e6+5;
const int INF = 0x3f3f3f3f;
int a[maxn],n;
int nex[maxn];
void get_nex() //next 前缀数组
{
int k = nex[0] = -1;
int i = 0;
while(i<n)
{
if(k<0 || a[i] == a[k])
{
i++;
k++;
nex[i] = k;
}
else k = nex[k];
}
}
int main()
{
// fin;
IO;
cin>>n;
for(int i=0;i<n;i++) cin>>a[i];
reverse(a,a+n);
get_nex();
int k,p,ansk = n,ansp = n;
for(int i=0;i<n;i++)
{
p = n-i-nex[n-i],k = i; //理解好题意!
//cout<<k<<" "<<p<<endl;
if(k+p<ansk+ansp){
ansk = k,ansp = p;
}
else if(k+p == ansk+ansp && p<ansp){
ansk = k,ansp = p;
}
}
cout<<ansk<<" "<<ansp<<endl;
return 0;
}