題目傳送門

題意：

• 給你n個數組成的序列，讓你求從k+1位置開始，序列以p個數為週期。最小的k+p是多少。

做法：

• kmp 演算法的一個應用，即最小迴圈節是i-next[i] (吃了沒文化的虧_(:з」∠)_ 

AC程式碼：

#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 字首陣列
{
    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; //理解好題意！
        //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;
}