题意：给定n个数，这n个数在m的范围之内。v[i]=(v[i-1]+v[i-2]+v[i-3])%m+1;   然后求一个最短连续子序列使得序列内包括1-k个数。

思路：枚举。枚举以v[i]结尾的最短序列。用ct维护已经找到的k个数中的个数，queue跳转区间。

code：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <sstream>
#include <string>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <bitset>using namespace std;typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;const int INF=0x3fffffff;
const int inf=-INF;
const int N=1000000;
const int M=2005;
const int mod=1000000007;
const double pi=acos(-1.0);#define cls(x,c) memset(x,c,sizeof(x))
#define cpy(x,a) memcpy(x,a,sizeof(a))
#define fr(i,s,n) for (int i=s;i<=n;i++)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lrt  rt<<1
#define rrt  rt<<1|1
#define middle int m=(r+l)>>1
#define lowbit(x) (x&-x)
#define pii pair<int,int>
#define mk make_pair
#define IN freopen("in.txt","r",stdin);
#define OUT freopen("out.txt","w",stdout);int n,m,k;
int v[N],vt[N];
int main()
{int T,ca=1,ct,ans;scanf("%d",&T);while (T--){scanf("%d %d %d",&n,&m,&k);cls(vt,0);ans=INF;ct=0;v[1]=1,v[2]=2,v[3]=3;fr(i,4,n) v[i]=(v[i-1]+v[i-2]+v[i-3])%m+1;queue<int>q;fr(i,1,n){if (v[i]>=1&&v[i]<=k){q.push(i);if (vt[v[i]]++==0) ct++;while (ct==k){int p=q.front();q.pop();ans=min(ans,i-p+1);if (--vt[v[p]]==0) ct--;}}}printf("Case %d: ",ca++);if (ans==INF) puts("sequence nai");else printf("%d\n",ans);}
}