uva 11536——Smallest Sub-Array

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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);    }}

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