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题意:
给你a,b两个序列
和一个p ,求有多少个 q恰好满足 b1,b2,b3....bm 就是 a[q],a[q+p],a[q+2p]......a[q+(m-1)p];
题解:
将a序列,每隔p位置分成一组,这样最多有p组,个数和是n
将每组和b序列跑KMP计算答案
#include<bits/stdc++.h> using namespace std; #pragma comment(linker, "/STACK:102400000,102400000") #define ls i<<1 #define rs ls | 1 #define mid ((ll+rr)>>1) #define pii pair<int,int> #define MP make_pair typedef long long LL; const long long INF = 1e18; const double Pi = acos(-1.0); const int N = 1e6+10, M = 1e6, mod = 1e9+7, inf = 2e9; int T,n,m,p,s[N],t[N],ans = 0; vector<int > P[N]; int nex[N]; int main() { int cas = 1; scanf("%d",&T); while(T--) { scanf("%d%d%d",&n,&m,&p); for(int i = 1; i <= n; ++i) scanf("%d",&t[i]); for(int i = 1; i <= m; ++i) scanf("%d",&s[i]); for(int i = 0; i < p; ++i) P[i].clear(); memset(nex,0,sizeof(nex)); ans = 0; int k = 0; for(int i=2; i<=m; i++) { while(k>0&&s[k+1]!=s[i]) k = nex[k]; if(s[k+1]==s[i])k++; nex[i] = k; } if(p == 1) { k = 0; for(int i=1; i<=n; i++) { while(k>0&&s[k+1]!=t[i]) k = nex[k]; if(s[k+1]==t[i]) k++; if(k==m) { k = nex[k]; ans++; } } printf("Case #%d: ",cas++); printf("%d\n",ans); } else { for(int i = 1; i <= n; ++i) P[i % p].push_back(t[i]); for(int i = 0; i < p; ++i) { k = 0; for(int j = 0; j < P[i].size(); ++j) { while(k>0&&s[k+1]!=P[i][j]) k = nex[k]; if(s[k+1]==P[i][j]) k++; if(k==m) { k = nex[k]; ans++; } } } printf("Case #%d: ",cas++); printf("%d\n",ans); } } }
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原文地址:http://www.cnblogs.com/zxhl/p/5931233.html