HDU 5918 Sequence I (KMP)

题意：给定一个序列，a1, a2, a3 ..., an 还有一个序列是 b1, b2, b3 .. bm，问你有多个 q，使得 aq, aq+p, aq+2p, ... aq+(m-1)p。

析：很容易看出来，就是每隔 p算一个序列有多少个匹配。KMP 裸版。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
//#include <tr1/unordered_map>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
//using namespace std :: tr1;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e6 + 5;
const LL mod = 10000000000007;
const int N = 1e6 + 5;
const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
int a[maxn], b[maxn], nxt[maxn];
int p, ans;

void Kmp(int n, int *a, int m, int *b){
    nxt[0] = -1;
    for(int i = 1, j = -1; i < n; nxt[i++] = j){
        while(~j && a[j+1] != a[i]) j = nxt[j];
        if(a[j+1] == a[i])  ++j;
    }

    for(int k = 0; k < p && k+(p-1)*n <= m; ++k){
        for(int j = -1, i = k; i < m; i += p){
            while(~j && a[j+1] != b[i])  j = nxt[j];
            if(a[j+1] == b[i])  ++j;
            if(j == n-1)  ++ans, j = nxt[j];
        }
    }
}

int main(){
    int T;  cin >> T;
    for(int kase = 1; kase <= T; ++kase){
        scanf("%d %d %d", &n, &m, &p);
        for(int i = 0; i < n; ++i)   scanf("%d", a+i);
        for(int i = 0; i < m; ++i)   scanf("%d", b+i);
        ans = 0;
        Kmp(m, b, n, a);
        printf("Case #%d: %d
", kase, ans);
    }
    return 0;
}