HDU - 4821 String (长春赛区I题)

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题意：求有多少个子串满足条件：

1. 长度为M*L

2. 每个长度为L的小子串不能都完全相同的情况

思路：当时没做出来，HASH处理，看了个思路是：

hash方法设一个种子base 打表出nbase[i]表示base的i次方

从S最后一个字符开始打表 hash[i]=hash[i+1]*base+str[i]-'a'+1 即将i位以后的串hash成一个unsigned long long

然后每个len长度的小串的hash值即为 hash[i]-hash[i+len]*nbase[len]

最后使用map搞一下就可以AC

临摹了一个

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <map>typedef unsigned long long ull;using namespace std;const int MAXN = 100010;const ull base = 31;ull nbase[MAXN],hash[MAXN];int n,len,ans,slen;char str[MAXN];map<ull, int> mp;int main() {ull tmp;nbase[0] = 1;for (int i = 1; i <= MAXN; i++)nbase[i] = nbase[i-1] * base;while (scanf("%d%d", &n, &len) != EOF) {scanf("%s", str);slen = strlen(str);hash[slen] = 0;for (int i = slen-1; i >= 0; i--) hash[i] = hash[i+1]*base+str[i]-'a'+1;ans = 0;for (int i = 0; i < len && i+n*len <= slen; i++) {mp.clear();for (int j = i; j < i+n*len; j += len) {tmp = hash[j] - hash[j+len]*nbase[len];mp[tmp]++;}if (mp.size() == n)ans++;for (int j = i+n*len; j+len <= slen; j += len) {tmp = hash[j-n*len] - hash[j-(n-1)*len]*nbase[len];mp[tmp]--;if (mp[tmp] == 0)mp.erase(tmp);tmp = hash[j] - hash[j+len]*nbase[len];mp[tmp]++;if (mp.size() == n)ans++;}}printf("%d\n", ans);}return 0;}

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