字符串Hash

给出N个只有小写字母的字符串，求其中的不同的字符串的个数

#include<iostream>
#include<string>
#include<algorithm>
#include<vector>
using namespace std;

const int MOD = 1000000007;
const int P = 10000019;
vector<int> ans;
long long hashFunc(string str) {
	long long H = 0;
	for (int i = 0; i < str.length(); i++) {
		H = (H * P + str[i] - 'a') % MOD;
	}
	return H;
}
int main() {
	string str;
	while (getline(cin, str), str != "#") {
		long long id = hashFunc(str);
		ans.push_back(id);
		}
	sort(ans.begin(), ans.end());
	int count = 0;
	for(int i = 0; i < ans.size(); i++){
		if (i == 0 || ans[i] != ans[i - 1]) {
			count++;
		}
	}
	cout<<count<<endl;
	return 0;
}

输入两个长度不超过1000的字符串，求他们的最长公共子串的长度。

计算出各个子串的Hash值，与长度保存成pair对。比较

#include<iostream>
#include<cstdio>
#include<string>
#include<vector>
#include<map>
#include<algorithm>
using namespace std;
typedef long long LL;
const LL MOD = 100000007;
const LL P = 10000019;
const LL MAXN = 1010;
LL powP[MAXN], H1[MAXN] = { 0 }, H2[MAXN] = { 0 };
vector<pair<int, int> > pr1, pr2;

void init(int len) {
	powP[0] = 1;
	for (int i = 1; i <= len; i++) {
		powP[i] = (powP[i - 1] * P) % MOD; //每一步取模，防止溢出
	}
}

void calH(LL H[], string& str) {
	H[0] = str[0];
	for (int i = 1; i < str.length(); i++) {
		H[i] = (H[i - 1] * P + str[i]) % MOD; //直接用str[i]

	}
}

int calSingleSubH(LL H[], int i, int j) {
	if (i == 0) return H[j];
	return((H[j] - H[i - 1] * powP[j - i + 1]) % MOD + MOD) % MOD;
}

void calSubH(LL H[], int len, vector<pair<int, int>>& pr) {
	for (int i = 0; i < len; i++) {
		for (int j = i; j < len; j++) {
			int hashValue = calSingleSubH(H, i, j);
			pr.push_back(make_pair(hashValue, j - i + 1));
		}
	}
}

int getMax() {
	int ans = 0;
	for (int i = 0; i < pr1.size(); i++) {
		for (int j = 0; j < pr2.size(); j++) {
			if (pr1[i].first == pr2[j].first) {//hash值相等
				ans = max(ans, pr1[i].second);
			}
		}
	}
	return ans;
}
int main() {
	/*
	ILoveYou
	YouDontLoveMe
	*/
	string str1, str2;
	getline(cin, str1);
	getline(cin, str2);
	init(max(str1.length(), str2.length()));
	calH(H1, str1);
	calH(H2, str2);
	calSubH(H1, str1.length(), pr1);
	calSubH(H2, str2.length(), pr2);
	printf("ans = %d\n", getMax());
	return 0;
}