每日一题 - 240116 - P3370 【模板】字符串哈希

• P3370 【模板】字符串哈希

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• TAG - $算法-【STL、进制哈希】$

• 时间复杂度 - $O(\ast )$

• // – STL – //

• set

//
#include 
using namespace std;
// #define int long long

void solve() {
    set<string> st;

    int n;
    scanf("%d", &n);
    while (n--) {
        string ai;
        cin >> ai;
        st.insert(ai);
    }
    printf("%d\n", st.size());
}

signed main() {
    int t = 1;
    // scanf("%d", &t);
    while (t--) solve();

    return 0;
}

• map

// map
#include 
using namespace std;
// #define int long long

void solve() {
    map<string, int> mp;

    int n;
    scanf("%d", &n);
    while (n--) {
        string ai;
        cin >> ai;
        mp[ai]++;
    }
    printf("%d\n", mp.size());
}

signed main() {
    int t = 1;
    // scanf("%d", &t);
    while (t--) solve();

    return 0;
}

• // – 进制哈希 – //
• 自然溢出

// 自然溢出
#include 
using namespace std;
#define int unsigned long long 

const int base = 131;   // (int)'z' = 122 < 131
const int N = 10005;
int mp[N];

int f_hash(string s) {
    int ans = 0;
    for (int i = 0; i < s.size(); i++) {
        ans = ans * base + s[i];
    }
    return ans;
}

void solve() {
    int n;
    scanf("%llu", &n);
    for (int i = 1; i <= n; i++) {
        string s;
        cin >> s;
        mp[i] = f_hash(s);
    }

    sort(mp + 1, mp + n + 1);

    int ans = 1;
    for (int i = 2; i <= n; i++) ans += (mp[i] != mp[i - 1]);
    printf("%llu\n", ans);
}

signed main() {
    int t = 1;
    // scanf("%llu", &t);
    while (t--) solve();

    return 0;
}

• 单 mod

// 单 mod
#include 
using namespace std;
#define int unsigned long long

// const 
const int base = 131, mod = 1e9 + 7;
const int N = 10005;
int mp[N];

int f_hash(string s) {
    int ans = 0;
    for (int i = 0; i < s.size(); i++) {
        ans = (ans * base + s[i]) % mod;
    }
    return ans;
}

void solve() {
    int n;
    scanf("%llu", &n);
    for (int i = 1; i <= n; i++) {
        string s;
        cin >> s;
        mp[i] = f_hash(s);
    }

    sort(mp + 1, mp + n + 1);
    
    int ans = 1;
    for (int i = 2; i <= n; i++) ans += (mp[i] != mp[i - 1]);
    printf("%llu\n", ans);
}

signed main() {
    int t = 1;
    // scanf("%llu", &t);
    while (t--) solve();

    return 0;
}

• 双 mod

// 双 mod
#include 
using namespace std;
#define int unsigned long long 

const int base = 131, mod1 = 1e9 + 7, mod2 = 1e9 + 9;
const int N = 10005;
struct A {
    int h1, h2;
} mp[N];

A f_hash(string s) {
    int ans1 = 0, ans2 = 0;
    for (int i = 0; i < s.size(); i++) {
        ans1 = (ans1 * base + s[i]) % mod1;
        ans2 = (ans2 * base + s[i]) % mod2;
    }
    return (A){ans1, ans2};
}

bool cmp(const A& x, const A& y) {
    return x.h1 == y.h1 ? x.h2 < y.h2 : x.h1 < y.h1;
}

void solve() {
    int n;
    scanf("%llu", &n);
    for (int i = 1; i <= n; i++) {
        string s;
        cin >> s;
        mp[i] = f_hash(s);
    }

    sort(mp + 1, mp + n + 1, cmp);

    int ans = 1;
    for (int i = 2; i <= n; i++) ans += !(mp[i].h1 == mp[i - 1].h1 && mp[i].h2 == mp[i - 1].h2);
    printf("%llu\n", ans);
}

signed main() {
    int t = 1;
    // scanf("%llu", &t);
    while (t--) solve();

    return 0;
}

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实现细节

• base - 131 13331
• mod - 1e9 + 7 1e9 + 9 19 个 1
• ios::sync_with_stdio(false); cin.tie(nullptr);

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参考示意图

• `

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参考链接

• 【算法学习】字符串哈希（Hash）
• //
• 题解 P3370 【【模板】字符串哈希】
• 哈希
• //
• 质数发生器和校验器
• long long /unsigned long long 输入
• //
• 初学算法是否有必要把用cin cout 的习惯改成scanf printf呢?
• Fast input/output for Competitive Programming
• // – 拓展 – //
• 生日悖论
• 生日攻击

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作者 | 乐意奥AI