AOJ1370: Hidden Anagrams(hash)
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2022-10-07 07:51:53
题意 "题目链接" Sol 直接对出现的次数hash即可,复杂度$O(26n^2)$ 一开始没判长度条件疯狂wa cpp include // define int long long define LL long long define ull unsigned long long using n ......
题意
sol
直接对出现的次数hash即可,复杂度\(o(26n^2)\)
一开始没判长度条件疯狂wa
#include<bits/stdc++.h> //#define int long long #define ll long long #define ull unsigned long long using namespace std; const int maxn = 4001, mod = 1e9 + 7, inf = 1e9 + 10; const double eps = 1e-9; template <typename a, typename b> inline bool chmin(a &a, b b){if(a > b) {a = b; return 1;} return 0;} template <typename a, typename b> inline bool chmax(a &a, b b){if(a < b) {a = b; return 1;} return 0;} template <typename a, typename b> inline ll add(a x, b y) {if(x + y < 0) return x + y + mod; return x + y >= mod ? x + y - mod : x + y;} template <typename a, typename b> inline void add2(a &x, b y) {if(x + y < 0) x = x + y + mod; else x = (x + y >= mod ? x + y - mod : x + y);} template <typename a, typename b> inline ll mul(a x, b y) {return 1ll * x * y % mod;} template <typename a, typename b> inline void mul2(a &x, b y) {x = (1ll * x * y % mod + mod) % mod;} template <typename a> inline void debug(a a){cout << a << '\n';} template <typename a> inline ll sqr(a x){return 1ll * x * x;} inline int read() { char c = getchar(); int x = 0, f = 1; while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();} while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar(); return x * f; } int n, m; char s1[maxn], s2[maxn]; map<ull, bool> mp; ull base = 29; int num[27]; ull get() { ull now = 0; for(int i = 0; i < 26; i++) now = now * base + num[i]; return now; } signed main() { scanf("%s", s1 + 1); scanf("%s", s2 + 1); n = strlen(s1 + 1); m = strlen(s2 + 1); for(int len = min(n, m); len >= 1; len--) { mp.clear(); memset(num, 0, sizeof(num)); for(int i = 1; i <= n; i++) { num[s1[i] - 'a']++; if(i > len) num[s1[i - len] - 'a']--; if(i >= len) mp[get()] = 1; } memset(num, 0, sizeof(num)); for(int i = 1; i <= m; i++) { num[s2[i] - 'a']++; if(i > len) num[s2[i - len] - 'a']--; if(i >= len && mp[get()]) { cout << len << '\n'; return 0; } } } puts("0"); return 0; }