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IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#485726#6299. Binary StringBalintRTL 0ms0kbC++203.3kb2024-07-21 02:49:112024-07-21 02:49:13

Judging History

你现在查看的是最新测评结果

  • [2024-07-21 02:49:13]
  • 评测
  • 测评结果:TL
  • 用时:0ms
  • 内存:0kb
  • [2024-07-21 02:49:11]
  • 提交

answer

#include <bits/stdc++.h>
using namespace std;

typedef unsigned uint;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
typedef vector<pii> vpii;
typedef complex<double> cpx;
template <typename T> using minPq = priority_queue<T, vector<T>, greater<T>>;
#define ms(a, x) memset(a, x, sizeof(a))
#define pb push_back
#define fs first
#define sn second
#define ALL(v) begin(v), end(v)
#define SZ(v) ((int) (v).size())
#define lbv(v, x) (lower_bound(ALL(v), x) - (v).begin())
#define ubv(v, x) (upper_bound(ALL(v), x) - (v).begin())
template <typename T> inline void UNIQUE(vector<T> &v){sort(ALL(v)); v.resize(unique(ALL(v)) - v.begin());}
const int INF = 0x3f3f3f3f;
const ll LLINF = 0x3f3f3f3f3f3f3f3f;
const double PI = acos(-1);
#define FR(i, n) for(int i = 0; i < (n); i++)
#define FOR(i, a, b) for(int i = (a); i < (b); i++)
#define FORR(i, a, b) for(int i = (a); i >= (b); i--)
#define dbg(x) {cerr << #x << ' ' << x << endl;}
#define dbgArr(arr, n) {cerr << #arr; FR(_i, n) cerr << ' ' << (arr)[_i]; cerr << endl;}
template <typename T, typename U>
ostream& operator<<(ostream &os, pair<T, U> p){return os << "(" << p.fs << ", " << p.sn << ")";}

const int MN = 1e6 + 5;
int n;
string str;
int arr[MN], resArr[MN], qu[MN*2], ql, qr;

int getItsToEq(){
    ql = qr = 0;
    FR(i, n){
        qu[qr++] = i;
        qr -= min(arr[i], qr);
    }
    FR(i, qr) qu[i] -= n;

    int res = 0;
    FR(i, n){
        qu[qr++] = i;
        if(arr[i]){
            qr -= arr[i]-1;
            if(qr <= ql) for(;;);
            res = max(res, i-qu[qr-1]);
            qr--;
        }
        if(qr > ql) for(;;);
        assert(ql <= qr);
        while(ql < qr && qu[ql] < (i-n+1)) ql++;
        resArr[i] = !(ql < qr && qu[ql] == i-n+1);
    }

    // dbg(res);
    return res;
}

const int MOD = 1e9 + 7;
const int X = 590294819;
int pws[MN], hashes[MN];

int getNumRepeats(){
    // dbgArr(resArr, n);
    FR(i, n) hashes[i+1] = ((ll) hashes[i]*X + resArr[i]) % MOD;
    FOR(k, 1, n) if(n % k == 0){
        ll h1 = hashes[k];
        for(int l = k; l < n; l += k){
            ll h2 = hashes[l+k] - (ll) hashes[l]*pws[k];
            if((h1 - h2) % MOD) goto loopEnd;
        }
        return n/k;
        loopEnd:;
    }
    return 1;
}

int solve(){
    cin >> str;
    int num1 = count(ALL(str), '1');
    if(num1 == 0 || num1 == SZ(str)) return 1;
    if(num1 > SZ(str)/2) for(char &ch : str) ch ^= 1;
    else reverse(ALL(str));
    rotate(str.begin(), find(ALL(str), '0'), str.end());
    rotate(str.begin(), find(ALL(str), '1'), str.end());
    assert(str.back() == '0');

    char lst = -1;
    n = 0;
    FR(i, SZ(str)){
        if(str[i] == '1'){
            n += lst != '1';
            arr[n-1]++;
        }
        else {
            n += lst == '0';
        }
        lst = str[i];
    }

    // dbgArr(arr, n);

    int init = getItsToEq();
    int numReps = getNumRepeats();
    // dbg(numReps);
    return init + SZ(str)/numReps;
}

int main(){
    cin.sync_with_stdio(0); cin.tie(0);
    pws[0] = 1;
    FR(i, MN-1) pws[i+1] = (ll) pws[i] * X % MOD;

    int t; cin >> t;
    while(t--){
        cout << solve() << '\n';
        fill_n(arr, n, 0);
    }
}

Details

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Test #1:

score: 0
Time Limit Exceeded

input:

3
1
001001
0001111

output:


result: