UVa 12174:shuffle (sliding window)

Title Link: Https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=844&page=show_ problem&problem=3326

Test instructions: The music player you are using has a so-called shuffle function that randomly disrupts the order of the songs to play. Assuming that there is a song of S, then the first song will be randomly sorted, all played and then re-randomly sorted, continue to play, and so on. Note that the S song will not be reordered until it finishes playing. In this way, every s song in the playback record is an arrangement of 1~s. Given a playback record of n (1≤s,n≤100000) (1 \le s, n \le 100000) length (not necessarily from the very beginning) of Xi (1≤xi≤s) x_i (1 \le x_i \le s), your task is to count the number of times the next random order has occurred Number of possibilities. For example, s=4, the playback record is 3,4,4,1,3,2,1,2,3,4, it is not difficult to find that there is only one possibility: the first two is a paragraph of the last two songs, followed by two complete segments, so the answer is 1; When s=3, there are two possible ways to play records 1,2,1: The first is a segment, The last two is another paragraph; the first two is a paragraph and the final one is another. The answer is 2. (This paragraph is excerpted from the "Algorithmic Competition Primer (2nd edition)")

Analysis:
The idea of sliding the window, maintaining the length of the window to see if it is legal, the last enumeration may be the answer to judge.

Code:

This # include <fstream> #include <iostream> #include <cstring> #include <algorithm> #include < stack> #include <sstream> #include <string> #include <map> #include <cmath> #include < queue> #include <vector> #include <set> #include <string> #include <vector> using namespace St

D

const int MAXN = 100000 + 5;
int T, ans, CNT, S, N;
int A[MAXN * 3], NUM[MAXN], X[MAXN * 2];

BOOL Flag;
    int main () {scanf ("%d", &t); for (int C = 0; C < T;
        ++c) {ans = 0;
        Memset (A,-1, sizeof (a));
        memset (x, 0, sizeof (x));
        memset (num, 0, sizeof (num));
        scanf ("%d%d", &s, &n);
        CNT = s;
        for (int i = s; i < S + N; ++i) scanf ("%d", &a[i]);  for (int i = s; i < S + S + N; ++i) {if (a[i-s]! =-1) {--num[a[i-
                S]]; if (num[a[i-s]] = = 0)--cnT
            } else--cnt;
                if (a[i]! =-1) {++num[a[i]];
            if (num[a[i]] = = 1) ++cnt;
            } else ++cnt;
        if (cnt = = s) x[i-s] = 1;
            } for (int i = 0; i < s; ++i) {flag = true;
                    for (int j = i; J < S + N; j + = s) if (!x[j]) {flag = false;
                Break
        } if (flag) ++ans;
    } printf ("%d\n", ans);
} return 0;
 }