【BZOJ5285】【HNOI2018】寻宝游戏

Description

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Solution

对于结果的每一位iii考虑如果该位在第j" role="presentation" style="position: relative;">jjj次操作上为111，那么bi" role="presentation" style="position: relative;">bibib_i的第jjj位为1" role="presentation" style="position: relative;">111，否则为000。
对于一种操作，如果第j" role="presentation" style="position: relative;">jjj次操作为与操作，那么xxx等于1" role="presentation" style="position: relative;">111，否则为000。
如果最终结果的第i" role="presentation" style="position: relative;">iii位上是111，那么x<bi" role="presentation" style="position: relative;">x<bix<bix，否则x≥bix≥bix\ge b_i。
然后排序后直接处理每个询问求出xx<script type="math/tex" id="MathJax-Element-994">x</script>的范围即可。

Code

/************************************************* Au: Hany01* Date: Apr 20th, 2018* Prob: [BZOJ5285][HNOI2018] 寻宝游戏* Email: hany01@foxmail.com
************************************************/#include<bits/stdc++.h>using namespace std;typedef long long LL;
typedef pair<int, int> PII;
#define File(a) freopen(a".in", "r", stdin), freopen(a".out", "w", stdout)
#define rep(i, j) for (register int i = 0, i##_end_ = (j); i < i##_end_; ++ i)
#define For(i, j, k) for (register int i = (j), i##_end_ = (k); i <= i##_end_; ++ i)
#define Fordown(i, j, k) for (register int i = (j), i##_end_ = (k); i >= i##_end_; -- i)
#define Set(a, b) memset(a, b, sizeof(a))
#define Cpy(a, b) memcpy(a, b, sizeof(a))
#define x first
#define y second
#define pb(a) push_back(a)
#define mp(a, b) make_pair(a, b)
#define ALL(a) (a).begin(), (a).end()
#define SZ(a) ((int)(a).size())
#define INF (0x3f3f3f3f)
#define INF1 (2139062143)
#define Mod (1000000007)
#define debug(...) fprintf(stderr, __VA_ARGS__)
#define y1 wozenmezhemecaiatemplate <typename T> inline bool chkmax(T &a, T b) { return a < b ? a = b, 1 : 0; }
template <typename T> inline bool chkmin(T &a, T b) { return b < a ? a = b, 1 : 0; }inline int read()
{register int _, __; register char c_;for (_ = 0, __ = 1, c_ = getchar(); c_ < '0' || c_ > '9'; c_ = getchar()) if (c_ == '-') __ = -1;for ( ; c_ >= '0' && c_ <= '9'; c_ = getchar()) _ = (_ << 1) + (_ << 3) + (c_ ^ 48);return _ * __;
}const int maxn = 1005, maxm = 5005;int n, m, q, pos[maxm];//For bitset
bitset<maxm> s[maxn], S;bool operator < (bitset<maxn> A, bitset<maxn> B) {For(i, 1, n) if (A[i] ^ B[i])if (A[i]) return 0; else return 1;return 0;
}struct BB
{bitset<maxn> b;int id, val;inline bool operator < (const BB A) const { return b < A.b; }inline void getval() {val = 0;For(i, 1, n) val = (val * 2 % Mod + b[i]) % Mod;}}b[maxm];inline bool getbit()
{register char c;for (c = getchar(); !isdigit(c); c = getchar()) ;return c ^ 48;
}bitset<maxm> getbits()
{static bitset<maxm> bt;Fordown(i, m, 1) bt[i] = getbit();return bt;
}
int main()
{
#ifdef hany01File("bzoj5285");
#endif//InitInputn = read(), m = read(), q = read();For(i, 1, n) s[i] = getbits();Get array b, which stands for if the i_th bit is 0 or 1, the value of x should < or > b[i], and sort them.For(i, 1, m) {For(j, 1, n) b[i].b[n - j + 1] = s[j][i];b[i].id = i, b[i].getval();}sort(b + 1, b + 1 + m);For(i, 1, m) pos[b[i].id] = i;//Answer questionspos[m + 1] = m + 1;For(i, 1, n) b[m + 1].b[i] = 1;b[m + 1].getval(), ++ b[m + 1].val;while (q --){S = getbits();register int Min = m + 1, Max = 0;For(i, 1, m) S[i] ? chkmin(Min, pos[i]) : chkmax(Max, pos[i]);if (Min <= Max) puts("0");else printf("%d\n", (b[Min].val - b[Max].val + Mod) % Mod);}return 0;
}
//红颜未老恩先断，斜倚薰笼坐到明。
//    -- 白居易《后宫词》