当前位置： 移动技术网 > IT编程>开发语言>C/C++ > loj#6436. 「PKUSC2018」神仙的游戏(生成函数)

loj#6436. 「PKUSC2018」神仙的游戏(生成函数)

2019年03月14日  | 移动技术网IT编程  | 我要评论

电源功率计算器,清明节是法定假日吗,小莫騒麦歌曲

题意

sol

生成函数题都好神仙啊qwq

我们考虑枚举一个长度\(len\)。有一个结论是如果我们按\(n - len\)的余数分类，若同一组内的全为\(0\)或全为\(1\)(?不算)，那么存在一个长度为\(len\)的border。

有了这个结论后我们考虑这样一种做法：把序列看成两个串\(a, b\)，若\(a_i = 0, b_j = 1\)，那么对于所有的\(k | (|i - j|)\), \(n-k\)都不会成为答案。

考虑怎么快速算不合法的\((i, j)\)。对于多项式乘法得到的多项式的第\(k\)项，实际上是由所有的\(a_i * a_j(i+j=k)\)相乘得到的。我们把序列\(b\)翻转一下，这时候得到的第\(k\)项实际上就是由\(a_i * a_{n - j}\)得到的。

然后枚举一个数看一下他的倍数是否\(>0\)就行了

#include<bits/stdc++.h> 
#define pair pair<int, int>
#define mp(x, y) make_pair(x, y)
#define fi first
#define se second
#define ll long long 
#define ull unsigned long long 
#define fin(x) {freopen(#x".in","r",stdin);}
#define fout(x) {freopen(#x".out","w",stdout);}
using namespace std;
const int maxn = 8e6 + 10, inf = 1e9 + 1;
const double eps = 1e-9, pi = acos(-1);
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, a[maxn], b[maxn], vis[maxn];
char s[maxn];
namespace poly {
    int rev[maxn], gpow[maxn], a[maxn], b[maxn], c[maxn], lim, inv2;
    const int g = 3, mod = 1004535809, mod2 = 1004535808;
    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;}
    int fp(int a, int p, int p = mod) {
        int base = 1;
        for(; p > 0; p >>= 1, a = 1ll * a * a % p) if(p & 1) base = 1ll * base *  a % p;
        return base;
    }
    int inv(int x) {
        return fp(x, mod - 2);
    }
    int getlen(int x) {
        int lim = 1;
        while(lim <= x) lim <<= 1;
        return lim;
    }
    int getorigin(int x) {//¼æëãô­¸ù 
        static int q[maxn]; int tot = 0, tp = x - 1;
        for(int i = 2; i * i <= tp; i++) if(!(tp % i)) {q[++tot] = i;while(!(tp % i)) tp /= i;}
        if(tp > 1) q[++tot] = tp;
        for(int i = 2, j; i <= x - 1; i++) {
            for(j = 1; j <= tot; j++) if(fp(i, (x - 1) / q[j], x) == 1) break;
            if(j == tot + 1) return i;
        }
        return -1;
    }
    void init(/*int p,*/ int lim) {
        inv2 = fp(2, mod - 2);
        for(int i = 1; i <= lim; i++) gpow[i] = fp(g, (mod - 1) / i);
    }
    void ntt(int *a, int lim, int opt) {
        int len = 0; for(int n = 1; n < lim; n <<= 1) ++len; 
        for(int i = 1; i <= lim; i++) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (len - 1));
        for(int i = 0; i <= lim; i++) if(i < rev[i]) swap(a[i], a[rev[i]]);
        for(int mid = 1; mid < lim; mid <<= 1) {
            int wn = gpow[mid << 1];
            for(int i = 0; i < lim; i += (mid << 1)) {
                for(int j = 0, w = 1; j < mid; j++, w = mul(w, wn)) {
                    int x = a[i + j], y = mul(w, a[i + j + mid]);
                    a[i + j] = add(x, y), a[i + j + mid] = add(x, -y);
                }
            }
        }
        if(opt == -1) {
            reverse(a + 1, a + lim);
            int inv = fp(lim, mod - 2);
            for(int i = 0; i <= lim; i++) mul2(a[i], inv);
        }
    }
    void mul(int *a, int *b, int n, int m) {
        memset(a, 0, sizeof(a)); memset(b, 0, sizeof(b));
        int lim = 1, len = 0; 
        while(lim <= n + m) len++, lim <<= 1;
        for(int i = 0; i <= n; i++) a[i] = a[i]; 
        for(int i = 0; i <= m; i++) b[i] = b[i];
        ntt(a, lim, 1); ntt(b, lim, 1);
        for(int i = 0; i <= lim; i++) b[i] = mul(b[i], a[i]);
        ntt(b, lim, -1);
        for(int i = 0; i <= n + m; i++) b[i] = b[i];
        memset(a, 0, sizeof(a)); memset(b, 0, sizeof(b));
    }
    void inv(int *a, int *b, int len) {//b1 = 2b - a1 * b^2 
        if(len == 1) {b[0] = fp(a[0], mod - 2); return ;}
        inv(a, b, len >> 1);
        for(int i = 0; i < len; i++) a[i] = a[i], b[i] = b[i];
        ntt(a, len << 1, 1); ntt(b, len << 1, 1);
        for(int i = 0; i < (len << 1); i++) mul2(a[i], mul(b[i], b[i]));
        ntt(a, len << 1, -1);
        for(int i = 0; i < len; i++) add2(b[i], add(b[i], -a[i]));
        for(int i = 0; i < (len << 1); i++) a[i] = b[i] = 0;
    }
    void dao(int *a, int *b, int len) {
        for(int i = 1; i < len; i++) b[i - 1] = mul(i, a[i]); b[len - 1] = 0;
    }
    void ji(int *a, int *b, int len) {
        for(int i = 1; i < len; i++) b[i] = mul(a[i - 1], fp(i, mod - 2)); b[0] = 0;
    }
    void ln(int *a, int *b, int len) {//g(a) = \frac{a}{a'} qiudao zhihou jifen 
        static int a[maxn], b[maxn];
        dao(a, a, len); 
        inv(a, b, len);
        ntt(a, len << 1, 1); ntt(b, len << 1, 1);
        for(int i = 0; i < (len << 1); i++) b[i] = mul(a[i], b[i]);
        ntt(b, len << 1, -1); 
        ji(b, b, len << 1);
        memset(a, 0, sizeof(a)); memset(b, 0, sizeof(b));
    }
    void exp(int *a, int *b, int len) {//f(x) = f_0 (1 - lnf_0 + a) but code ..why....
        if(len == 1) return (void) (b[0] = 1);
        exp(a, b, len >> 1); ln(b, c, len);
        c[0] = add(a[0] + 1, -c[0]);
        for(int i = 1; i < len; i++) c[i] = add(a[i], -c[i]);
        ntt(c, len << 1, 1); ntt(b, len << 1, 1);
        for(int i = 0; i < (len << 1); i++) mul2(b[i], c[i]);
        ntt(b, len << 1, -1);
        for(int i = len; i < (len << 1); i++) c[i] = b[i] = 0;
    }
    void sqrt(int *a, int *b, int len) {
        static int b[maxn];
        ln(a, b, len);
        for(int i = 0; i < len; i++) b[i] = mul(b[i], inv2);
        exp(b, b, len); 
    }
};
using namespace poly; 
bool flag[maxn];
signed main() {
    scanf("%s", s);
    n = strlen(s); int lim = getlen(n); init(4 * lim);
    for(int i = 0; i < n; i++) a[i] = (s[i] == '0'), b[i] = (s[n - i - 1] == '1');
    mul(a, b, lim, lim);
    ll ans = 1ll * n * n;
    for(int i = 1; i <= n; i++) {
        ans ^= 1ll * (n - i) * (n - i);
        for(int j = i; j < n; j += i)
            if(b[n - j - 1] || b[n + j - 1]) 
                {ans ^= 1ll * (n - i) * (n - i); break;}
    }   
    cout << ans;
    return 0;
}

您可能感兴趣的文章:

如对本文有疑问，请在下面进行留言讨论，广大热心网友会与你互动！！ 点击进行留言回复

相关文章:

验证码：
最近更新的文章
大家感兴趣的文章
移动技术网