luogu P2000 拯救世界

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嘟嘟嘟


题目有点坑，要你求的多少大阵指的是召唤kkk的大阵数 * lzn的大阵数，不是相加。


看到这个限制条件，显然要用生成函数推一推。
比如第一个条件“金神石的块数必须是6的倍数”，就是\(1 +x ^ 6 + x ^ {12} + \ldots\)，也就是\(\frac{1 - x ^ {6n}}{1 - x ^ 6}\)。当\(x \in (-1, 1)\)时，就变成了\(\frac{1}{1 - x ^ 6}\)。
剩下的同理。
然后把这10个条件都乘起来，一顿化简，答案就是\(\frac{(n + 1) * (n + 2) * (n + 3) *(n + 4)}{24}\)。


本来想快乐的写高精，但是\(n = 1e5\)还非得用fft。
于是就写了一发，不开O2会TLE飞，开了后TLE最后一个点。然后把fft的预处理改成bin哥的写法后就过了。

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define enter puts("") 
#define space putchar(‘ ‘)
#define Mem(a, x) memset(a, x, sizeof(a))
#define In inline
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const db PI = acos(-1);
const int maxn = 4e6 + 5;
inline ll read()
{
    ll ans = 0;
    char ch = getchar(), last = ‘ ‘;
    while(!isdigit(ch)) last = ch, ch = getchar();
    while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + ch - ‘0‘, ch = getchar();
    if(last == ‘-‘) ans = -ans;
    return ans;
}
inline void write(ll x)
{
    if(x < 0) x = -x, putchar(‘-‘);
    if(x >= 10) write(x / 10);
    putchar(x % 10 + ‘0‘);
}

char a1[maxn];
int n, m, a[maxn], b[maxn];

int len = 1;
struct Comp
{
    db x, y;
    In Comp operator + (const Comp& oth)const
    {
        return (Comp){x + oth.x, y + oth.y};
    }
    In Comp operator - (const Comp& oth)const
    {
        return (Comp){x - oth.x, y - oth.y};
    }
    In Comp operator * (const Comp& oth)const
    {
        return (Comp){x * oth.x - y * oth.y, x * oth.y + y * oth.x};
    }
    friend In void swap(Comp& a, Comp& b)
    {
        swap(a.x, b.x); swap(a.y, b.y);
    }
}c[maxn], d[maxn], omg[maxn], inv[maxn];
int r[maxn];
In void init()
{
    omg[0] = inv[0] = (Comp){1, 0};
    omg[1] = inv[len - 1] = (Comp){cos(2 * PI / len), sin(2 * PI / len)};
    for(int i = 2; i < len; ++i) omg[i] = inv[len - i] = omg[i - 1] * omg[1];
}
In void fft(Comp* a, Comp* omg)
{
    for(int i = 0; i < len; ++i) if(i < r[i]) swap(a[i], a[r[i]]);
    for(int l = 2; l <= len; l <<= 1)
    {
        int q = l >> 1;
        for(Comp* p = a; p != a + len; p += l)
            for(int i = 0; i < q; ++i)
            {
                Comp tp = omg[len / l * i] * p[i + q];
                p[i + q] = p[i] - tp, p[i] = p[i] + tp;
            }
    }
}

In void mul(int* a, int* b)
{
    int tot = max(n, m); len = 1;
    while(len < (tot << 1)) len <<= 1;
    int lim = 0;
    while((1 << lim) < len) ++lim;
    for(int i = 0; i < len; ++i) r[i] = (r[i >> 1] >> 1) | ((i & 1) << (lim - 1));
    for(int i = 0; i < len; ++i) c[i] = d[i] = (Comp){0, 0};
    for(int i = 0; i < n; ++i) c[i] = (Comp){a[i], 0};
    for(int i = 0; i < m; ++i) d[i] = (Comp){b[i], 0};
    init();
    fft(c, omg), fft(d, omg);
    for(int i = 0; i < len; ++i) c[i] = c[i] * d[i];
    fft(c, inv);
    for(int i = 0; i <= len; ++i) a[i] = 0;
    for(int i = 0; i < len; ++i)
    {
        a[i] += (int)(c[i].x / len + 0.5);  
        if(a[i] >= 10) a[i + 1] += a[i] / 10, a[i] %= 10;
    } 
    n = len;
    while(n - 1 && !a[n - 1]) --n;
//  for(int i = n - 1; i >= 0; --i) printf("%d", a[i]); enter;
}
In void add(int* a, int x, int& n)
{
    a[0] += x;
    for(int i = 0; i < n; ++i)
        if(a[i] >= 10) a[i + 1] += a[i] / 10, a[i] %= 10;
        else break;
    ++n;
    while(n - 1 && !a[n - 1]) --n;
//  for(int i = n - 1; i >= 0; --i) printf("%d", a[i]); enter;
}
In void div(int* a, int x)
{
    static int ret[maxn];
    reverse(a, a + n);
    int tp = 0, cnt = 0;
    for(int i = 0; i < n; ++i)
    {
        tp = tp * 10 + a[i];
        ret[++cnt] = tp / x;
        tp %= x;
    }
    int sta = 1;
    while(sta < cnt && !ret[sta]) ++sta;
    for(int i = sta; i <= cnt; ++i) write(ret[i]); enter;
}

int main()
{
//  freopen("random.in", "r", stdin);
//  freopen("ac.out", "w", stdout);
    scanf("%s", a1);
    m = n = strlen(a1);
    for(int i = 0; i < n; ++i) b[i] = a[i] = a1[n - i - 1] - ‘0‘;
    add(a, 1, n); add(b, 1, m);
    for(int i = 2; i <= 4; ++i)
    {
        add(b, 1, m);
        mul(a, b);
    }
    div(a, 24);
    return 0;
}

luogu P2000 拯救世界

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原文地址：https://www.cnblogs.com/mrclr/p/10325758.html