BZOJ4827: [Hnoi2017]礼物(FFT 二次函数)

题意

题目链接

sol

越来越菜了。。裸的fft写了1h。。

思路比较简单，直接把

\(\sum (x_i - y_i + c)^2\)

拆开

发现能提出一坨东西，然后与c有关的部分是关于c的二次函数可以直接算最优取值

剩下的要求的就是\(max (\sum x_i y_i)\)

画画图就知道把y序列倒过来就是个裸的fft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 fin(x) {freopen(#x".in","r",stdin);}
#define fout(x) {freopen(#x".out","w",stdout);}
using namespace std;
const int maxn = 1e6 + 10, mod = 1e9 + 7, inf = 1e9 + 10;
const double eps = 1e-9, pi = acos(-1);
template <typename a, typename b> inline bool chmin(a &a, b b){if(a > b) {a = b; return 1;} return 0;}
template <typename a, typename b> inline bool chmax(a &a, b b){if(a < b) {a = b; return 1;} return 0;}
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;}
template <typename a> inline void debug(a a){cout << a << '\n';}
template <typename a> inline ll sqr(a x){return 1ll * x * x;}
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;
int a[maxn], b[maxn], c[maxn], t[maxn], nx, ny, rev[maxn];
double sx, sy;
struct com {
    double x, y;
}a[maxn], b[maxn], c[maxn];
com operator * (const com a, const com b) {
    return (com) {a.x * b.x - a.y * b.y, a.x * b.y + a.y * b.x};
}
com operator + (const com a, const com b) {
    return (com) {a.x + b.x, a.y + b.y};
}
com operator - (const com a, const com b) {
    return (com) {a.x - b.x, a.y - b.y};
}
void fft(com *a, int lim, int opt) {
    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) {
        com wn = (com) {cos(pi / mid), opt * sin(pi / mid)};
        for(int i = 0, r = mid << 1; i <= lim; i += r) {//tag
            com w = (com) {1, 0};
            for(int j = 0; j < mid; j++, w = w *  wn) {
                com x = a[i + j], y = w * a[i + j + mid];
                a[i + j] = x + y;
                a[i + j + mid] = x - y;
            }
        }
    }
    if(opt == -1) {
        for(int i = 0; i <= lim; i++) a[i].x /= lim;
    }
}
ll check(int c) {
    ny = 0;
    memcpy(b, t, sizeof(t));
    for(int i = 0; i < n; i++) b[i] += c, b[i + n] = b[i], ny += b[i] * b[i];
    int m = 2 * n - 1;
    reverse(b, b + m + 1);
    memset(a, 0, sizeof(a));
    memset(b, 0, sizeof(b));
    memset(c, 0, sizeof(c));
    n--;
    for(int i = 0; i <= n; i++) a[i].x = a[i];
    for(int i = 0; i <= m; i++) b[i].x = b[i];
    int lim = 1, len = 0;
    while(lim <= n + m) lim <<= 1, len++;
    for(int i = 1; i <= lim; i++) rev[i] = (rev[i >> 1] >> 1) + ((i & 1) << (len - 1));
    fft(a, lim, 1); fft(b, lim, 1);
    for(int i = 0; i <= lim; i++) c[i] = a[i] * b[i];
    fft(c, lim, -1);
    
    n++;
    ll ret = 0;
    for(int i = 0; i <= m; i++) chmax(ret, c[i].x + 0.5); 
    return -2 * ret + nx + ny;
}   
signed main() {
    n = read(); m = read();
    for(int i = 0; i < n; i++) a[i] = read(), sx += a[i], nx += a[i] * a[i];
    for(int i = 0; i < n; i++) b[i] = read(), sy += a[i];
    memcpy(t, b, sizeof(b));
    int c = - (sx - sy) / n;
    ll ans = check(c);
    ans = min(ans, min(check(c - 1), check(c + 1)));
    cout << ans;
    return 0;
}