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。
代码语言：javascript复制#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;
}
