bzoj2962
线段树+卷积
这个东西直接算不太好,但是合并两段结果却很方便,假设c[i]表示选i个数乘积的和,那么$a[i]=\sum_{j=0}^{i}{b[j]*c[i-j]}$
线段树维护即可
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include<bits/stdc++.h> using namespace std; const int N = 2e5 + 5; const double pi = acos(-1); int n, len, ans, suma, sumb, mx = -1e9, m; struct data { double x, y; data() {} data(double _x, double _y) : x(_x), y(_y) {} data friend operator - (const data &a, const data &b) { return data(a.x - b.x, a.y - b.y); } data friend operator + (const data &a, const data &b) { return data(a.x + b.x, a.y + b.y); } data friend operator * (const data &a, const data &b) { return data(a.x * b.x - a.y * b.y, a.x * b.y + a.y * b.x); } } a[N], b[N]; void fft(data *a, int len, int f) { int n = 1 << len; for(int i = 0; i < n; ++i) { int t = 0; for(int j = 0; j < len; ++j) { if(i & (1 << j)) { t |= 1 << (len - j - 1); } } if(i < t) { swap(a[i], a[t]); } } for(int l = 2; l <= n; l <<= 1) { int m = l >> 1; data w = data(cos(pi / m), f * sin(pi / m)); for(int i = 0; i < n; i += l) { data t = data(1, 0); for(int k = 0; k < m; ++k, t = t * w) { data x = a[k + i], y = t * a[i + k + m]; a[k + i] = x + y; a[i + m + k] = x - y; } } } } int main() { scanf("%d%d", &n, &m); for(int i = 1; i <= n; ++i) { scanf("%lf", &a[i].x); suma += a[i].x; } for(int i = 1; i <= n; ++i) { scanf("%lf", &b[i].x); b[i + n].x = b[i].x; sumb += b[i].x; } int c = floor((double)(sumb - suma) / n + 0.5); for(int i = 1; i <= n; ++i) { a[i].x += c; ans += a[i].x * a[i].x + b[i].x * b[i].x; } reverse(b + 1, b + 2 * n + 1); for(; 1 << len <= 2 * n; ++len); fft(a, len, 1); fft(b, len, 1); for(int i = 0; i < 1 << len; ++i) { a[i] = a[i] * b[i]; } fft(a, len, -1); for(int i = n + 1; i <= 2 * n + 1; ++i) { a[i].x /= (1 << len); mx = max(mx, (int)(a[i].x + 0.1)); } printf("%d\n", ans - 2 * mx); return 0; }