hihocoder 1388 &&2016 ACM/ICPC Asia Regional Beijing Online Periodic Signal

#1388 : Periodic Signal

时间限制:5000ms

单点时限:5000ms

内存限制:256MB

描述

Profess X is an expert in signal processing. He has a device which can send a particular 1 second signal repeatedly. The signal is A_{0 ...} A_n-1 under n Hz sampling.

One day, the device fell on the ground accidentally. Profess X wanted to check whether the device can still work properly. So he ran another n Hz sampling to the fallen device and got B_{0 ...} B_n-1.

To compare two periodic signals, Profess X define the DIFFERENCE of signal A and B as follow:

You may assume that two signals are the same if their DIFFERENCE is small enough.
Profess X is too busy to calculate this value. So the calculation is on you.

输入

The first line contains a single integer T, indicating the number of test cases.

In each test case, the first line contains an integer n. The second line contains n integers, A_{0 ...} A_n-1. The third line contains n integers, B_{0 ...} B_n-1.

T≤40 including several small test cases and no more than 4 large test cases.

For small test cases, 0<n≤6⋅10³.

For large test cases, 0<n≤6⋅10⁴.

For all test cases, 0≤A_i,B_i<2²⁰.

输出

For each test case, print the answer in a single line.

样例输入

2
9
3 0 1 4 1 5 9 2 6
5 3 5 8 9 7 9 3 2
5
1 2 3 4 5
2 3 4 5 1

样例输出

80
0
这题啊，我觉得暴力可做，刚开始超时了，又做了一点优化还是不行啊。
然后我觉得这个k的取值，和排完序的前m项有很大的关系，然后取m在不超时和wa的范围之间。。。这个看运气，竟然AC了
我的思路是排序a， b数组，按住其中一个数组不动，在前m个数内，用a1的下标减去b的下标，取一个得到k的最大值就好
正规做法竟然是fft。。不会啊
我的方法是歪门邪道。。看看就好，不要采纳

#include <iostream>
#include <sstream>
#include <fstream>
#include <string>
#include <vector>
#include <deque>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <algorithm>
#include <functional>
#include <utility>
#include <bitset>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <cstdio>
#include <cstring>
#define FOR(i, a, b)  for(int i = (a); i <= (b); i++)
#define RE(i, n) FOR(i, 1, n)
#define FORP(i, a, b) for(int i = (a); i >= (b); i--)
#define REP(i, n) for(int i = 0; i <(n); ++i)
#define SZ(x) ((int)(x).size )
#define ALL(x) (x).begin(), (x.end())
#define MSET(a, x) memset(a, x, sizeof(a))
using namespace std;


typedef long long int ll;
typedef pair<int, int> P;
ll read() {
    ll x=0,f=1;
    char ch=getchar();
    while(ch<'0'||ch>'9') {
        if(ch=='-')f=-1;
        ch=getchar();
    }
    while(ch>='0'&&ch<='9') {
        x=x*10+ch-'0';
        ch=getchar();
    }
    return x*f;
}
const double pi=3.14159265358979323846264338327950288L;
const double eps=1e-6;
const int mod = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const int MAXN = 1005;
const int xi[] = {0, 0, 1, -1};
const int yi[] = {1, -1, 0, 0};

int N, T;
ll a[120002], b[120002];
ll c[60002], d[60002];
struct asort {
    int num;
    ll date;
} sa[60002], sb[60002];
bool cmp(asort a, asort b) {
    return a.date > b.date;
}
int main() {
    //freopen("in.txt", "r", stdin);
    int t, n, k;
    scanf("%d", &t);

    while(t--) {
        ll sum = 0;
        scanf("%d", &n);

        for(int i = 0; i < n; i++) a[i] = read();
        for(int i = 0; i < n; i++) b[i] = read();
        for(int i = n; i < 2*n ; i++) {
            a[i] = a[i-n];
            b[i] = b[i-n];
        }
        for(int i = 0; i < n; i++) {
            sum += a[i]*a[i];
            sum += b[i]*b[i];
            sa[i].num = i, sa[i].date = a[i];
            sb[i].num = i, sb[i].date = b[i];
        }
        sort(sa, sa+n, cmp);
        sort(sb, sb+n, cmp);
        int m = min(n, 10000);
        ll res = 0;
        for(int ai = 0; ai < m; ai++) {
            ll ans = 0;
            int i = (sb[0].num - sa[ai].num + n)%n;
            for(int j = i; j < n+i; j++) {
                ans += (a[j-i]*b[j]) <<1;
            }
            if(ans > res) {
                res = ans;
                k = i;
            }
        }
        printf("%lld\n", sum - res);
        //  printf("%d\n", k);
    }
    return 0;
}