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=,f=;
     char ch=getchar();
     while(ch<''||ch>'') {
         if(ch=='-')f=-;
         ch=getchar();
     }
     while(ch>=''&&ch<='') {
         x=x*+ch-'';
         ch=getchar();
     }
     return x*f;
 }
 const double pi=.14159265358979323846264338327950288L;
 const double eps=1e-;
 const int mod = 1e9 + ;
 const int INF = 0x3f3f3f3f;
 const int MAXN = ;
 const int xi[] = {, , , -};
 const int yi[] = {, -, , };
 
 int N, T;
 ll a[], b[];
 ll c[], d[];
 struct asort {
     int num;
     ll date;
 } sa[], sb[];
 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 = ;
         scanf("%d", &n);
 
         for(int i = ; i < n; i++) a[i] = read();
         for(int i = ; i < n; i++) b[i] = read();
         for(int i = n; i < *n ; i++) {
             a[i] = a[i-n];
             b[i] = b[i-n];
         }
         for(int i = ; 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, );
         ll res = ;
         for(int ai = ; ai < m; ai++) {
             ll ans = ;
             int i = (sb[].num - sa[ai].num + n)%n;
             for(int j = i; j < n+i; j++) {
                 ans += (a[j-i]*b[j]) <<;
             }
             if(ans > res) {
                 res = ans;
                 k = i;
             }
         }
         printf("%lld\n", sum - res);
         //  printf("%d\n", k);
     }
     return ;
 }