Periodic Signal

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 A0 ... An-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 B0 ... Bn-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, A0 ... An-1. The third line contains n integers, B0 ... Bn-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≤Ai,Bi<220.

输出

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

样例输入

293 0 1 4 1 5 9 2 65 3 5 8 9 7 9 3 251 2 3 4 52 3 4 5 1

样例输出

80
0
#include <iostream>#include <cmath>#include <vector>#include <cstdlib>#include <cstdio>#include <climits>#include <ctime>#include <cstring>#include <queue>#include <stack>#include <list>#include <algorithm>#include <map>#include <set>#define LL long long#define Pr pair<int,int>#define fread(ch) freopen(ch,"r",stdin)#define fwrite(ch) freopen(ch,"w",stdout)using namespace std;const int INF = 0x3f3f3f3f;const int mod = 1e9+7;const double eps = 1e-3;const int maxn = 61234;struct Line{    LL l,r,val;    bool operator <(const struct Line a)const    {        return l < a.l;    }} ln[maxn];LL bit[maxn];LL st,en,t;int n,m,tp;LL to[maxn];map <LL,int> mp;int id,iid;int Lowbit(int x){    return x&(-x);}void Add(int x,LL ad){    x++;    while(x <= id+1)    {        bit[x] += ad;        x += Lowbit(x);    }}LL Sum(int x){    LL ans = 0;    x++;    while(x)    {        ans += bit[x];        x -= Lowbit(x);    }    return ans;}int Search(LL x){    int l,r;    l = 0,r = id;    int ans = -1;    while(l <= r)    {        int mid = (l+r)>>1;        if(to[mid] <= x)        {            ans = mid;            l = mid+1;        }        else r = mid-1;    }    return ans;}LL solve(){    LL mx = 0;    memset(bit,0,sizeof(bit));    sort(ln,ln+tp);    int l,r;    l = 0,r = 0;    for(int i = 0; i <= id; ++i)    {        int rr = Search(to[i]+t);        while(r < tp && mp[ln[r].l] <= rr)        {            Add(mp[ln[r].r],ln[r].val);            ++r;        }        while(l < tp && mp[ln[l].l] < i)        {            Add(mp[ln[l].r],-ln[l].val);            ++l;        }        mx = max(mx,Sum(rr));//        printf("%d,%d %lld\n",i,rr,mx);//        printf("%d %d\n",l,r-1);    }    return mx;}int main(){    LL t2,x,y,val,sum;    while(~scanf("%lld%lld",&t,&t2))    {        mp.clear();        iid = 0;        scanf("%lld",&st);        en = st+t2;        scanf("%d%d",&n,&m);        tp = 0;        sum = 0;        for(int i = 0; i < n; ++i)        {            scanf("%lld%lld%lld",&x,&y,&val);            x += y;            if(st <= x && x <= en)            {                ln[tp].l = x+y;                ln[tp].r = (en-x)/(2*y)*(2*y)+ln[tp].l;                ln[tp++].val = val;                sum += val;                to[iid++] = ln[tp-1].l;                to[iid++] = ln[tp-1].r;//                printf("%lld %lld\n",ln[tp-1].l,ln[tp-1].r);            }        }        for(int i = 0; i < m; ++i)        {            scanf("%lld%lld%lld",&x,&y,&val);            ln[tp].l = x+y;            if(st <= x+2*y && x+2*y <= en)            {                ln[tp].r = (en-x)/(2*y)*(2*y)+ln[tp].l;            }            else ln[tp].r = x+y;            ln[tp++].val = val;            sum += val;            to[iid++] = ln[tp-1].l;            to[iid++] = ln[tp-1].r;       }        sort(to,to+iid);        id = -1;        for(int i = 0; i < iid; ++i)        {            if(!i || to[i] != to[i-1])            {                ++id;                to[id] = to[i];                mp[to[i]] = id;            }        }        printf("%lld\n",sum-solve());    }    return 0;}