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.
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
800
#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; }
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