1269-consecutive Sum

1269-consecutive Sum

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Time Limit:3 second (s)

Memory limit:64 MB

Little Jimmy is learning what to add integers. As in decimal the digits is 0 to 9, it makes a bit hard for him to understand the summation of all pair of digits. Since addition of numbers requires the knowledge of adding digits. So, he mother gave him a software that can convert a decimal integers to their binary and a binary to its corresponding deci Mal. So, Jimmy's idea was to convert the numbers into binaries, and then he adds them and turns the result back to decimal using The software. It's easy-to-add in binary, since-need to know how to add (0, 0), (0, 1), (1, 0), (1, 1). Jimmy doesn ' t has the idea of carry operation, so he thinks that

1 + 1 = 0

1 + 0 = 1

0 + 1 = 1

0 + 0 = 0

Using These operations, he adds the numbers in binary. So, according to his calculations,

3 (011) + 7 (111) = 4 (100)

Now is given an array of n integers, indexed from 0 to n-1, and you has to find both indices C3>i J in the array (0≤i≤j < n), such that the summation (according to Jimmy) of all integers between Indices I and Jin the array, is maximum. And you also has to find a indices, p Q in the array (0≤p≤q < n), such that the summation (ACC Ording to Jimmy) of any integers between indices p and Q in the array, is minimum. You are only maximum and minimum integers.

Input

Input starts with an integer T (≤10), denoting the number of test cases.

Each case is starts with a line containing an integer n (1≤n≤50000). The next line contains n space separated non-negative integers, denoting the integers of the given array. Each of the integer fits into a is signed integer.

Output

For each case, print the case number, the maximum and minimum summation that can be made using Jimmy's addition.

Sample Input	Output for Sample Input
2 5 6 8 2) 4 2 5 3 8 2) 6 5	Case 1:14 2 Case 2:15 1

Note

The Dataset is huge, and the use faster I/O methods.

Test instructions: The maximum and minimum values of the difference of the continuous interval are obtained;

Idea: trie tree, greedy;

First, the prefix XOR, and then first add 0 to the tire tree, 0 XOR or any value for that value itself, and then we know pre[i]^pre[j];j<i;

Ans[j+1]^ans[j+2]...^ans[i]; that is, the interval [i,j], then we add the preceding prefix tire, and then we in the tree greedy selection, if it is to find the largest, the greedy choice is different from the current position, because it is the highest bit

Down, so we guarantee the high number to be as large as possible, if it does not exist with the current bit the same, then can only go the same. Know how greedy the biggest, the smallest and this is similar.

Complexity is (N*log (1<<31-1));

1#include <stdio.h>2#include <algorithm>3#include <string.h>4#include <stdlib.h>5#include <iostream>6 using namespacestd;7typedefLong LongLL;8LL ans[500005];9 intid[ +];Ten structnode One { ANode *p[2]; - node () -     { theMemset (P,0,sizeof(P)); -     } - }; -Node *Root; + void inch(int*v) - { +     inti,j,k; ANode *c=Root; at      for(i= -; i>=0; i--) -     { -         intPre=V[i]; -         if(c->p[v[i]]==NULL) -         { -c->p[v[i]]=Newnode (); in         } -C=c->P[v[i]]; to     } + } -LL MA (int*v) the { *LL sum=0; $LL l=1;Panax NotoginsengNode*c=Root; -     inti; the      for(i= -; i>=0; i--) +     { A         intPre=V[i]; the         if(C->p[(pre+1)%2]==NULL) +         { -C=c->P[pre]; $sum*=2; $sum+=0; -         } -         Else the         { -c=c->p[(pre+1)%2];Wuyisum*=2; thesum+=1; -         } Wu     } -     returnsum; About } $LL mi (int*v) - { -LL sum=0; -LL l=2; ANode*c=Root; +     inti; the      for(i= -; i>=0; i--) -     { $         intPre=V[i]; the         if(c->p[pre]==NULL) the         { thec=c->p[(pre+1)%2]; thesum*=2; -sum+=1; in         } the         Else the         { AboutC=c->P[pre]; thesum*=2; thesum+=0; the         } +     } -     returnsum; the }Bayi voidDel (node *c) the { the      for(intI=0; i<2; i++) -     { -         if(c->p[i]!=NULL) theDel (c->p[i]); the     } the      Free(c); the } - intMainvoid) the { the     inti,j,k; thescanf"%d",&k);94     ints; the     intn,m; the      for(s=1; s<=k; s++) the     {98scanf"%d",&n); About          for(i=1; i<=n; i++) -         {101scanf"%lld",&ans[i]);102ans[i]^=ans[i-1];103         }104root=Newnode (); theLL maxx=0;106LL minn=1e12;107memset (ID,0,sizeof(ID));108         inch(ID);109          for(i=1; i<=n; i++) the         {111memset (ID,0,sizeof(ID)); theLL kk=Ans[i];113             intCnt=0; the              while(KK) the             { theid[cnt++]=kk%2;117Kk/=2;118             }119maxx=Max (Maxx,ma (id)); -minn=min (Minn,mi (id));121             inch(ID);122 }del (root);123printf"Case %d:", s);124printf"%lld%lld\n", Maxx,minn); the}return 0;126}

1269-consecutive Sum