1394 to thunderbolt

Test instructions: Input 0 ~ n-1 Total n number, each time take the first number to the end of the sequence to form a new sequence, ask these series in reverse order the number of the least.Example: Input 0,3,2,4,1,5First, the number of reverse order, first put 0 to the last 3,2,4,1,5,0 and then the number of reverse order, go down, take the inverse number of the minimum value.Use the line segment tree to find the inverse pair of the originally entered sequence.There are two kinds of thinking in

Test instructionsGive you a sequence of n numbers each time you put the first number at the end to get a new sequence there are n sequences to find which sequence in these sequences contains the smallest total number of reverse order (the minimum total number of reverse order)Analysis:Using bit to find out the inverse number of the initial number, the first number is put in reverse order, and the positive sequence of the order.#include #includeSet>#include#include#include#include#include#include

) - { +tree[rt].num=tree[rt1].num+tree[rt1|1].num; A } at - voidBuildintLintRintRT) - { -Tree[rt].l=l; -Tree[rt].r=R; -tree[rt].num=0; in if(L==R)return; - intm=Tree[rt].mid (); toBuild (l,m,rt1); +Build (m+1,r,rt1|1); - } the * intQueryintLintRintRT) $ {Panax Notoginseng if(tree[rt].l==ltree[rt].r==R) - returnTree[rt].num; the intm=Tree[rt].mid (); + if(rm) A returnQuery (l,r,rt1); the Else if(l>m) + returnQuery (l,r,rt1|1); - Else $

we know the solution of this sub-problem: for example, X is the final winner, then according to the above table to turn this x back is not exactly the solution of n personal situation?! The formula to change back is very simple, I believe everyone can push out: X ' = (x+k) mod n;So as long as we repeat this process, we can get the first person's number, because this person's final number is 0 (he alone):0-> (0+k)%2-> ((0+k)%2+k)%3->f[n]= (f[n-1]+k)%n,f[1]=0; F[i] Indicates that there is an I pe

Problem DescriptionThe Inversion number of a given number sequence A1, A2, ..., the number of pairs (AI, aj) that SA Tisfy i For a given sequence of numbers a1, A2, ..., an, if we move the first m >= 0 numbers to the end of the seqence, we'll Obtain another sequence. There is totally n such sequences as the following:A1, A2, ..., An-1, an (where m = 0-the initial seqence)A2, A3, ..., an, A1 (where m = 1)A3, A4, ..., an, A1, A2 (where m = 2)...An, A1, A2, ..., an-1 (where m = n-1)You is asked to

Test instructions is the number of reverse order.First, there is no brain line tree to find out the inverse number of the original sequence.And then:But the first number of a series is put to the last. This repeats n-1 times. Find the smallest number of reverse order in the new series.If the first number is a[i] then after the end, the inverse number of the new series is the inverse number of the original sequence minus the number of smaller than a[i], plus the number larger than a[i].namely ans

Problem descriptionthe inversion number of a given number sequence A1, A2,..., an is the number of pairs (AI, AJ) that satisfy I For a given sequence of numbers A1, A2 ,..., an, if we move the first m> = 0 numbers to the end of the seqence, we will obtain another sequence. there are totally N such sequences as the following:A1, A2,..., An-1, an (where m = 0-the initial seqence)A2, A3,..., An, A1 (where M = 1)A3, A4,..., An, A1, A2 (where m = 2)...An, A1, A2,..., An-1 (where M = N-1)You are asked

The main idea: to give you a graph of n points, to find the slim (maximum edge minus minimum) as small as possible spanning treeIdea: Sort after the violent enumeration interval can be//see if it explodes. int! array will not be one dimension less! //You must be careful about the conditions of the winning.#include using namespacestd;#defineLL Long Long#defineAll (a) A.begin (), A.end ()#definePB Push_back#defineMk Make_pair#defineFi first#defineSe SecondConst intINF =0x3f3f3f3f;Const intMAXN = -

numbers of the smallest number of reverse-order numbers in the array of all transformations. Puzzle: The tree-like array to find the most initial reverse number, you can transform all the reverse order number, the first is a[i], then when it is put to the last go, the whole sequence in reverse order number of n-a[i], less a[i]-1.#include #include#includestring.h>#include#include#includeusing namespacestd;Const intN =5005;intA[n],b[n],c[n],n;intLowbit (intx) { returnx (-x);}voidUpdateintIdxin

Reprint please indicate this article link http://blog.csdn.net/yangnanhai93/article/details/40506571Title Link: http://ac.jobdu.com/problem.php?pid=1394Problem Analysis:This problem is a four-star problem. But it feels a little too simple.The first message to us is that it needs to be sequential, so we'll definitely sort the original array, and then we'll find out how many are missing. The problem is a bit similar to the question of finding the biggest stock in n days, I just care about my curre

Still practicing the use of the tree array:The main topic: give the number of n, these numbers can be deleted from the back and then put to the front to form a new sequenceCan be obtained in the N case, find out in these n cases which kind of the number of reverse order of the smallestproblem-solving ideas: first to find the first sequence of the reverse number, and then use a very clever way to find the next sequence of the number of reverse, until all to find outSequence 4 5 2 1 3 6, this sequ

Reprint please indicate this article link http://blog.csdn.net/yangnanhai93/article/details/40506571Title Link: http://ac.jobdu.com/problem.php?pid=1394Problem Analysis:This problem is a four-star problem, but it feels a little too simple.The first message to us is that it needs to be sequential, so we'll definitely sort the original array, and then we'll find out how many are missing. This topic to think will find and find the biggest stock in n days the problem is a bit similar, that is, I onl

Logon Timeout network connection interrupted (including Thunder 7.2, Thunder 7.9, the Thunder Prestige Edition, all network applies) 1. It is recommended to restart the Thunderbolt. 2. Close the kill soft, firewall, 360 and other flow monitoring software. 3. Unplug the network cable or do not connect to the router, reconnect the Internet and try again later. 4. To the computer here Delete c→program Files→co

return ; - intMid= (first+end) >>1; -Build (rt1, first,mid); -Build (rt1|1, mid+1, end); + } - intQuary (intRtintFirstintEndintLeftintRight ) + { A if(Leftend) at returnSum[rt]; - intMid= (first+end) >>1; - intans=0; - if(leftmid) -Ans+=quary (rt1, first,mid,left,right); - if(right>mid) inAns+=quary (rt1|1, mid+1, end,left,right); - returnans; to } + voidUpdateintRtintFirstintEndinta) - { the if(first==end) * { $sum[rt]++;Panax Notoginseng

the number of reverse order in its own sequence, and then determine whether each element is placed in the last one when the number of reverse order is less than the number of reversed sequence, successive updates. #include #includestring.h>#includeusing namespacestd;Const intinf=0x3f3f3f3f;Const intn=5010;structnode{intLeft , right, num;} no[4*N];intMin;voidBulid (intLeftintRightintroot) { intmid; No[root].left=Left ; No[root].right=Right ; No[root].num=0; if(left = right)return ; Mid= (left

initial sequence number, the reverse number of the next sequence is handed out 1. When traversing to the current sequence element, the query from this element to the n-1 interval of the reverse and, that is, the element is greater than the element has appeared several (inverse reverse, but the result The node and its parent node are then updated.　　Traversing to the end, you can find out the reverse order of the sequence. 2. Since it is starting from 0 consecutive numbers for reverse order, so,

arrangement of length n (composed of 0 to n-1), each time you can put the first element to the last surface, then will produce the reverse number, a permutation of the number of reverse is equal to the permutation of all the numbers in the inverse number of the sum. So, this problem, can be solved with the line segment tree, we first establish an empty tree, that is, the sum==0 of each node in the tree, representing all 0, and then, each time we insert a number, in the corresponding position of

Test instructions: There is a sequence of length n, the number of the sequence is a 0~n-1 composition, and then this sequence can be considered as a ring, then there is n a sequence of n length, ask the number of the smallest in the n sequence.First, the initial sequence of the reverse number of the calculation, and then move each start number A to the back, the reverse number of changes is larger than a number of numbers plus one, less than a small number of reverse order minus one, so accordin

DescriptionThe inversion number of a given number sequence A1, A2, ..., is the number of pairs (AI, AJ) that satisfy I For a given sequence of numbers a1, A2, ..., an, if we move the first m >= 0 numbers to the end of the seqence, we'll Obtain another sequence. There is totally n such sequences as the following:A1, A2, ..., An-1, an (where m = 0-the initial seqence)A2, A3, ..., an, A1 (where m = 1)A3, A4, ..., an, A1, A2 (where m = 2)...An, A1, A2, ..., an-1 (where m = n-1)You is asked to write

], the number of inverse sequence will be reduced t[i], corresponding increase n (t[i]+1)If the violence, the estimate will time out, so just use the number of line segments, note that line tree is just a tool,!!On the code:#include #include using namespace STD;Const intmaxn=5005;intNUM[MAXN];struct{intL, R, Num;} tree[4*MAXN];voidBuildintRootintLintR) {tree[root].l=l; Tree[root].r=r; tree[root].num=0;if(L = = r)return;intMid= (L + R) >>1; Build (root1, L, mid); Build (root1|1, mid+1, r);}voi