1394 motherboard

*/ - if(lR) {Wuyi returnTree[rt].value; the } - Else{ Wu intM= (TREE[RT].L+TREE[RT].R) >>1; - ints1=0, s2=0; About if(l1, l,r); $ if(r>m) S2=query (rt1|1, l,r); - returns1+S2; - } - } A + intMain () the { - intN; $ while(~SCANF ("%d",N)) { theBuild (1,1, n); the intsum=0; the for(intI=0; i){ thescanf"%d",a[i]); -a[i]++; inSum+=query (1, a[i]+1, n); theUpdate (1, A[i]); the } About intres =

, from above, a[1]=0.Each subsequent deletion is to repeat the last process, the final left of the person must be in the position No. 0, then you can always push up to find out where the winner begins.1#include 2#include 3 using namespacestd;4 5 Const intMAXN =10005;6 intA[MAXN];7 8 intMain ()9 {Ten intN, K, m; One while(Cin >> n >> k >> m n k m) A { -Memset (A,0,sizeof(a)); - for(inti =2; I ) theA[i] = (A[i-1] + k)%i; - intAns = (A[n-1] + m)%N; -cout 1Endl; -

. Sort the query by the right endpoint from small to large.2. Record the position of each number in the last occurrence. When the number reappears, it removes the count from its last occurrence position.Implementation:The subject is a double pointer.#include using namespacestd;Const intN (1e5+5);intN, M, Q, A[n], pos[n], bit[n], ans[n];voidAddintXintv) { for(; xx);}intSumintx) { intres=0; for(; x; res+=bit[x], x-=x-x); returnRes;}structp{intL, R, id; P (intLintRintID): L (L), R (r), ID (

Minimum Inversion numberTime limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)Total submission (s): 14879 Accepted Submission (s): 9082Problem 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

and, that is, the number of new reverse order. The complexity of time is nlogn.Finally, each number as the first element appears in the sequence of the number of reverse order, to find the minimum value can be easily introduced, each time the number of replacement head, the new number of reverse order (the original number of reverse) minus (the number of numbers) plus (the total number minus the number of the numerical subtraction).#include using namespaceStd;ConstintN =5555;intNum[N2];intHehe[

number of nodes inserted before inserting (if present, it will be reversed with 4) v9=5Insert 2 in the segment tree, asking the interval [2,9] The number of nodes inserted before inserting (if present, it will be reversed with 2) v10=7Add v1+......+v10 = 22, this is the reverse number of 1 3 6 9 0 8 5 7 4 2. is to insert a number into a segment tree, asking for a few more than this number before each insert. is actually asking how many numbers are inserted in a range!#include Copyright NOTICE

Else returnCalc (sum (lp,l,mid), SUM (rp,mid+1, R)); - } Wu - intMain () { About intT; $ while(SCANF ("%d", n)! =EOF) { - for(inti =1; I "%d",A[i]), a[i]; -Build_tree (1,1, n); - A intAns =0; + for(inti =1; I ){ theAns + = SUM (1, a[i]+1, n); Update (1, A[i],1); - //printf ("ans =%d\n", ans); $ } the the intres =ans; the for(inti =1; I ){ theAns + = n2*a[i]-1; -res =min (res,ans); in //p

title Link: http://acm.hdu.edu.cn/showproblem.php?pid=1394violence (TIME:453MS)#include Segment Tree (TIMES:93MS)#include Copyright NOTICE: This article for Bo Master original article, without Bo Master permission not reproduced. HDU 1394 Minimum Inversion Number (violence/Line tree)

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