class 2 div 2 enclosure

Bizon the champion isn't just charming, he also is very smart. While some of us were learning the multiplication table, Bizon the champion had fun in his own manner. Bizon the champion paintedN? ×?MMultiplication table, where the element on the intersection ofI-Th row andJ-Th Column equalsI·J(The rows and columns of the table are numbered starting from 1). Then he was asked: what number in the table isK-Th largest number? Bizon the champion always answered correctly and immediately. Can you repe

The meaning of this problem is to give you a photo of the end-to-end, the current photo in 1, can only transfer a unit, the transfer time is a, the picture may be reversed, the picture will need to be the time to B, look at the photo of the time is 1, want to see as many photos, ask the number is how much, We can preprocess the time required to view each photo, and then enumerate the time needed from the left, looking at the time required from the right, and two points looking at the time requir

S2) {if(S1.W==S2.W)returnS1.d>s2.d; returns1.wS2.W;}intN,m,vis[n];vectorint,pairint,int> > >ans;intMain () {intflag=0; scanf ("%d%d",n,m); for(intI=1; i) {scanf ("%d%d",a[i].w,A[I].D); A[i].id=i; } sort (A+1, a+m+1, CMP); intaim=n-1; intans1=1, ans2=2, l=2, r=2; vis[1]=vis[2]=1; for(intI=1; i) { if(a[i].d==0)

} theCaptype Maxflow (intSintTintN) { +Captype ans=0; -Snode=s; Enode=T; $ while(SearchPath (n)) { $ans+=Rest[enode]; - intv=Enode; - while(v!=SNode) { the intu=Pre[v]; -f[u][v]+=Rest[enode];WuyiF[v][u]-=rest[enode];//give a chance to return thev=u; - } Wu } - returnans; About } $ intMain () { -Freopen ("C:/users/fuliujun/desktop/input.txt","R", stdin); -Freopen ("C:/users/fuliujun/desktop/output.txt","W", stdout); - inta,b,m,n,a1[ -],b1[ -]

Eat (R). For K = 2 and length 2 Marmot can eat (RR) and (WW). For K = 2 and length 3 Marmot can eat (RRR), (RWW) and ( WWR). For K = 2 and length 4 Marmot can eat, for example, (wwww) or (rwwr), but F or example he can ' t eat (wwwr). Test Instructions : There is

(); } while(ch>='0'ch'9') {x=x*Ten+ch-'0'; ch=GetChar (); }returnx*F;}//****************************************Const intn=200005;#defineMoD 10000007#defineINF 10000007#defineMAXN 10000intf[200005],s[n],k=0, kk=0;intMain () {ll ans=0, A; ll N=read (); for(intI=1; i) {scanf ("%i64d",a); Ans+=A; if(a>0) {f[++k]=A; } Elses[++kk]=-A; } if(ans>0) {cout" First"Endl; } Else if(ans0) cout"Second"Endl; Else { for(intI=1; I) { if(F[i]>s[i]) {cout" First"return 0;} Else if(

Water a-uncowed Forces#include 　　(two points) + greedy b-more cowbellTest instructions: n items in a maximum of K boxes, each box up to two, ask the size of the box is the smallest number.Analysis: Can be two-enumeration volume size, then determine whether to meet the conditions need to greedy, as the solution says, if k > N, then the volume is the largest single. Otherwise there must be n-k a box must put two items (solution equation), then the first choice combination volume small, that is, th

Spongebob is already tired trying to reason his weird actions and calculations, so he simply asked you to find all pairs O f N and M, such that there is exactlyx distinct squares in the table consisting ofn rows and m Columns. For example, in a3?x?5 table there is squares with Side one,8 squares with side and 3 squares with side three. The total number of distinct squares in a 3?x?5 table is 15?+?8?+?3?=?26. InputThe first line of the input contains a single integer x (1?≤

/* ***********************************************author:maltubemail: [email Protected]blog: htttp://www.xiang578.com************************************************ * *#include #include #include #include #include #include #include #include #include #include #include #include //#include #define REP (i,a,n) for (int i=a;i#define PER (i,a,n) for (int i=n-1;i>=a;i--)#define PB Push_backusing namespace STD;typedef vectorint>VI;typedef Long LongllConstll mod=1000000007;Const intn=2048;Chars[100000+T

is 2 need a special sentence.The problem may be that the data is too watery to see others using the most brutal method of judging primes: O (T*SQRT (n)) can also be too.The code is as follows:#include #include #include #include #include #include #include #include using namespace STD;BOOLIsPrime (intNUM) {if(num = =2|| num = =3) {return true; }if(num%6!=1 num%6!=5) {return false; } for(inti =5; I*i 6)

Polycarp loves geometric progressions very much. Since He is only three years old and he loves only the progressions of length three. He also has a favorite integer k and a sequence a, consisting of n integers. He wants to know how many subsequences of length three can is selected from a, so that they form a geometric Progression with common ratio k.A subsequence of length three is a combination of three such indexes i1, i2, i3 , that 1≤ i1 i2 i3≤ n. That's, a subsequence of length three i

+1] + = (p[j][k]-1)/2; P[J][K]=1; //cout//cout//cout } } } } intCNT =0; for(intj=1; j) { for(intk=1; k) { if(p[j][k]>=1) CNT++; }} coutEndl; } return 0;}View CodeIf you push to the n+1 layer, the number of ==1 is counted.#include #include#include#include#includeSet>#include#includeusing namespaceStd;typedefLong Longll;Doublep[ A][ A];intMain ()

put them all together to marry home.There are only two things to do when a doll is set:1: Put a single big Doll out of a string or a doll.2: Take a single doll off the outermost layer of a doll.How many times do I have to ask the youngest to marry the doll home?Problem Solving Ideas:Is the simulation, is the Mini.1#include 2#include 3#include 4#include 5#include 6#include 7 using namespacestd;8 9 Const int

Test (s)Input52 132 32 13Output0.5000000.0000001.5000001.3333331.500000Input62 11 2 202 21 2-333Output0.50000020.50000014.33333312.33333317.50000017.000000NoteIn the second sample, the sequence becomesSave these numbers with the St array, size indicates how many numbers, and the add array represents the value that each number in the St array should incrementThen real-time recording and/********************

(u->next[i]); u->sum+=u->next[i]->sum; } }}voidSeach (node *root) {node *p=root; while(1) {k-=p->flag;if(k0){return; }if(p->next[0]!=null) {if(p->next[0]->sum>=k) {str[cnt]=' A '; p=p->next[0]; cnt++;Continue; }Elsek-=p->next[0]->sum; } str[cnt]=' B '; p=p->next[1]; cnt++; }}intMain () {intI, J, Len, H; while(scanf('%s ', s)!=eof) {scanf("%d", k); len=strlen(s);memset(OK,0,sizeof(OK)); for(i=0; i1;if(i1s[i]==s[i+1]) ok[i][i+1]=1;if(i2s[i]==s[i+

: 123234545 except for the first lamp, the other lights are only passed through even several times.The total number of operations is the same as n-n parity. If the sum of 1 is different from the parity of N, there is no solution, and if the parity is the same, there is a solution. 1#include 2#include 3 intT,n,count,a;4 intMainintargcChar**argv)5 {6scanf"%d",t);7 while(t--)8 {9scanf"%d",n);TenCount=0; One for(intI=0; i) A { -

You ' ve got array a[1],? A[2],?...,? a [N], consisting ofn integers. Count the number of ways to split all the elements of the array into three contiguous parts so, the sum of elements in Each of the same.More formally, need to find the number of such pairs of indices i,? J (2?≤? I? ≤? j. ≤? n?-? 1), that.InputThe first line contains integer n (1?≤? N? ≤?5 105), showing how many numbers is in the ar

*now= (!c?Cont1:cont2); Now[contcomp]++; for(intI=0; I) {flag=dfs (g[u][i],!c); } returnFlag; }}intMain () {///freopen ("/home/files/cppfiles/in", "R", stdin);Cin>>n>>m; for(intI=0; i){ intf,t; scanf ("%d%d",f,t); G[F].PB (t); G[T].PB (f); } if(m==0) {cout"3"1) * (n2)/6Endl; return 0; } BOOLflag=1; for(intI=1; iif(G[i].size () >1) flag=0; if(flag) {cout"2"2) *mEndl; return 0

Codeforces Round #306 (Div. 2) (ABCE) A. Two Substrings time limit per test 2 seconds memory limit per test 256 megabytes You are given stringS. Your task is to determine if the given stringSContains two non-overlapping substrings AB and BA (the substrings can go in any order ).Input The only line of input contains a stringSOf length between 1 and 105 consisti