dec vt220

Question: https://www.lydsy.com/JudgeOnline/problem.php? Id = 1607 #include Bzoj 1607 [usaco DEC] patting heads shoot-enumerative multiple

Problem 3:threatening Letter [J. Kuipers, 2002]FJ have had a terrible fight with he neighbor and wants to send Hima nasty Wants to remain anonymous. As so many Beforehim has done, he plans-to-cut out printed letters and paste Themonto a sheet of paper. He has a infinite number of the most recentissue of the The Moo York times that have N (1 A look at the substring associated with the suffix that set. The idea is to try to find a longer substring in the text string that matches the current string

The main question: in the number of N, for each of the number, in the other (n-1) the number of several is his factor.The puzzle: Considering that the range of the AI is not too large, you can use a bucket to count the number of 1~max (AI), and then sift the 1~max in multiples of MAX. 100,000 numbers in 1000000 barrels are very sparse, ignoring the number of occurrences of 0 is very important for time efficiency.Code:1#include 2#include 3#include 4#include 5#include 6 using namespacestd;7 #defin

)) ans+=bit[x-1]; * } $ }Panax Notoginseng //printf ("Now%d x%d flag%d Zero%d =%d\n", Now,x,flag,zero,ans); -c[now][x][flag][zero]=ans; the returnans; + } A the voidSolve (LL x,ll tmp) + { -memset (D,0,sizeof(d)); $memset (c,-1,sizeof(c)); $ intL=0; LL y=x; - while(y) - { theD[++l]= (int) (y%Ten); -Y/=Ten;Wuyi } the for(intI=0; i -; i++) ret[i]=x%bit[i]+1; - for(intI=0; i9; i++) sum[i]+=g (i, -,1,1)*tmp; Wu } - About intMain () $ { - //freopen

to build a 2.00-length road between farm 1 and farm 2, and between farm 3 and farm 4 A road with a length of 2.00. In this way, the total length of the road to be built is 4.00, and this is the road in all schemes The smallest total length. Originally, I was about to finish the question = As a result, when I updated Windows 7, I restarted the Qaq blog Park and didn't save it... I wanted to cry. Forget it .. Write another copy. Question: When we see that all vertices are connected and the tot

Some cows need to be cleaned. There are n (n Idea: The data volume of 1 W is not very large. In addition, the time limit is 5 s, so we will return with N ^ 2. Sort the ox by the start of the time period. Set f [I] to the minimum time spent on cleaning the first ox. Then f [I] = min (F [I], F [J] + cost [I]) (F [I]. st The last is the initial value and the answer. As the question says that there are cows cleaning every moment, the initial value of F is the maximum value. The initial value of t

the second and third cows;The second cows pats no cows; etc.HINTSourceSilverSolutionThe original test instructions wrong, here made a small amendment.Using a method like a prime sieve, when a number is sifted, it indicates which cows hold the number.Perceptual understanding, do not understand the code.(as the only question on page 7th is not the title of the title I would not Qaq)1#include 2 using namespacestd;3 inta[100005], b[1000005], ans[1000005];4 intMain ()5 {6 intN, mx =0;7scanf"%d",

=sta[l++]; for(inti=head2[u];i!=-1; i=edge2[i].next) {intto=edge2[i].to;inch[to]--;if(!inch[to]) Sta[++r]=to; } }returnr== -;}intTail[n];intMain () {init (); scanf"%d", n);intLa=0; for(intI=1; i1; scanf'%s ', all+la+1);intLen=strlen (all+la+1); Tail[i]=la+len; la+=len+1; } for(intI=1; i0; for(intj=fir[i];jif(!find (hasha,all[j]-' A '+1)) Ins (hasha,all[j]-' A '+1); hasha=hasha*Base+ALL[J]; } ins (hasha,-1); }intans=0; for(intI=1; iinch,0,sizeof(inch)); Ull hasha=0; for(intj=fir[

must be numbered 1~k at least one point to build the diagram hierarchy, the lower layer to copy the above, and then 1~k point from the upper level of the edge of the right side of 0, run Floyd I really bi the dog opened a 200* An array of 200 re is constantly criticized by the yellow giant1#include 2#include 3#include 4#include 5 #defineLL Long Long6 #defineINF 1000000000007 using namespacestd;8 intN,m,k,q,tot;9 Long Longans;Ten Long Longdist[410][410]; One inline LL read () A { -LL x=0, f=1;Ch

Endl; - #endif Wu intj=1; -F (I,1, l) f[i]=INF; About intst=0, ed=0; $f[0]=0; -q[ed++]=0; - for(intI=2; i2){ - while(i>cow[j].x jN) { A intlast=i; +I=max (i, (cow[j].y+1)/2*2), J + +; the for(inti=last;i2) - if(f[i-2*a]!=INF) { $ while(St1]]>f[i-2*a]) ed--; theq[ed++]=i-2*A; the } the } the while(St2*B) st++; - if(f[i-2*a]!=INF) { in while(St1]]>f[i-2*a]) ed

Description John suffered a major loss: the cockroach ate all his hay and left a herd of hungry cows.　　He rode a carriage in C (1≤c≤50000), and went to the house to buy some hay.　　Dayton has H (1≤h≤5000) bales of hay, each pack has its volume Vi (l≤vi≤c). John can only buy the whole package, how many volumes of hay will he be able to ship back? Input line 1th enters C and H, and then line H enters a VI. The maximum volume of available hay for the Output. Sample Input7 3//Total volume of 7, with

The topic describes a problem, it should be the number of each note, how many individual notes on the number is his approximate.Sieve method.Record the number of occurrences of each number, and then this number is pushed back to the number of his multiples.1#include 2 #defineMax (A, b) a>b?a:b3 Const intMAXN =1000005;4 5 intCNT[MAXN],ANS[MAXN],A[MAXN];6 intMain ()7 {8 intN;SCANF ("%d",n);9 intMmax =0;Ten for(inti =1; ii) { Onescanf"%d",a[i]); Acnt[a[i]]++; -Mmax =Max (mmax,a[i]); -

Title: Given some cows, each cow has an initial position and speed, if a cow can catch up with the back of the speed will be the same as the back of the head, ask T minutes after the formation of how many small groupsOn the inefficiency of sorting algorithms and how to avoid using sorting algorithms and the importance of reading questions carefullyThe speed of a cow is not affected by the cow behind it.So we sweep backwards, and if the current bull doesn't catch up with the slowest cow in the sm

Create or Replace procedure p_getstring (p_finalstring out varchar2, p_rulestring in number, p_sourcestring in VARCHAR2) as V_num number:=1; V_resoucenum number:=1; V_getnum number; V_getsting VARCHAR2 (2000); V_errorstring exception; V_errorrule exception;begin If Length (p_sourcestring) 　　Based on 23,423 pieces, intercept the field ' Abdecsdsadsadsad ' to ab/dec/sdsa/ds/ads output

Long,int>PII;8 structnode9 {Ten intu, V, W, next; One}a[5000005]; A inttot, N, M, Rxa, Rxc, RYA, RYC; - intdis[1000005]; - intRP, T, first[1000005]; the intdone[1000005]; - voidAddedge (intStintEndintval) - { -a[++tot].u = ST;A[TOT].V = END;A[TOT].W =Val; +A[tot].next =first[st];first[st] =tot; - } + intDij () A { atPriority_queue Q; - for(inti =0; I inf; -Q.push (Make_pair (0,0)); -dis[0] =0; - while(!q.empty ()) - { in intU =q.top (). Second;q.pop (); - if(Do

Advertising#include int main(){ puts("转载请注明出处[vmurder]谢谢"); puts("网址：blog.csdn.net/vmurder/article/details/43970671");}--silverF[I][J] indicates that the minimum value of J has been skipped to the first IThen the violence shifted from the past.Its time complexity is 125 million, but the constant is far less than 1.--bronzeAs in the Silver group, just change the size of the array, then m directly assigns a value of 1.Silver Group Code:#include #include #include #include #include #define N 5

Bare shortest path...Creating a graph is still subtle... But it doesn't feel fast...Create a vertex at each time point. Then, we can build a graph in two steps:(1) I point in time connects a directed edge to the I-1 point in time(2) If there is a ox [l, r], then l-1 connects an edge to rThe last answer is dis [T].Think about it and think it's very clever... But it's slow...1 /************************************** * *********************** 2 Problem: 3389 3 User: rausen 4 Language: C

Flip the string back to the back of the original string, with a separator in the middle, each time greedy selection Rank The little one.It's actually a template for practicing a suffix array.#include #include #include #include #include #include #include #include #include #include #define INF 0x7FFFFFFF#define LL Long Long#define N 1000005using namespace STD;CharGet () {CharC=getchar (); for(;c>' Z '|| c' A '; C=getchar ());returnC;}CharS[n];intT1[n],t2[n],cc[n],sa[n

Bzoj4394 [usaco 2015 Dec] BessieDescription After eating too much fruit in Farmer John's kitchen, Bessie the cow is getting some very strange dreams! In her most recent dream, she is trapped in a maze in the shape of an N x M grid of tiles (1 ≤ N, M ≤ 1,000 ). she starts on the top-left tile and wants to get to the bottom-right tile. when she is standing on a tile, she can potentially move to the adjacent tiles in any of the four cardinal directions c

]=read (); ne[i]=first[a[i]];first[a[i]]=i;cc[i]++; - } + for(intI=1; i){ A if(Ok[i])Continue; atok[i]=1;d fn[i]=++tot;dd[tot]=i;intto=A[i]; - while(!Ok[to]) { -dfn[to]=++tot;dd[tot]=to ; -ok[to]=1; to=A[to]; - } -sz=tot-dfn[to]+1; in for(inti=tot;i>=dfn[to];i--){ - intli=dd[i];cc[li]=0; an[li]=sz; to for(intJ=FIRST[LI];J;J=NE[J])if(Dfn[j]1; + } - while(ht) { the intx=q[h];h++;ok[x]=1; *