hack cf

],2); dp[i][j][1] = Factor (Map[i][j],5); } } for(inti =0; I i) { for(intj =0; J j) { for(intK =0; K 2; ++k) {if(i = =0 J = =0)Continue; if(i = =0) {Dp[i][j][k]+ = dp[i][j-1][k]; STEP[I][J][K]='R'; } Else if(J = =0) {Dp[i][j][k]+ = dp[i-1][j][k]; STEP[I][J][K]='D'; } Else{Dp[i][j][k]+ = min (dp[i-1][j][k],dp[i][j-1][k]); STEP[I][J][K]= dp[i-1][j][k] 1][K]?'D':'R'; } } } } if(Min (dp[n-1][n-1][0],dp[n-1][n-1][1]) >1Zero) {printf ("1\n")

Test instructions: A game, a children's shoes in the range of 1~n guess a number, b children's shoes to ask a set, a children's shoes to answer each number in the collection, he guessed whether the number can be divided, b to get a guess by these answers to the number, the minimum number of guesses to guess?Solution: A number can be multiplied by the exponential Shan of several prime numbers, so the answer to all numbers can be obtained by asking the exponential power of all prime numbers less t

The main topic: there is a person who needs to eat ai kilograms of meat every day, and the price of the meat is pi, ask n days he will need at least how much money to buy meat.Analysis: Note that he can buy meat in advance to put it.The code is as follows: #include #includestring.h>#include#include#includeusing namespacestd;Const intMAXN = -;Const intOO = 1e9+7;intMain () {intN; while(SCANF ("%d", n)! =EOF) { intAi, pi, ans=0, minp=Oo; while(n--) {scanf ("%d%d", ai, pi); MINP=min

The second time ~ I began to think is with a greedy DP, who knows the wrong. Later only to understand that violence two points + memory DP#include 　　CF #EDU R1 E

The equivalent to give you some points, for you to delete up to no more than K, so that you can use a rectangular edge length as an integer, parallel to the XY axis, so that the area of the rectangle is the smallest.Class with a pen to draw, and then suddenly thinking on the open, if the game is also good ~ ~ first by x, y respectively, because K is small, and, when deleted is sure to delete the outermost point, so, can be left and right to enumerate what points, sorted array to simulate this pr

This is actually a simple simulation, but our first turn is over, later seniors Rejudge, we again tle, this look not only delayed our time, also fluctuation to our mood, the original time is 2s, (the original OJ is 2s), and then changed to 1s, I used the O (n*n) cycle directly timed out , this is not scary, we are misled by this idea, has been optimized on the basis of O (N*n), made several times .... Finally know that is not so complicated, the problem of the sample is very special, after the d

(LL) Cm (n%p,m%p,p) * (LL) Lucas (n/p,m/p,p)%p; $ } the the voidInit () the { theinv[1] =1 ; - inti; in for(i =2; I 1000010; i++) theInv[i] =INV (i,mod); the } About the intMain () the { the //freopen ("In.txt", "R", stdin); + intT; -scanf"%d", T); the intCase =0 ;Bayi init (); the while(t--) the { -case++ ; -scanf"%d%d%d", n, m, k); the if(n = =1) the { theprintf"Case #%d:%d\n", case, m); the Continue ; - } the inti; th

(intD,num TN,intBLong Longl) { if(D> (n+1) >>1) ) {STA[COU].N=TN; Sta[cou].Base=C; Sta[cou++].route=l; return; } for(intI=0;i3;++i) DFS1 (d+1, tn+rem[d][i],b+ (I? Rnum[d]:0), (l2)|i);}voidDFS2 (intD,num TN,intBLong Longl) { if(d>N) {state TS; Ts.n.a=-Tn.a; TS.N.B=-tn.b; intP=lower_bound (Sta,sta+cou,ts)-STA; if(PBase+b>BBB) {BBB=STA[P].Base+b; LLL= (sta[p].route2* (n/2)))|M; } return; } for(intI=0;i3;++i) dfs2 (d+1, tn+rem[d][i],b+ (I? Rnum[d]:0), (l2)|i);}voidShow () {intr

In a side length of 10^6 square, it can be divided into 1000 segments by dividing the x-axis. When an odd number is scanned from the bottom up, the even number is scanned from the top down. To estimate the longest length, first of all, we know that there are 10^6 points, then the y-mail direction moves up to 10^3*10^6. For the x-axis direction, if all are within one segment, move the 10^3*10^6 up to a maximum of 2000*10^3 if not in one segment. Must be smaller than the limit length.#include

≤2000, 0≤ p ≤ 1). Numbers n and t are integers, numberp is real, given with exactly TS after the decimal point.OutputPrint a single real number-the expected number of people who'll be standing on the escalator after T s Econds. The absolute or relative error mustn ' t exceed 6. Sample Test (s)input1 0.50 1Output0.5input1 0.50 4Output0.9375input4 0.20 2Output0.4Simple DPDP (I,J) indicates the probability that a J person would go in the first minute.Expected =∑J*DP (T,J)Note: There are 2

Judgment question, a bit around, notice in the answer is not only a moment of judging conditions to be judged by B array#include #include#includeusing namespacestd;intb[110000],f[110000];mapint,int>Map,a;intMain () {intn,i,m; CIN>> N >>m; for(i=1; ii) {cin>>F[i]; if(! Map[f[i]]) map[f[i]]=i; Else{A[f[i]]=true; } } for(i=1; ii) {cin>>B[i]; if(!Map[b[i]]) {cout"Impossible"Endl; return 0; } } for(i=1; ii) {if(A[b[i]]) {cout"Ambiguity"Endl; return 0; }} cout"Possible"Endl; for(i=1; i

;118}Else if(Length-cmx==mn uans) {119Ans =u; - }121 122Rep (I,0, SZ) {123v =Vc2[u][i];124 if(v = =FA) the Continue;126 127 if(mx[v]+1==Mx[u]) { -TMP = MAX (PMX, Mx_[u]) +1;129}Else { theTMP = MAX (PMX, Mx[u]) +1;131 } the 133 DFS2 (V, U, tmp);134 }135 }136 137 intMain () {138Ios::sync_with_stdio (false);139 #ifndef Online_judge $Freopen ("data.in","R", stdin);141Freopen ("Data.out","W", stdout);142 #endif143 144 intu, v, x;145 146scanf"%d%d",

　　Feel the whole bzoj do not have a future, a few Uncle Gan SRM, I will open a pit to mouth Hu Yipo cf/srm in order to quickly roll thick. (because it is the mouth of the Hu do not set the counter, in fact, I am lazy.)"SRM645" easy to sort by the sorting line directly to sweep it again."SRM645" Medium if the direction vector of each point is the same then it may be able to reach otherwise certainly not. ① If even step, there are two kinds of vector sy

);}BOOLCheck2 () { for(inti =1; I Str. Size (); i++) {if(Str[I] = =StrI1]) {Str= Spilt (I-1, i);return true; } }return false;}BOOL Check(intP) {intCNT =0; for(inti =2; I Str. Size (); i++) {if(i-cnt = = p) {Str= Spilt (CNT, I-1); CNT = i;return true; }if(Str[i]! =StrI2]) {cnt = i-1; } }if(Str. Size ()-cnt = = p) {Str= Spilt (CNT,Str. Size ()-1);return true; }return false;}intMain () {intT_case; scanf"%d", t_case); for(intI_case =1; I_case Str; while(Check2 () | |Check(8)); while(Check(7))

) {scanf ("%d",Node[i].hei); } node[0].hei =0; Node[n].hei=0; for(intI=1; i) {NODE[I].L= i-1; while(I >0 Node[node[i].l].hei >=Node[i].hei) {NODE[I].L=NODE[NODE[I].L].L; } } for(inti=n-1;i>0; i--) {NODE[I].R= i +1; while(I Node[i].hei) {NODE[I].R=NODE[NODE[I].R].R; }} memset (ans,-1,sizeofans); Sort (Node+1, node+n,cmp); for(intI=1; i){ intpos = NODE[I].R-NODE[I].L-1; Ans[pos]=Max (Ans[pos],node[i].hei); } for(inti=n-1;i>0; i--) {Ans[i]= Max (ans[i],ans[i+1]); } for(int

conditions is fulfilled: Each bracket is either isn't colored any color, or is colored red, or is colored blue. For any pair of matching brackets exactly one of the them is colored. In other words, for any bracket the following are true:either it or the matching bracket that corresponds to it is colored . No. Neighboring colored brackets has the same color. Find the number of different ways to color the bracket sequence. The ways should meet the above-given conditions. The

is:After the first visit of cells (2,?2) the ice on it cracks and when you step there for the second time, your Chara Cter falls through the ice as intended.Test instructions: The map of N*m, ' X ' represents a cracked ice, '. ' Represents the complete ice, the cracked ice will be broken once again, the complete ice will be trampled to become a cracked ice, now tell the starting point and the end, ask whether from the starting point can go to the end and the end of the ice to break down. Can't

A-combination LockThe main topic: To have N (0-9) ring password lock, a few strings s1->s2 the minimum number of moves;Topic Analysis:Simple simulation;Code:const int N=100007;char s1[n],s2[n];int Main () { int N; while (scanf ("%d", n) ==1) { int ans=0; scanf ("%s%s", s1+1,s2+1); for (int i=1;iB-school MarksThe main topic: N number, known as K, the median >=y; and all the number sumTopic Analysis:Because sum is less than X, try to fill in decimals. Mid left Fill 1,

is:After the first visit of cells (2,?2) the ice on it cracks and when you step there for the second time, your Chara Cter falls through the ice as intended.Links: Http://codeforces.com/problemset/problem/540/CTest instructions: x is the Pit, '. ' It's ice, and the ice goes through once and then into a pit. Ask if you can start from Sx,sy and fall into the pit at Ex,ey.Practice: BFs A bit, x if it is not the end, it must not go. If it is an endpoint, and it is X, you can return directly. If it

Software Introduction: Just enter your QQ number can make your own brush gun, brush drill software no longer need to envy others software author qq:646234584 "Let's sell CF plug-in to make a lot of money, this soft no use can only make money!!Software of the luxury interface, please look down!!Builder interface650) this.width=650; "src=" http://s3.51cto.com/wyfs02/M01/6E/67/wKioL1V7qabzMpdSAAEzNI8wCEM446.jpg "title=" 5.jpg " alt= "Wkiol1v7qabzmpdsaaez