hearthstone goblin

problem DescriptionCDFPYSW loves playing a card game called "Hearthstone". Now he had N cards, he wants to split these cards into 4 piles.Let's assume the number of cards in each pile is A1, A2, A3, A4. It must be satisfied that:a1 * K1 = a2 + a3 + a4 A2 * k2 = a1 + A3 + a4 a3 * K3 = A1 + a2 + A4 A1, a2, A3, A4 Must be positive Because cdfpysw are clever, there must be a-to split there card. Can you tell CDFPYSW?InputThe first line was an integer T,

,0,sizeof(DP)); LL p,n,m; scanf ("%i64d%i64d%i64d",p,n,m); LL N=n+m; for(LL i=n;i) scanf ("%i64d",Val[i]); dp[0]=1; for(LL st=0;st1) { if(dp[st]==0)Continue; LL Dam=0, num_a=0, num_b=0; for(LL i=n;i) { if(st (1; } if(dam>=p)Continue; for(LL i=0; i) { if(st (1; } if(num_a+1Continue; for(LL i=0; i) { if(st (1Continue; Dp[st+(1Dp[st]; }} LL ans=0, all=Fac[n]; for(LL st=

of your turn, you draw a-card from the top of the the card deck. You can use any of the cards in your hands until to run out of it. Your task is to calculate the probability so you can win in this turn, i.e., can deal at least P damage to Your enemy.Inputthe first line is the number of test cases T (tThen come three positive integers P (pOutputfor each test case, output the probability as a reduced fraction (i.e., the greatest common divisor of the numerator and denominator is 1). If the answer

Fifth session of the provincial race: HearthstoneCombinatorial mathematics.n Races, M tables (n>=m). Each race is a table, and each table is used at least once.The idea behind the question has been how to fill the M table with n positions.is actually the Tao Stirling number model, the direct set formula M!*{n m}#include #includestring.h>#defineL 1000000007intMain () {intn,m; Long Longa[101]; while(SCANF ("%d%d", n,m)! =EOF) {a[0]=0; for(intI=1; i) A[i]=1; for(intI=3; i) for(intj=i

] = 1;for (int i = 2; I Swap (c, p);memset (Dp[c], 0, sizeof dp[c]);For (int j = 1; J Dp[c][j] = dp[c][j-1] + dp[p][i-j + 1];if (Dp[c][j] >= MOD)Dp[c][j]-= MOD;}}int ans = 0;for (int i = 1; I if (ans + = dp[c][i]) >= mod) ans-= mod;printf ("%d\n", (ans return 0;}-------------------------------------------------------------------------- 1925: [Sdoi2010] Goblin Tribe Time Limit:Ten Sec Memory Limit:MB Submit:965 Solved:580 [Submit] [Sta

J-Goblin WarsTime limit:432MS Memory Limit:1572864KB 64bit IO Format:%lld A mp %llu SubmitStatusPracticeSpoj amr11jDescriptionThe Wizards and Witches of Hogwarts School of witchcraft found Prof. Binn ' s History of Magic lesson to being no less boring t Han you found your own classes. Recently Binns have been droning on about Goblin wars, and which Goblin civili

Description legend a long time ago, the Earth lived a mysterious creature: Goblin. Goblins like to live in the Endless Mountains. Specifically, an n-length mountain range h can be divided into n segments from left to right, each with a unique height of hi, where hi is a positive integer between 1 and N. If a mountain range is higher than all the mountains adjacent to it, the mountain range is a mountain. The mountains at the edge have only one contigu

break it up and readThis is a very casual sort of name algorithm, it's me I'll call him slime sort sunglass (▔,▔) denyPrincipleThe sort of goblin is also a sort of exchange. It only makes a comparison, in this round comparison, encountered the comparison of the previous element large on the back of a continuation of the comparison, encountering a smaller than the previous face value of the exchange, and move forward one bit.Complexity ofFor the queue

the descent into ascending by the nature of the beginning. If the initial value range is j+1 ~ I then through N-xi + 1 conversion, the beginning of the range of values becomes 1 ~ i-j. (If the previous representation is rising, there will be a problem here.) So it is obvious that f[i-1][i-j], as to why the i-1, because the length of the sequence is only i-1. That's it.1#include 2#include 3 #defineRep (i,j,k) for (register int i = j; I 4 using namespacestd;5 6InlineintRead () {7 ints =0, t =

= 0]After Exchange [1 2 4 6 5 9][i = 1][i = 1] Compare 1 and 2, do not exchange, as long as not exchange on the self-increment 1[i = 2] Compare 2 and 4, do not exchange, as long as not exchange on the self-increment 1[i = 3] Compare 4 and 6, do not exchange, as long as not exchange on the self-increment 1[i = 4] Compare 6 and 5, exchange, as long as the exchange is reduced by 1Before Exchange [1 2 4 6 5 9][i = 4]After Exchange [1 2 4 5 6 9][i = 3][i = 3] Compare 4 and 5, do not exchange, as

Spoj Goblin Wars (simple bfs), spojbfsJ-Goblin WarsTime Limit:432 MSMemory Limit:1572864KB64bit IO Format:% Lld % lluSubmitStatusPracticeSPOJ AMR11J Description The wizards and witches of Hogwarts School of Witchcraft found Prof. binn's History of Magic lesson to be no less boring than you found your own history classes. recently Binns has been droning on about Goblin