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common multiple { for(j=1; j +; j + +) {Lcm[i][j]=LCM (I,J); } } while(SCANF ("%d%d%d", n,m,k)! =EOF) { intCnt=0; //because least common multiple m is known, the AI must be his factor. for(i=1; i) { if(m%i==0) num[cnt++]=i; } //Dp[now][i][j]now represents the number of solutions for the current state, and for I, least common multiple J. Recursion k times out the answer. intnow=0; //memset (dp[nom],0,sizeof (dp[nom)); for(

details in the sample and hint below.Sample Input4 2 23) 2 2Sample Output12Test instructionsGive n,m,k, ask K number of and is n, least common multiple for m of case there are severalIdeas:Because least common multiple is m, you can know that these numbers are necessarily m factors, then we just need to select all of these factors, take these factors to the backpack can beDp[i][j][k] indicates that the number of I has been placed, and the case of J, common multiple K has severalBut the problem,

+1][j+1]+=c[i][j],c[i+1][j]+=C[i][j]; - while(t--) { -N=read (); K=read (); cnt=0; -memset (Size,0,sizeof(size)); -memset (Vis,0,sizeof(Vis)); -Rep (I,1, N) v[i]=read (); inRep (I,1, N)if(!Vis[i]) { -cnt++;intj=i; to Dosize[cnt]++,vis[j]=1, J=v[j]; while(j!=i); + } -Memset (F,0,sizeof(f)); thef[1][0]=1.0;intCur=0; *Rep (I,1, CNT) { $Rep (J,0, cur) rep (k0,1, Size[i]) f[i+1][j+k0]+=f[i][j]*C[size[i]][k0];Panax Notoginsengcur+=Size[i]; - } theprintf"%.6lf\n", f[cn

that each row has 1, which is the premise. And to get the answer, we can enumerate I .columnis 0 (total C (m,i) selection), then consider the remaining (m-i) positions of each line, weyou can nowto arbitrarily place the 2^ (m-i), but in order to ensure that each line has 1 to reduce 1, this timefor (2^ (m-i)-1).Then there are n rows, so sum[i]= (2^ (m-i)-1) ^n, and then we're going fromThere are some things out there that areonly column I is all 0 (note that just I is not the only column)so let

%MoD; //tmp = KSC (tmp,i);Up (ret,tmp); TMP= (sum[3]-ask (3, POS) +mod)%MoD; TMP= (mod-tmp)%MoD; Up (RET,TMP); /*********/} init (); PTR=m; for(intI=n; I --i) {intPOS; for(; ptr>iptr;--ptr) {POS= Lower_bound (p+1, p+1+CNT,B[PTR])-p; Add (0Pos1); ++sum[0]; Add (1, pos,ptr); Up (sum[1],ptr); Add (2, Pos,b[ptr]); Up (sum[2],b[ptr]); Add (3, pos,1ll*b[ptr]*ptr%mod); Up (sum[3],1ll*b[ptr]*ptr%MoD); } /**j>i,b[j]>a[i]**/POS= Upper_bound (p+1, p+1+cnt,a[i])-p; --POS; if(pos!=CNT

/************************************************* author:running_time* Created time:2015/10/28 Wednesday 20:20:09* Fi Le name:h.cpp ************************************************/#include 　　DP (optimized) uvalive 6073 Math Magic