2016 "Baidu Star"-Preliminary round (Astar round2a)

title Link:http://acm.hdu.edu.cn/search.php?field=problem&key=2016%26quot%3B%B0%D9%B6%C8%D6%AE%D0%C7% 26quot%3b+-+%b3%f5%c8%fc%a3%a8astar+round2a%a3%a9&source=1&searchmode=source

1001:

Matrix Fast Power

#include <iostream>#include<algorithm>#include<cstdio>#include<cstring>#include<cmath>#include<queue>#include<vector>using namespacestd;Const intN = 5e5+ -, M = 1e2+Ten, mod =1000000007, INF =1e9;typedefLong Longll;structMatix {ll arr[m][m];}; ll X,m,k,c;    Matix Mul (Matix A,matix b,ll hang, ll lie) {Matix ans; memset (Ans.arr,0,sizeof(Ans.arr));  for(intI=1; i<=hang;i++) {       for(intt=1; t<=lie;t++)         for(intj=1; j<=lie;j++) {Ans.arr[i][t]+ = (a.arr[i][j]*b.arr[j][t]); Ans.arr[i][t]%=K; }    }    returnans;} Matix Pow (Matix Ans,matix a,ll x) { while(x) {if(x&1) Ans=mul (Ans,a,2,2); A=mul (A,a,2,2); X/=2; }    returnans;}intMain () {ll x, y, Z; intt,cas=1; scanf ("%d",&T);  while(t--) {scanf ("%i64d%i64d%i64d%i64d",&x,&m,&k,&c);        Matix fir,sec;        Matix now; now.arr[1][1]=x; now.arr[1][2]=x; memset (Fir.arr,0,sizeof(Fir.arr)); sec.arr[1][1]=Ten; sec.arr[1][2]=0; sec.arr[2][1]=1; sec.arr[2][2]=1; fir.arr[1][1]=1; fir.arr[2][2]=1; Fir=pow (fir,sec,m-1); Fir=mul (Now,fir,2,2); printf ("Case #%d:\n", cas++); if(fir.arr[1][1]==c) puts ("Yes"); ElsePuts"No"); }    return 0;}

View Code

1005:

Pretreatment +dfs

Preprocessing each length, the answer of each length

Then find out the answer to the corresponding length Dfs

#include <iostream>#include<algorithm>#include<cstdio>#include<cstring>#include<cmath>#include<queue>#include<vector>using namespacestd;Const intN = 1e3+ -, M = 1e6+Ten, mod =1000000007, INF =1e9;typedefLong LongLl;ll len[n],dp[n],l,r;ll dfs (ll x) {if(x<=0)return 0; if(x<=2)returnx; if(x==3)return 2; ll ans=0; intpos = Upper_bound (len+1, len+ +, x)-Len-1; if(len[pos]==x)returnDp[pos]; if(x-len[pos]==1)returndp[pos]+1; LL ret= dp[pos]+1; ll Shen= x-len[pos]-1; returnRet+shen-dp[pos] + DFS (len[pos)-shen);}intMain () {dp[1] =1; len[1] =1;  for(intI=2; i<= A; i++) {Len[i]= len[i-1]*2+1; }     for(intI=2; i<= A; i++) {Dp[i]= dp[i-1] +1+ len[i-1]-dp[i-1]; }    intT; scanf ("%d",&T);  while(t--) {scanf ("%i64d%i64d",&l,&s); printf ("%i64d\n", DFS (R)-dfs (l1)); }    return 0;}

View Code

1006

Priority queue

First, according to the topology, the penetration is 0 of the press in, the priority position of the large front ...

#include <iostream>#include<algorithm>#include<cstdio>#include<cstring>#include<cmath>#include<queue>#include<vector>using namespacestd;Const intN = 5e5+ -, M = 1e6+Ten, mod =1000000007, INF =1e9;typedefLong Longll;intK,c,x,d[n],n,m,vis[n];vector<int>G[N];intAns[n];intMain () {intT; intCAS =1; scanf ("%d",&T);  while(t--) {scanf ("%d%d",&n,&m);  for(intI=1; i<=n;i++) g[i].clear ();  for(intI=1; i<=n;i++) Vis[i] =0, d[i] =0;  for(intI=1; i<=m;i++) {            intb; scanf ("%d%d",&a,&b);            G[a].push_back (b); D[B]++; } priority_queue<int>Q;  for(intI=1; i<=n;i++) {            if(d[i]==0) Q.push (i); }        intCNT =0 ;  while(!Q.empty ()) {            intK =Q.top (); ans[++CNT] =K; //cout<<k<<endl;Q.pop ();  for(intI=0; I<g[k].size (); i++) {D[g[k][i]]--; if(d[g[k][i]]==0&&!Vis[g[k][i]])                    {Q.push (g[k][i]); Vis[g[k][i]]=1; }            }        }        intMi =1e9; LL AA=0;  for(intI=1; i<=cnt;i++) {mi=min (ans[i],mi); AA= AA +mi; } printf ("%i64d\n", AA); }    return 0;}

View Code

2016 "Baidu Star"-Preliminary round (Astar round2a)