Bestcoder Round #69 (Div.2)

A.geometry

#include <cstdio>#include<vector>#include<cstring>#include<iostream>#include<algorithm>using namespacestd;intMain () {//freopen ("In.txt", "R", stdin);    intT;SCANF ("%d",&T);  while(t--){        intx, y; scanf ("%d%d",&x,&y); cout<<abs (2*x*y) <<Endl; }    return 0;}

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B.tree

Idea: The closest point to each point is necessarily itself, the distance is unique. So you just have to pass all the points that are 0 from each other and check the set to shrink the group. Initially wanted to write a Dfs timeout, and then changed and check set, because the brain residue reasons WA a few hair.

#include <cstdio>#include<vector>#include<cstring>#include<iostream>#include<algorithm>using namespacestd;intpar[100100];intcnt[100100];intFindintx) {    returnPar[x]= (X==par[x]?X:find (par[x]));}intMain () {//freopen ("In.txt", "R", stdin);    intT;SCANF ("%d",&T);  while(t--){        intN;SCANF ("%d",&N);  for(intI=0; i<=n;i++) {Par[i]=i; Cnt[i]=1; }         for(intI=0; i<n-1; i++){            intu,v,w; scanf ("%d%d%d",&u,&v,&W); if(!W) {Par[v]=find (U); CNT[PAR[V]]+=Cnt[v]; }        }         for(intI=1; i<=n;i++) Cnt[i]=Cnt[par[find (i)]; intans=cnt[1];  for(intI=2; i<=n;i++) ans^=Cnt[i]; printf ("%d\n", ans); }    return 0;}

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C.graph

Idea: Before doing a similar, randomly select a point on the chart, and so on the probability to go to the point adjacent to him, seeking K step after his probability at each point. The probability transfer equation for this problem is the same as the previous one, just because the point is less than 20, so it becomes the form of the matrix, and then it is good to calculate by the matrix fast power. It is worth mentioning that the quick power template has problems that lead to WA. There is also a theorem that does not know number theory. Thanks for the chicory in the BC Group.

#include <cstdio>#include<vector>#include<cstring>#include<iostream>#include<algorithm>using namespaceStd;typedefLong Longll;Constll mod=1e9+7;Const intmaxn= -;//Order of matricesBOOLg[ -][ -];inttot[ -];structmatrix{ll M[MAXN][MAXN]; Matrix () {memset (M,0,sizeof(m));  for(intI=0; i<maxn;i++) M[i][i]=1; }    voidClear () {memset (M,0,sizeof(m)); }    voidShow () { for(intI=0; i<maxn;i++){             for(intj=0; j<maxn;j++) cout<<m[i][j]<<" "; cout<<Endl; }    }}; Matrix Mtmul (Matrix A,matrix B) {inti,j,k,tmp;    Matrix C;    C.clear ();  for(intI=0; i<maxn;i++)         for(intj=0; j<maxn;j++)             for(intk=0; k<maxn;k++) C.m[i][j]= (c.m[i][j]%mod+ (a.m[i][k]%mod) * (b.m[k][j]%mod))%MOD; returnC;} Matrix Mtpow (Matrix A,intk) {k--; Matrix ans=A;  while(k) {if(k&1) ans=Mtmul (ans,a); A=Mtmul (a,a); K>>=1; }    returnans;} ll Quick_pow (ll X,intm) {ll sum=1;  while(m) {if(m&1) sum= ((sum%mod) * (x%mod))%MOD; X= ((x%mod) * (x%mod))%MOD; M>>=1; }    returnsum;}intMain () {//freopen ("In.txt", "R", stdin);    intn,m;  while(cin>>n>>m) {        intu,v;        Matrix M;        M.clear (); memset (G,0,sizeof(G)); memset (Tot,0,sizeof(tot));  while(m--) {cin>>u>>v; Tot[u-1]++; G[v-1][u-1]=true; }         for(intI=0; i<n;i++)             for(intj=0; j<n;j++)                if(G[i][j]) {M.m[i][j]= ((M.m[i][j])%mod+quick_pow (tot[j],1e9+5)%mod)%MOD; }        intQ; CIN>>p;  while(q--){            intu,k; CIN>>u>>K; Matrix ans=Mtpow (m,k); //ans.show ();             for(intI=0; i<n;i++) cout<<ans.m[i][u-1]<<" "; cout<<Endl; }    }    return 0;}

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Bestcoder Round #69 (Div.2)