HDU 4630-no Pain No Game (segment tree + offline processing)

Test instructions

Give you a sequence of n number a,q a query, each query to L,r, for the subscript i in [l,r] the interval of any two number of greatest common divisor in the maximum value

Analysis:

With hdu3333 experience, we sweep the sequence from left to right, and if the current number of approximations appears before, then this approximate may be the biggest answer. So we enumerate all the approximate numbers of the current number, maintain interval maximums with the segment tree, and query sequences are processed offline to ensure that queries are correct.

#include <map>#include<Set>#include<list>#include<cmath>#include<queue>#include<stack>#include<cstdio>#include<vector>#include<string>#include<cctype>#include<complex>#include<cassert>#include<utility>#include<cstring>#include<cstdlib>#include<iostream>#include<algorithm>using namespaceStd;typedef pair<int,int>Pii;typedefLong Longll;#defineLson l,m,rt<<1#definePi ACOs (-1.0)#defineRson m+1,r,rt<<1|1#defineAll 1,n,1#defineN 50010#defineRead Freopen ("In.txt", "R", stdin)Constll infll =0x3f3f3f3f3f3f3f3fll;Const intinf=0x7ffffff;Const intMoD =1000000007;//bef the position of the current number in the previous occurrenceintmaxv[n*4],n,bef[n],e,a[n],maxn[n];structque{intL,r,id;} Q[n];BOOLcmp (que x,que y) {returnx.r<Y.R;}voidPushup (intRT) {Maxv[rt]=max (maxv[rt<<1],maxv[rt<<1|1]);}voidBuildintLintRintRT) {Maxv[rt]=1; if(L==R)return; intM= (l+r) >>1;    Build (Lson); Build (Rson);}voidUpdateintPintVintLintRintRT) {    if(l==R) {Maxv[rt]=Max (MAXV[RT],V); return; }    intM= (l+r) >>1; if(p<=m) update (P,v,lson); ElseUpdate (P,v,rson); Pushup (RT);}intQueryintLintRintLintRintRT) {    if(l<=l&&r>=R) {        returnMaxv[rt]; }    intM= (l+r) >>1; intnum=0; if(l<=m) num=Max (Num,query (L,r,lson)); if(r>m) num=Max (Num,query (L,r,rson)); returnnum;}voidsolve () {sort (q+1, q+e+1, CMP); memset (BEF,0,sizeof(BEF)); Build (1N1); intb=1;  for(intI=1; i<=n;++i) {         for(intj=1; j*j<=a[i];j++){            if(a[i]%j==0){                if(Bef[j]) {update (BEF[J],J,1N1); }                inttmp=a[i]/J; if(bef[tmp]&&tmp!=j) Update (BEF[TMP],TMP,1N1); BEF[J]=i; BEF[TMP]=i; }        }         while(b<=e&&q[b].r==i) {            if(q[b].l==Q[B].R) maxn[q[b].id]=0; Else{Maxn[q[b].id]=query (Q[B].L,Q[B].R,1N1); } b++; }    }     for(intI=1; i<=e;++i) printf ("%d\n", Maxn[i]);}intMain () {intT; scanf ("%d",&t);  while(t--) {scanf ("%d",&N);  for(intI=1; i<=n;++i) scanf ("%d",&A[i]); scanf ("%d",&e);  for(intI=1; i<=e;++i) {scanf ("%d%d",&q[i].l,&Q[I].R); Q[i].id=i;    } solve (); }return 0;}

HDU 4630-no Pain No Game (segment tree + offline processing)