914D Bash and a Tough Math Puzzle

Transmission Door

Analysis

Using the segment tree to maintain the interval GCD, each query finds the first point that is not a multiple of x, and if there is an interval that gcd cannot be divisible by x, the interval is illegal.

Code

#include <iostream>#include<cstdio>#include<cstring>#include<string>#include<algorithm>#include<cctype>#include<cmath>#include<cstdlib>#include<queue>#include<ctime>#include<vector>#include<Set>#include<map>#include<stack>using namespacestd;intd[2000100],cnt,a[500100];inlineintgcdintXintY) {returny==0? X:GCD (y,x%y);} InlinevoidBuildintLeintRiintwh) {      if(le==RI) {D[WH]=A[le]; return; }      intMid= (Le+ri) >>1; Build (Le,mid,wh<<1); Build (Mid+1,ri,wh<<1|1); D[WH]=GCD (d[wh<<1],d[wh<<1|1]); return;} InlinevoidUpdateintLeintRiintWhintPlintk) {      if(le==RI) {D[WH]=K; return; }      intMid= (Le+ri) >>1; if(MID&GT;=PL) Update (le,mid,wh<<1, pl,k); ElseUpdate (mid+1,ri,wh<<1|1, pl,k); D[WH]=GCD (d[wh<<1],d[wh<<1|1]); return; } InlinevoidQintLeintRiintWhintXintYintk) {      if(le>=x&&ri<=y) {          if(d[wh]%k==0)return; if(cnt>=1) {CNT++; return; }      }      if(le==RI) {CNT++; return; }      intMid= (Le+ri) >>1; if(mid>=x) Q (le,mid,wh<<1, x,y,k); if(mid<y) Q (mid+1,ri,wh<<1|1, x,y,k); return;}intMain () {intn,m,i,j,k,x,y,z; scanf ("%d",&N);  for(i=1; i<=n;i++) scanf ("%d\n",&A[i]); Build (1N1); scanf ("%d",&m);  for(i=1; i<=m;i++) {scanf ("%d",&k); if(k==1) {scanf ("%d%d%d",&x,&y,&z); CNT=0; Q (1N1, x, Y, z); if(cnt>1) puts ("NO"); ElsePuts"YES"); }Else{scanf ("%d%d",&x,&y); Update (1N1, x, y); }      }      return 0;}

914D Bash and a Tough Math Puzzle