BZOJ4373: Arithmetic genius ⑨ and arithmetic progression

Description

Arithmetic genius ⑨ like to play with arithmetic progression very much.
One day, he gave you a sequence of length n, where the number of I is a[i].
He wants to test you, every time he will give an inquiry l,r,k, ask the interval [l,r] in the number from small to large, can form a tolerance of k arithmetic progression.
Of course, he will constantly revise one of these items.
In order not to be despised by him, you have to answer all the questions quickly and correctly.
Note: Only one number of the series is also arithmetic progression.

Input

The first line contains two positive integer n,m (1<=n,m<=300000), each representing the length of the sequence and the number of operations.
The second line contains n integers, which in turn represent each number in the sequence A[i] (0<=a[i]<=10^9).
Next m line, each line starts with a number OP,
If Op=1, then the next two integers, X, y (1<=x<=n,0<=y<=10^9), indicate A[X] is modified to Y.
If op=2, then the next three integers, l,r,k (1<=l<=r<=n,0<=k<=10^9), represent a query.
In the subject, the x,y,l,r,k are encrypted and need to be decrypted by different or the number of yes you have previously output.

Output

Output a number of lines, for each query, if you can form arithmetic progression, then output yes, otherwise output No.

Sample Input5 3
1 3 2) 5 6
2 1 5 1
1 5 4
2 1 5 1Sample OutputNo
YesWOC original line segment tree to open 4 times times space, then I noi when is how to come over?????? OrzQiancl Big Uncle Hash violence Dafa Good ... The http://qiancl.is-programmer.com/posts/192245.html is very detailed.

#include <cstdio>#include<cctype>#include<queue>#include<cmath>#include<cstring>#include<algorithm>#defineRep (i,s,t) for (int i=s;i<=t;i++)#defineDwn (i,s,t) for (int i=s;i>=t;i--)#defineren for (int i=first[x];i;i=next[i])using namespacestd;Const intBuffersize=1<< -;Charbuffer[buffersize],*head,*Tail;inlineCharGetchar () {if(head==tail) {        intL=fread (Buffer,1, Buffersize,stdin); Tail= (Head=buffer) +l; }    return*head++;} InlineintRead () {intx=0, f=1;CharC=Getchar ();  for(;! IsDigit (c); C=getchar ())if(c=='-') f=-1;  for(; IsDigit (c); C=getchar ()) x=x*Ten+c-'0'; returnx*F;} typedef unsignedLong Longull;Const intmaxn=300010; ull SUMV[MAXN*4],sum,mn;intminv[maxn*4],lastans;voidBuildintOintLintr) {if(l==r) Minv[o]=read (), sumv[o]= (ull) minv[o]*Minv[o]; Else {        intMid=l+r>>1,lc=o<<1, rc=lc|1; Build (Lc,l,mid); Build (Rc,mid+1, R); Minv[o]=min (MINV[LC],MINV[RC]); Sumv[o]=sumv[lc]+SUMV[RC]; }}voidUpdateintOintLintRintPOS) {    if(l==r) Minv[o]=read () ^lastans,sumv[o]= (ull) minv[o]*Minv[o]; Else {        intMid=l+r>>1,lc=o<<1, rc=lc|1; if(pos<=mid) Update (LC,L,MID,POS); ElseUpdate (rc,mid+1, R,pos); Minv[o]=min (MINV[LC],MINV[RC]); Sumv[o]=sumv[lc]+SUMV[RC]; }}voidQueryintOintLintRintQlintqr) {    if(QL&LT;=L&AMP;&AMP;R&LT;=QR) sum+=sumv[o],mn=min (Mn, (ull) minv[o]); Else {        intMid=l+r>>1,lc=o<<1, rc=lc|1; if(ql<=mid) query (LC,L,MID,QL,QR); if(qr>mid) query (rc,mid+1, R,QL,QR); }}intMain () {intN=read (), m=read (); Build (1,1, N);  while(m--) {        if(read () = =1) Update (1,1, N,read () ^Lastans); Else {            intL=read () ^lastans,r=read () ^lastans,v=read () ^Lastans; Ull Len=r-l+1; mn=2e9;sum=0; Query (1,1, N,l,r); ull Res=len*mn* (mn+ (len-1) *v) +len* (len-1)*(2*len-1)/6*v*v; if(Res==sum) puts ("Yes"), lastans++; ElsePuts"No"); }    }    return 0;}

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BZOJ4373: Arithmetic genius ⑨ and arithmetic progression