One day Kefa the parrot was walking down the street as he was on the way home from the restaurant when he saw something glittering by the road. As he came nearer he understood that it was a watch. He decided to take it to the pawnbroker to earn some money.
The pawnbroker said that each watch contains a serial number represented by a string of digits from 0 to 9, and the more quality checks this number passes, the higher is the value of the watch. The check is defined by three positive integers l, r and d. The watches pass a check if a substring of the serial number from l to r has period d. Sometimes the pawnbroker gets distracted and Kefa changes in some substring of the serial number all digits to c in order to increase profit from the watch.
The seller has a lot of things to do to begin with and with Kefa messing about, he gave you a task: to write a program that determines the value of the watch.
Let us remind you that number x is called a period of string s (1 ≤ x ≤ |s|), if si = si + x for all i from 1 to |s| - x.
The first line of the input contains three positive integers n, m and k (1 ≤ n ≤ 105, 1 ≤ m + k ≤ 105) — the length of the serial number, the number of change made by Kefa and the number of quality checks.
The second line contains a serial number consisting of n digits.
Then m + k lines follow, containing either checks or changes.
The changes are given as 1 l r с (1 ≤ l ≤ r ≤ n, 0 ≤ c ≤ 9). That means that Kefa changed all the digits from the l-th to the r-th to be c.
The checks are given as 2 l r d (1 ≤ l ≤ r ≤ n, 1 ≤ d ≤ r - l + 1).
For each check on a single line print "YES" if the watch passed it, otherwise print "NO".
3 1 2
112
2 2 3 1
1 1 3 8
2 1 2 1
NO
YES
6 2 3
334934
2 2 5 2
1 4 4 3
2 1 6 3
1 2 3 8
2 3 6 1
NO
YES
NO
In the first sample test two checks will be made. In the first one substring "12" is checked on whether or not it has period 1, so the answer is "NO". In the second one substring "88", is checked on whether or not it has period 1, and it has this period, so the answer is "YES".
In the second statement test three checks will be made. The first check processes substring "3493", which doesn't have period 2. Before the second check the string looks as "334334", so the answer to it is "YES". And finally, the third check processes substring "8334", which does not have period 1.
这里的hash函数定义H[i]=s[n-1]*x^(n-1-i)+s[n-2]*x^(n-2-i)+...+s[i]，那么对于一段长度为L的子串s[i]~s[i+L-1]，它的hash值就可以表示为
H[i]-H[i+L]这个值只和它本身有关。
现在是在线段树上维护的话，按照定义有H[root] = H[lch]+H[rch]*x^(len(lch))。
查询一个子串的hash值类似处理。
题目的要求有一个全部设置为c，这个操作只需要预处理出各种长度下的1+x+x^2+..+x^len值。
第一次写，各种写挂。没模素数，单hash挂在75组数据，双hash换了几组key才过。。。事实证明还是模素数稳。
这题还可以memset memcmp 可以水过去。。。这种常数很小的复杂度算出来除个10和一般的差不多。
#include<bits/stdc++.h> using namespace std; typedef unsigned long long ull; const int Key[2] = {666,233};//{799817,451309};//{ 1000003,999997 }; const int maxn = 1e5+50; char s[maxn];//定义成了char 型的slen 导致RE int slen; struct Seg { int len; ull Hash[2]; int Set; }tr[maxn<<2]; ull sHash[maxn][2];//set related ull pKey[maxn][2]; inline void push_down(Seg&u,Seg&c1,Seg&c2) { if(~u.Set){ c1.Set = c2.Set = u.Set; c1.Hash[0] = c1.Set*sHash[c1.len][0]; c2.Hash[0] = c2.Set*sHash[c2.len][0]; c1.Hash[1] = c1.Set*sHash[c1.len][1]; c2.Hash[1] = c2.Set*sHash[c2.len][1]; u.Set = -1; } } inline void push_up(Seg&u,Seg&c1,Seg&c2) { u.Hash[0] = c1.Hash[0] + c2.Hash[0]*pKey[c1.len][0]; u.Hash[1] = c1.Hash[1] + c2.Hash[1]*pKey[c1.len][1]; } #define lid (id<<1) #define rid (id<<1|1) void build(int id = 1,int l = 1,int r = slen) { tr[id].Set = -1; if(l == r) { tr[id].len = 1; tr[id].Hash[0] = tr[id].Hash[1] = s[l]-'0'; }else { int M = (l+r)>>1,lc = lid, rc = rid; build(lc,l,M); build(rc,M+1,r); tr[id].len = tr[lc].len+tr[rc].len; push_up(tr[id],tr[lc],tr[rc]); } } int ql,qr,val; void updata(int id = 1,int l = 1, int r = slen) { if(ql<=l&&r<=qr) { tr[id].Set = val;//设置标记的同时就应该把值修改好 tr[id].Hash[0] = val*sHash[tr[id].len][0]; tr[id].Hash[1] = val*sHash[tr[id].len][1]; }else { int M = (l+r)>>1, lc = lid, rc = rid; push_down(tr[id],tr[lc],tr[rc]); if(ql<=M) updata(lc,l,M); if(qr>M) updata(rc,M+1,r); push_up(tr[id],tr[lc],tr[rc]); } } #define PB push_back #define MP make_pair #define fi first #define se second pair<ull,ull> query(int id = 1,int l = 1, int r = slen) { if(ql<=l&&r<=qr) { return pair<ull,ull>(tr[id].Hash[0],tr[id].Hash[1]); }else { int M = (l+r)>>1, lc = lid, rc = rid; push_down(tr[id],tr[lc],tr[rc]); if(qr<=M) return query(lc,l,M); if(ql>M) return query(rc,M+1,r); auto lq = query(lc,l,M), rq = query(rc,M+1,r); int llen = M-max(ql,l)+1;//取决于左边，要取max。。。 return pair<ull,ull>(lq.fi+rq.fi*pKey[llen][0],lq.se+rq.se*pKey[llen][1]); } } void init() { pKey[0][0] = pKey[0][1] = 1ull; for(int i = 1; i < slen; i++){ pKey[i][0] = pKey[i-1][0]*Key[0]; pKey[i][1] = pKey[i-1][1]*Key[1]; } sHash[1][0] = sHash[1][1] = 1ull; for(int i = 2; i <= slen; i++){ sHash[i][0] = sHash[i-1][0]+pKey[i-1][0]; sHash[i][1] = sHash[i-1][1]+pKey[i-1][1]; } build(); } int main() { //freopen("in.txt","r",stdin); int m,k; scanf("%d%d%d",&slen,&m,&k); if(!k) return 0; scanf("%s",s+1); init(); k += m; while(k--){ int op,l,r,x; scanf("%d%d%d%d",&op,&l,&r,&x); if(op == 1){ ql = l; qr = r; val = x; updata(); }else { int dlen = r-l-x; if(dlen<0) { puts("YES"); continue; }//长度为0 ql = l; qr = l+dlen; auto h1 = query(); ql = l+x; qr = l+x+dlen; puts(h1 == query()?"YES":"NO"); } } return 0; }