Codeforces Round #321 (Div. 2) E

终于补好了。

题目链接: http://codeforces.com/contest/580/problem/E

E. Kefa and Watch

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 s_i  =  s_i + x for all i from 1 to |s|  -  x.

Input

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).

Output

For each check on a single line print "YES" if the watch passed it, otherwise print "NO".

Sample test(s)

Input

3 1 2
112
2 2 3 1
1 1 3 8
2 1 2 1

Output

NO
YES

Input

6 2 3
334934
2 2 5 2
1 4 4 3
2 1 6 3
1 2 3 8
2 3 6 1

Output

NO
YES
NO

Note

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.

大意是：给一个字符串，修改一个区间都为一个值，区间询问区间的两端是否完全相等。

当然HASH啦，但是类似自然溢出会被卡（CF有特殊的卡HASH的技巧，自然溢出就是用unsigned long long 这种，相当于mod 2^64)

这里用双hash，貌似很难很难卡。

HASH1=10^9+9,HASH2=10^9+7

然后就是区间修改，和区间询问，区间修改学的差，看了好久才懂

  1 #include<bits/stdc++.h>
  2 
  3 using namespace std;
  4 
  5 #define N 100005
  6 #define mod1 1000000009
  7 #define mod2 1000000007
  8 
  9 
 10 typedef long long ll;
 11 const int  c1=;
 12 const int  c2=;
 13 int pow_c1[N],pow_c2[N],sum_c1[N],sum_c2[N];
 14 
 15 struct node{
 16     int l,r;
 17     int h1,h2;
 18     int lazy;
 19     node(){
 20     lazy=-;
 21     }
 22     void make_lazy(int x)
 23     {
 24         lazy=x;
 25         h1=1ll*sum_c1[r-l]*x%mod1;
 26         h2=1ll*sum_c2[r-l]*x%mod2;
 27     }
 28 }tree[N<<];
 29 
 30 void lazy_sons(int rt)
 31 {
 32     if (tree[rt].lazy!=-)
 33     {
 34         tree[rt<<].make_lazy(tree[rt].lazy);
 35         tree[rt<<|].make_lazy(tree[rt].lazy);
 36         tree[rt].lazy=-;
 37     }
 38 }
 39 
 40 void unite(node &a,node b,node &c){
 41      a.h1=(1ll*b.h1*pow_c1[c.r-c.l+]%mod1+c.h1)%mod1;
 42      a.h2=(1ll*b.h2*pow_c2[c.r-c.l+]%mod2+c.h2)%mod2;
 43 }
 44 char s[N];
 45 void build(int l,int r,int rt)
 46 {
 47     tree[rt].l=l,tree[rt].r=r;
 48     tree[rt].lazy=-;
 49     if (l==r)
 50     {
 51         tree[rt].h1=tree[rt].h2=(s[l]-''+);
 52         return;
 53     }
 54     int mid=(l+r)>>;
 55     build(l,mid,rt<<);
 56     build(mid+,r,rt<<|);
 57     unite(tree[rt],tree[rt<<],tree[rt<<|]);
 58 }
 59 void update(int l,int r,int val,int rt)
 60 {
 61     if (tree[rt].l==l&&tree[rt].r==r)
 62     {
 63         tree[rt].make_lazy(val);
 64         return;
 65     }
 66     lazy_sons(rt);
 67     int mid=(tree[rt].l+tree[rt].r)>>;
 68     if (r<=mid) update(l,r,val,rt<<);
 69     else if (l>mid) update(l,r,val,rt<<|);
 70     else {
 71         update(l,mid,val,rt<<);
 72         update(mid+,r,val,rt<<|);
 73     }
 74     unite(tree[rt],tree[rt<<],tree[rt<<|]);
 75 }
 76 node ans;
 77 void query(int l,int r,int rt)
 78 {
 79     if (tree[rt].l==l&&tree[rt].r==r){
 80         unite(ans,ans,tree[rt]);
 81         return;
 82     }
 83     lazy_sons(rt);
 84     int mid=(tree[rt].l+tree[rt].r)>>;
 85     if (r<=mid) query(l,r,rt<<);
 86     else if (l>mid) query(l,r,rt<<|);
 87     else {
 88         query(l,mid,rt<<);
 89         query(mid+,r,rt<<|);
 90     }
 91 }
 92 int main()
 93 {
 94     pow_c1[]=sum_c1[]=;
 95     pow_c2[]=sum_c2[]=;
 96 
 97     for (int i=;i<N;i++)
 98     {
 99         pow_c1[i]=1ll*pow_c1[i-]*c1%mod1;
         pow_c2[i]=1ll*pow_c2[i-]*c2%mod2;
         sum_c1[i]=(1ll*sum_c1[i-]+pow_c1[i])%mod1;
         sum_c2[i]=(1ll*sum_c2[i-]+pow_c2[i])%mod2;
     }
     int n,m,k;
     scanf("%d%d%d",&n,&m,&k);
 
     scanf("%s",s+);
     build(,n,);
     m+=k;
     int tp,l,r,c;
     int aa,bb,cc,dd;
     while (m--)
     {
         scanf("%d%d%d%d",&tp,&l,&r,&c);
         if (tp==) update(l,r,c+,);
         else
         {
             ans.h1=ans.h2=;
             if (l<=r-c) query(l,r-c,);
             aa=ans.h1;
             bb=ans.h2;
             ans.h1=ans.h2=;
             if (l+c<=r) query(l+c,r,);
             cc=ans.h1,dd=ans.h2;
             if (aa==cc&&bb==dd) cout<<"YES"<<endl;
             else cout<<"NO"<<endl;
 
         }
     }
     return ;
 }