codeforces580E. Kefa and Watch

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原创最新推荐文章于 2026-09-08 14:46:47 发布 611 阅读 AI 写同款· GEO 优化 ›

〔重庆理工大学〕大学物理实验--变温霍耳效应及其参数测定每个老师要求不同，实验报告内容和数据仅供参考。阅读详情

传送门：http://codeforces.com/problemset/problem/580/E

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.

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.

思路：首先对于第一种操作，可以用线段树维护哈希值，覆盖就打标记。

而对于询问，怎样快速判定一个字符串是否循环节为D呢？

充要条件就是len<=d（这个要记得写特判）或者s[1,len-d]==s[d,len]

为什么可以这样判呢，这很简单

我们把字符串分成n段，每段长为d，错开一段后比较若相等，那每段就相等了，于是循环节为d

于是用线段树维护一下就可以了。

#include<cstdio>
#include<cstring>
#include<algorithm>
#define fi first
#define se second
#define PI pair<long long,long long>
#define mp(a,b) make_pair(a,b)
#define ls (p<<1)
#define rs ((p<<1)|1)
#define mid ((l+r)>>1)
const int mod1=573292817,mod2=1e9+7,base1=233,base2=2333,maxn=100010,maxt=maxn<<3;
typedef long long ll;
using namespace std;
int n,m,k;char s[maxn];
ll pow1[maxn],pow2[maxn],sp1[maxn],sp2[maxn];
struct node{ll v1,v2;int cov,len;};
struct Segment_Tree{
	node t[maxt];
	void add_cov(int p,int c){
		t[p].cov=c;
		t[p].v1=c*sp1[t[p].len-1]%mod1;
		t[p].v2=c*sp2[t[p].len-1]%mod2;
	}
	void down(int p){if (t[p].cov!=-1) add_cov(ls,t[p].cov),add_cov(rs,t[p].cov),t[p].cov=-1;}
	node connect(node a,node b){
		int len=a.len+b.len;
		ll v1=(a.v1*pow1[b.len]+b.v1)%mod1;
		ll v2=(a.v2*pow2[b.len]+b.v2)%mod2;
		return (node){v1,v2,-1,len};
	}
	void update(int p){
		t[p].v1=(t[ls].v1*pow1[t[rs].len]%mod1+t[rs].v1)%mod1;
		t[p].v2=(t[ls].v2*pow2[t[rs].len]%mod2+t[rs].v2)%mod2;
	}
	void build(int p,int l,int r){
		t[p].len=r-l+1,t[p].cov=-1;
		if (l==r){t[p].v1=t[p].v2=s[l]-'0';return;}
		build(ls,l,mid),build(rs,mid+1,r);
		update(p);
	}
	void modify(int p,int l,int r,int a,int b,int c){		
		if (l==a&&r==b){add_cov(p,c);return;}
		down(p);
		if (b<=mid) modify(ls,l,mid,a,b,c);
		else if (a>mid) modify(rs,mid+1,r,a,b,c);
		else modify(ls,l,mid,a,mid,c),modify(rs,mid+1,r,mid+1,b,c);
		update(p);
	}
	node query(int p,int l,int r,int a,int b){
		if (l==a&&r==b){return t[p];}
		down(p);node ans;
		if (b<=mid) ans=query(ls,l,mid,a,b);
		else if (a>mid) ans=query(rs,mid+1,r,a,b);
		else ans=connect(query(ls,l,mid,a,mid),query(rs,mid+1,r,mid+1,b));
		//printf("%d %d %d %d %lld %lld\n",l,r,a,b,ans.v1,ans.v2);
		return ans;
	}
	PI query(int a,int b){node ans=query(1,1,n,a,b);return mp(ans.v1,ans.v2);}
}T;

void init(){
	scanf("%d%d%d%s",&n,&m,&k,s+1),m=m+k;
	pow1[0]=pow2[0]=sp1[0]=sp2[0]=1;
	for (int i=1;i<=n+5;i++){
		pow1[i]=pow1[i-1]*base1%mod1;
		pow2[i]=pow2[i-1]*base2%mod2;
		sp1[i]=(sp1[i-1]+pow1[i])%mod1;
		sp2[i]=(sp2[i-1]+pow2[i])%mod2;
	}
	T.build(1,1,n);
}
void work(){
	for (int i=1,op,a,b,c;i<=m;i++){
		scanf("%d%d%d%d",&op,&a,&b,&c);
		if (op==1) T.modify(1,1,n,a,b,c);
		else{
			if (b-a+1<=c){puts("YES");continue;}
			else puts(T.query(a,b-c)==T.query(a+c,b)?"YES":"NO");
			//for (int i=a;i<=b-c;i++) printf("%lld",T.query(i,i));puts("");
			//for (int i=a+c;i<=b;i++) printf("%lld",T.query(i,i));puts("");
			//printf("%lld %lld\n",T.query(a,b-c).fi,T.query(a+c,b).fi);
		}
	}
}
int main(){
	init(),work();
	return 0;
}
/*
20 1 2
34075930750342906718
2 1 20 20
1 1 20 6
2 1 20 1
*/