Codeforces Round #321 (Div. 2) E. Kefa and Watch 线段树hash

E. Kefa and Watch

Time Limit: 1 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/580/problem/E

Description

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 Input

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

Sample Output

NO
YES

HINT

题意

给你一个字符集只有10的串 ，然后又两个操作

1.区间更新

2.查询lr区间的字符的周期是否为d

题解：

直接暴力线段树hash就好了

查询操作只有判断(l,l+len-d)和(l+len-d+1,r)是否一样就好了

可以用类似错位来解释

代码来自peterpan

代码：

//#pragma comment(linker, "/STACK:102400000,102400000")
#include<cmath>
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<queue>
#include<cstring>
using namespace std;
typedef long long ll;
#define input freopen("/Users/peteryuanpan/data.txt","r",stdin)

#define N 100010
int n, m, k, lens;
char s[N];

#define lch id<<1
#define rch id<<1|1
class HashTree{
public:
    int mod, p;
    ll powp[N], sump[N];
    void getpowp(){
        powp[] = ;
        sump[] = ;
        for(int i = ; i < N; i++){
            powp[i] = powp[i-] * p;
            if(powp[i] >= mod) powp[i] %= mod;

            sump[i] = sump[i-] + powp[i];
            if(sump[i] >= mod) sump[i] %= mod;
        }
    }
    ll v[N << ];
    int len[N << ];
    char lazy[N << ];
    void plant(int id,int l,int r){
        lazy[id] = '\0';
        if(l == r){
            v[id] = s[l];
            len[id] = ;
            return;
        }
        int mid = (l + r) >> ;
        plant(lch, l, mid);
        plant(rch, mid + , r);
        len[id] = len[lch] + len[rch];
        v[id] = v[lch] * powp[len[rch]] + v[rch];
        if(v[id] >= mod) v[id] %= mod;
    }
    void pushdown(int id){
        if(lazy[id] != '\0'){
            lazy[lch] = lazy[rch] = lazy[id];

            v[lch] = lazy[id] * sump[len[lch] - ];
            if(v[lch] >= mod) v[lch] %= mod;

            v[rch] = lazy[id] * sump[len[rch] - ];
            if(v[rch] >= mod) v[rch] %= mod;

            lazy[id] = '\0';
        }
    }
    void update(int id,int ql,int qr,int l,int r,char c){
        if(ql == l && qr == r){
            lazy[id] = c;
            v[id] = c * sump[len[id] - ];
            if(v[id] >= mod) v[id] %= mod;
            return;
        }
        pushdown(id);
        int mid = (l + r) >> ;
        if(qr <= mid) update(lch, ql, qr, l, mid, c);
        else if(mid < ql) update(rch, ql, qr, mid + , r, c);
        else update(lch, ql, mid, l, mid, c), update(rch, mid + , qr, mid + , r, c);

        v[id] = v[lch] * powp[len[rch]] + v[rch];
        if(v[id] >= mod) v[id] %= mod;
    }
    ll query(int id,int ql,int qr,int l,int r){
        if(ql == l && qr == r){
            return v[id];
        }
        pushdown(id);
        int mid = (l + r) >> ;
        if(qr <= mid) return query(lch, ql, qr, l, mid);
        else if(mid < ql) return query(rch, ql, qr, mid + , r);
        else{
            ll t1 = query(lch, ql, mid, l, mid);
            ll t2 = query(rch, mid + , qr, mid + , r);
            ll t = t1 * powp[qr - (mid + ) + ] + t2;
            if(t >= mod) t %= mod;
            return t;
        }
    }
}tree1, tree2;

bool equal(int l1, int r1, int l2, int r2){
    if(tree1.query(, l1, r1, , lens - ) != tree1.query(, l2, r2, , lens - )) return false;
    if(tree2.query(, l1, r1, , lens - ) != tree2.query(, l2, r2, , lens - )) return false;
    return true;
}

bool judge(int l, int r, int d){
    if(r - l +  == d) return true;
    int l2 = l + d, r2 = r;
    int len = r2 - l2 + ;
    int l1 = l, r1 = l1 + len - ;
    return equal(l1, r1, l2, r2);
}

int main(){
    //input;
    tree1.mod = 1e9 + , tree1.p = , tree1.getpowp();
    tree2.mod = 1e9 + , tree2.p = , tree2.getpowp();
    scanf("%d%d%d",&n,&m,&k);

    scanf("%s",s);
    lens = (int)strlen(s);
    tree1.plant(, , lens - ), tree2.plant(, , lens - );

    for(int i = ; i <= m + k; i++){
        int ty, l, r;
        scanf("%d%d%d",&ty,&l,&r);
        l--, r--;
        if(ty == ){
            char c[];
            scanf("%s",c);
            tree1.update(, l, r, , lens - , c[]);
            tree2.update(, l, r, , lens - , c[]);
        }
        else{
            int d;
            scanf("%d",&d);
            printf("%s\n", judge(l, r, d) ? "YES" : "NO");
        }
    }

    return ;
}