Codeforces Round #321 (Div. 2)E 线段树+字符串hash

E. Kefa and Watch
time limit per test1.5 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard 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 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 c (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”.

Examples
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.

这个题有几个坑点
第一个就是需要双关键字hash
第二个是在线段树更新的过程中需要去掉不需要的长度
第三个是lazy在用到的时候必须进行更新
其实也不算坑点
还是自己菜吧…

#include<iostream>
#include<cstring>
#include<string>
#include<map>
using namespace std;
int mo1 = 1000000009, mo2 = 1000000007;
unsigned long long p1[100011], s1[110001], p2[110001], s2[101001];
int bas = 17;
struct qq
{
    long long z, y, chang, z1, z2, lazy;
};
qq shu[600001];
string q;
void jian(int gen, int zuo, int you)
{
    shu[gen].z = zuo;
    shu[gen].y = you;
    shu[gen].lazy = -1;
    if (zuo == you)
    {
        shu[gen].z1 = (q[zuo-1] - '0') % mo1;
        shu[gen].z2 = (q[zuo-1] - '0') % mo2;
        return;
    }
    int mid = (zuo + you) / 2;
    jian(2 * gen, zuo, mid);
    jian(2 * gen + 1, mid + 1, you);
    shu[gen].z1 = (p1[shu[2 * gen + 1].y - shu[2 * gen + 1].z +1] * shu[2 * gen].z1 + shu[2 * gen + 1].z1) % mo1;
    shu[gen].z2 = (p2[shu[2 * gen + 1].y - shu[2 * gen + 1].z +1] * shu[2 * gen].z2 + shu[2 * gen + 1].z2) % mo2;
}
void gengxin(int gen, int zuo, int you, int wz, int wy, int bian)
{
    if (zuo >= wz&&you <= wy)
    {
        shu[gen].lazy = bian;
        shu[gen].z1 = bian*s1[you - zuo] % mo1;
        shu[gen].z2 = bian*s2[you - zuo] % mo2;
        return;
    }
    else if (zuo > wy || you < wz)return;
    else
    {
        int mid = (zuo + you) / 2;
        if (shu[gen].lazy != -1)
        {
            shu[2 * gen].lazy = shu[2 * gen + 1].lazy = shu[gen].lazy;
            shu[gen].lazy = -1;
            shu[2 * gen].z1 = (shu[2 * gen].lazy*s1[shu[2 * gen].y - shu[2 * gen].z]) % mo1;
            shu[2 * gen].z2 = (shu[2 * gen].lazy*s2[shu[2 * gen].y - shu[2 * gen].z]) % mo2;
            shu[2 * gen + 1].z1 = (shu[2 * gen].lazy*s1[shu[2 * gen + 1].y - shu[2 * gen + 1].z]) % mo1;
            shu[2 * gen + 1].z2 = (shu[2 * gen].lazy*s2[shu[2 * gen + 1].y - shu[2 * gen + 1].z]) % mo2;
        }
        gengxin(2 * gen, zuo, mid, wz, wy, bian);
        gengxin(2 * gen + 1, mid + 1, you, wz, wy, bian);
        shu[gen].z1 = (p1[shu[2 * gen + 1].y - shu[2 * gen + 1].z + 1] * shu[2 * gen].z1 + shu[2 * gen + 1].z1) % mo1;
        shu[gen].z2 = (p2[shu[2 * gen + 1].y - shu[2 * gen + 1].z + 1] * shu[2 * gen].z2 + shu[2 * gen + 1].z2) % mo2;
    }
}
int flag = 0;
pair<pair<long long,long long>, long long> xunw(int gen, int zuo, int you, int wz, int wy)
{
    if (zuo > wy || you < wz)return make_pair(make_pair(-1, -1),0);
    if (zuo >= wz&&you <= wy)return make_pair(make_pair(shu[gen].z1, shu[gen].z2),you-zuo+1);
    int mid = (zuo + you) / 2;
    if (shu[gen].lazy != -1)
    {
        shu[2 * gen].lazy = shu[2 * gen + 1].lazy = shu[gen].lazy;
        shu[gen].lazy = -1;
        shu[2 * gen].z1 = (shu[2 * gen].lazy*s1[shu[2 * gen].y - shu[2 * gen].z]) % mo1;
        shu[2 * gen].z2 = (shu[2 * gen].lazy*s2[shu[2 * gen].y - shu[2 * gen].z]) % mo2;
        shu[2 * gen + 1].z1 = (shu[2 * gen].lazy*s1[shu[2 * gen + 1].y - shu[2 * gen + 1].z]) % mo1;
        shu[2 * gen + 1].z2 = (shu[2 * gen].lazy*s2[shu[2 * gen + 1].y - shu[2 * gen + 1].z]) % mo2;
    }
    pair<pair<long long, long long>, long long>zz = xunw(2 * gen, zuo, mid, wz, wy);
    pair<pair<long long, long long>, long long>yy = xunw(2 * gen + 1, mid + 1, you, wz, wy);
    if (zz.first.first == -1)return yy;
    if (yy.first.first == -1)return zz;
    int youc = yy.second;
    return make_pair(make_pair((zz.first.first*p1[youc] + yy.first.first) % mo1, (zz.first.second*p2[youc] + yy.first.second) % mo2),zz.second+yy.second);
}
int main()
{
    int n, m, k;
    cin >> n >> m >> k;
    cin >> q;
    p1[0] = p2[0] = s1[0] = s2[0] = 1;
    for (int a = 1;a <= q.size();a++)
    {
        p1[a] = p1[a - 1] * bas%mo1;
        s1[a] = (s1[a - 1] + p1[a]) % mo1;
        p2[a] = p2[a - 1] * bas%mo2;
        s2[a] = (s2[a - 1] + p2[a]) % mo2;
    }
    int r, t, y, u;
    jian(1, 1, q.size());
    for (int a = 1;a <= m + k;a++)
    {
        scanf("%d%d%d%d", &r, &t, &y, &u);
        if (r == 1)gengxin(1, 1, q.size(), t, y, u);
        else
        {
            if (y - t + 1 == u)
            {
                cout << "YES" << endl;
                continue;
            }
            pair<pair<long long, long long>, long long> qwe = xunw(1, 1, q.size(), t, y - u);
            pair<pair<long long, long long>, long long> wer = xunw(1, 1, q.size(), t+u, y);
            if (qwe.first.first == wer.first.first&&qwe.first.second == wer.first.second)cout << "YES" << endl;
            else cout << "NO" << endl;
        }
    }
    return 0;
}