Lucky Queries (线段树的区间合并)

Lucky Queries

 CodeForces - 145E 

　　Petya loves lucky numbers very much. Everybody knows that lucky numbers are positive integers whose decimal record contains only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17, 467 are not.

Petya brought home string s with the length of n. The string only consists of lucky digits. The digits are numbered from the left to the right starting with 1. Now Petya should execute m queries of the following form:

• switch l r — "switch" digits (i.e. replace them with their opposites) at all positions with indexes from l to r, inclusive: each digit 4 is replaced with 7and each digit 7 is replaced with 4 (1 ≤ l ≤ r ≤ n);
• count — find and print on the screen the length of the longest non-decreasing subsequence of string s.

Subsequence of a string s is a string that can be obtained from s by removing zero or more of its elements. A string is called non-decreasing if each successive digit is not less than the previous one.

Help Petya process the requests.

Input

The first line contains two integers n and m (1 ≤ n ≤ 10⁶, 1 ≤ m ≤ 3·10⁵) — the length of the string s and the number of queries correspondingly. The second line contains n lucky digits without spaces — Petya's initial string. Next m lines contain queries in the form described in the statement.

Output

For each query count print an answer on a single line.

Examples

Input

2 3
47
count
switch 1 2
count

Output

2
1

Input

3 5
747
count
switch 1 1
count
switch 1 3
count

Output

2
3
2

Note

In the first sample the chronology of string s after some operations are fulfilled is as follows (the sought maximum subsequence is marked with bold):

1. 47
2. 74
3. 74

In the second sample:

1. 747
2. 447
3. 447
4. 774
5. 774

题意：n代表字符串的长度，m代表操作的次数。两种操作：count：输出该字符串中非递减子序列的最大长度（子序列不是子串，不需要连续）switch:表示将从l 到 r的所有字符改变，4变成7,7变成4

题解：

在线段树中维护一下值：

len4表示全是4的最长子序列。

len7表示全是7的最长子序列。

len47表示以4开头以7结尾的最长子序列。

对于操作2，因为反转之后对len4,len7,len47的值影响较大（主要还是len47），所以一般对这样的题都是直接维护两个值，一个是当前的，一个是反转后的，当进行反转操作时再交换就好了。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=1e6+10;
int n,m; 
    char s[maxn];
    char op[20];
struct node{
    int l;
    int r;
    int laz;
    int len4[2],len7[2],len47[2];
}e[maxn<<2];
void pushup(int cur,int c)
{
    e[cur].len4[c]=e[cur<<1].len4[c]+e[cur<<1|1].len4[c];//4的长度为左子树4的长度+右子树4的长度 
    e[cur].len7[c]=e[cur<<1].len7[c]+e[cur<<1|1].len7[c];
    int temp=e[cur<<1].len4[c]+max(e[cur<<1|1].len7[c],e[cur<<1|1].len47[c]);//len47=左子树len4+右子树len47 左子树len4+右子树len7  左子树len47+右子树len7中取较大的 
    e[cur].len47[c]=max(temp,e[cur<<1].len47[c]+e[cur<<1|1].len7[c]);
    return;
}
void change(int cur)
{
    e[cur].laz^=1;
    swap(e[cur].len4[0],e[cur].len4[1]);
    swap(e[cur].len7[0],e[cur].len7[1]);
    swap(e[cur].len47[0],e[cur].len47[1]);
return;
}
void pushdown(int cur)
{
    if(e[cur].laz)
    {
        //int mid=(e[cur].l+e[cur].r)/2;
        change(cur<<1);
        change(cur<<1|1);
        e[cur].laz=0;
    }
    return;
}
void build(int l,int r,int cur)
{
    e[cur].l=l;
    e[cur].r=r;
    e[cur].laz=0;
    if(l==r)
    {
        if(s[l]=='4')
        {
            e[cur].len4[0]=e[cur].len7[1]=1;//e[cur].len4[0]代表没改变前len4为1，e[cur].len7[1]=1表示如果把4的值改变为7，那么len7为1 len数组中0代表没改变，1代表改变后的值 
            e[cur].len4[1]=e[cur].len7[0]=0; 
        }
        else
        {
            e[cur].len7[0]=e[cur].len4[1]=1;
            e[cur].len7[1]=e[cur].len4[0]=0;
        }
        e[cur].len47[0]=e[cur].len47[1]=0;
        return;
    }
    int mid=(l+r)/2;
    build(l,mid,cur<<1);
    build(mid+1,r,cur<<1|1);
    pushup(cur,0);
    pushup(cur,1);
}
void update(int pl,int pr,int cur)
{
    if(pl<=e[cur].l&&e[cur].r<=pr)
    {
        change(cur);
        return;
    }
    pushdown(cur);
    int mid=(e[cur].l+e[cur].r)/2;
    if(pl<=mid)
        update(pl,pr,cur<<1);
    if(pr>mid)
        update(pl,pr,cur<<1|1);
    pushup(cur,0);
    pushup(cur,1);
}
int main()
{
    scanf("%d%d",&n,&m);
    scanf("%s",s+1);
    build(1,n,1);
    while(m--)
    {
        int x,y;
        scanf("%s",op);
        if(op[0]=='s')
        {
            scanf("%d%d",&x,&y);
            update(x,y,1);
        }
        else
        {
            printf("%d
",max(e[1].len4[0],max(e[1].len47[0],e[1].len7[0])));
        }
    }
}