Codeforces Round #104 (Div. 1) E. Lucky Queries

E. Lucky Queries

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 7 and 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.

Sample test(s)

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： 给出一个4,7序列，操作1，把一段里面的4表7,7变4.操作2询问最长不降子序列

直接线段树搞

Max表示最长不降子序列，Min表示最长不升子序列

num1表示4的个数，num2表示7的个数

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<vector>
#include<set>
#include<stack>
#include<map>
#include<ctime>
#include<bitset>
#define LL long long
#define ll long long
#define INF 0x3f3f3f3f
#define maxn 1000010
#define eps 1e-6
#define mod 1000000007
using namespace std;

struct node
{
    int num1,num2,Max,Min,len;
    bool rev;
}tree[maxn*3];

char a[maxn] ;
int ql,qr;

void up(int o)
{
    tree[o].len=tree[o<<1].len+tree[o<<1|1].len;
    tree[o].Max=max(tree[o<<1].Max,tree[o<<1|1].Max) ;
    tree[o].Min=max(tree[o<<1].Min,tree[o<<1|1].Min) ;
    tree[o].Max=max(tree[o<<1].Max+tree[o<<1|1].num2,tree[o].Max) ;
    tree[o].Max=max(tree[o<<1].num1+tree[o<<1|1].Max,tree[o].Max) ;
    tree[o].Min=max(tree[o<<1|1].Min+tree[o<<1].num2,tree[o].Min) ;
    tree[o].Min=max(tree[o<<1|1].num1+tree[o<<1].Min,tree[o].Min) ;
    tree[o].num1=tree[o<<1].num1+tree[o<<1|1].num1;
    tree[o].num2=tree[o<<1].num2+tree[o<<1|1].num2;
    tree[o].Max=max(tree[o].num1,tree[o].Max) ;
    tree[o].Max=max(tree[o].num2,tree[o].Max) ;
    tree[o].Min=max(tree[o].num2,tree[o].Min) ;
    tree[o].Min=max(tree[o].num1,tree[o].Min) ;
}
void SWAP(int o)
{
    swap(tree[o].num1,tree[o].num2) ;
    swap(tree[o].Max,tree[o].Min) ;
    tree[o].rev ^= 1;
}
void down(int o )
{
    if(tree[o].rev)
    {
        SWAP(o<<1) ;
        SWAP(o<<1|1) ;
        tree[o].rev=0;
    }
}
void build(int L,int R,int o)
{
    tree[o].rev=false;
    if(L==R)
    {
        if(a[L]=='4')tree[o].num1=1,tree[o].num2=0 ;
        else tree[o].num2=1,tree[o].num1=0;
        tree[o].Max=tree[o].Min=1;
        return ;
    }
    int mid=(L+R)>>1;
    build(L,mid,o<<1) ;
    build(mid+1,R,o<<1|1) ;
    up(o) ;
}
void update(int L,int R,int o )
{
    if(ql<=L&&qr>=R)
    {
        SWAP(o) ;
        return ;
    }
    int mid=(L+R)>>1 ;
    down(o) ;
    if(ql<=mid) update(L,mid,o<<1) ;
    if(qr>mid) update(mid+1,R,o<<1|1) ;
    up(o) ;
}

int main()
{
    int n,m,i,j;
    char ss[10] ;
    while(scanf("%d%d",&n,&m) != EOF)
    {
        scanf("%s",a+1) ;
        build(1,n,1) ;
        while(m--)
        {
            scanf("%s",ss) ;
            if(ss[0]=='c')
            {
                printf("%d
",tree[1].Max) ;
            }
            else
            {
                scanf("%d%d",&ql,&qr) ;
                update(1,n,1) ;
            }
        }
    }
    return 0 ;
}
/*
3 5
747
count
switch 1 1
count
switch 1 3
count
*/