Codeforces - 145ELucky 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.

Examples

input

Copy

2 3
47
count
switch 1 2
count

output

Copy

2
1

input

Copy

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

output

Copy

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：给出一个字符串，两种操作，第一种反正l~r 第二种查询 字符串长度(4在前7在后这种) 

思路：维护一下翻转和没翻转的 4 7 47的长度，最后结果肯定是由 4L+7R 47L+7R 4L+47R 这三种

代码：

#include<bits/stdc++.h>
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
using namespace std;
const int maxn = 1e6+5;
char s[maxn];
int len4[maxn<<2][2],len7[maxn<<2][2],len[maxn<<2][2];
bool fg[maxn<<2];
void push_up(int rt)
{
    int l=rt<<1,r=rt<<1|1;
    for(int i=0; i<2; i++)
    {
        len4[rt][i]=len4[l][i]+len4[r][i];
        len7[rt][i]=len7[l][i]+len7[r][i];
        len[rt][i]=max(len4[l][i]+len7[r][i],max(len[l][i]+len7[r][i],len4[l][i]+len[r][i]));
    }
}
void Reverse(int rt)
{
    swap(len4[rt][0],len4[rt][1]);
    swap(len7[rt][0],len7[rt][1]);
    swap(len[rt][0],len[rt][1]);
    fg[rt]=!fg[rt];
}
void push_down(int rt)
{
    if(!fg[rt]) return;
    Reverse(rt<<1);
    Reverse(rt<<1|1);
    fg[rt]=!fg[rt];
}
void build(int l,int r,int rt)
{
    if(l==r)
    {
        s[l]=='4'?len4[rt][0]++:len7[rt][0]++;
        len4[rt][1]=len7[rt][0];
        len7[rt][1]=len4[rt][0];
        return;
    }
    int m=(l+r)>>1;
    build(lson);
    build(rson);
    push_up(rt);
}
void updata(int l,int r,int rt,int L,int R)
{
       if(L<=l&&r<=R)
       {
           Reverse(rt);
           return;
       }
       push_down(rt);
       int m=l+r>>1;
       if(L<=m)
        updata(lson,L,R);
       if(R>m)
        updata(rson,L,R);
        push_up(rt);
}
int query(int rt)
{
    return max(len4[rt][0],max(len7[rt][0],len[rt][0]));
}
int main()
{
    int n,m;
    scanf("%d %d",&n,&m);
    scanf("%s",s+1);
    build(1,n,1);
   // cout<<"?"<<endl;
    while(m--)
    {
        char op[10];
        scanf("%s",op);
        if(op[0]=='c')
        {
            printf("%d\n",query(1));
        }
        else
        {
            int l,r;
            scanf("%d %d",&l,&r);
            updata(1,n,1,l,r);
        }
    }
}

 

 

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