Codeforces Round #345 (Div. 2) D. Image Preview __ two pointers and pretreat

D. Image Preview

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

Vasya's telephone contains n photos. Photo number 1 is currently opened on the phone. It is allowed to move left and right to the adjacent
photo by swiping finger over the screen. If you swipe left from the first photo, you reach photo n. Similarly, by swiping right from
the last photo you reach photo 1. It takes a seconds
to swipe from photo to adjacent.

For each photo it is known which orientation is intended for it — horizontal or vertical. Phone is in the vertical orientation and can't be rotated. It takes b second
to change orientation of the photo.

Vasya has T seconds to watch photos. He want to watch as many photos as possible. If Vasya opens the photo for the first time, he spends 1 second
to notice all details in it. If photo is in the wrong orientation, he spends b seconds on rotating it before watching it. If Vasya
has already opened the photo, he just skips it (so he doesn't spend any time for watching it or for changing its orientation). It is not allowed to skip unseen photos.

Help Vasya find the maximum number of photos he is able to watch during T seconds.

Input

The first line of the input contains 4 integers n, a, b, T (1 ≤ n ≤ 5·105, 1 ≤ a, b ≤ 1000, 1 ≤ T ≤ 109)
— the number of photos, time to move from a photo to adjacent, time to change orientation of a photo and time Vasya can spend for watching photo.

Second line of the input contains a string of length n containing symbols 'w'
and 'h'.

If the i-th position of a string contains 'w',
then the photo i should be seen in the horizontal orientation.

If the i-th position of a string contains 'h',
then the photo i should be seen in vertical orientation.

Output

Output the only integer, the maximum number of photos Vasya is able to watch during those T seconds.

Examples

input

4 2 3 10
wwhw

output

2

input

5 2 4 13
hhwhh

output

4

input

5 2 4 1000
hhwhh

output

5

input

3 1 100 10
whw

output

0

Note

In the first sample test you can rotate the first photo (3 seconds), watch the first photo (1 seconds), move left (2 second), rotate fourth photo (3 seconds), watch fourth photo (1 second). The whole process takes exactly 10 seconds.

Note that in the last sample test the time is not enough even to watch the first photo, also you can't skip it.

Source

Codeforces Round #345 (Div. 2) D. Image Preview

My Solution

This
is really a good problem for me ☺☺



pretreat whhw......  to array val[maxn] when read the test data

then use two pointers to find the
maximum number of photos Vasya is able to watch during those T seconds.

if a is
really small, we can go right and come back , go left, then we can go right again. But for the final result it is only come back once,

so we just
make it that  it is only come back once in the progress.

1.go right first

//if this progress end,and sum == T, it's a hard problem to solve,★★ please read my code for more details.

2.then r--, l--;the left pointer begin to move towards the left

3.if r == 0 when break the while above , we are supposed to run r = 0, l = 0,sum = 0; l-- to go left.

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn = 5*1e5 +8;
int val[maxn];

int main()
{
#ifdef LOCAL
freopen("a.txt", "r", stdin);
#endif // LOCAL
memset(val, 0, sizeof val); //initial the array will make the program running faster
int n, a, b, T;
char ph;
scanf("%d%d%d%d", &n, &a, &b, &T);
getchar();
for(int i = 0; i < n; i++){
scanf("%c", &ph);
if(i == 0) {if(ph == 'w') val[0] = b+1; else val[0] = 1;}
else{
if(ph == 'w') val[i] = b+a+1;
else val[i] = a+1;
}
//cout<<val[i]<<" ";
}
//cout<<endl;
//two pointers,
//If Vasya has already opened the photo, he just skips it
//(so he doesn't spend any time for watching it or for changing its orientation).
//but it will only get back once, unless it will wast time which is not necessary.
int l = 0, r = 0, sum = 0;
int ans = -1;
//go rights first
while(sum < T && r < n){
sum += val[r++]; //!from WA13 to Accepted
if(sum > T ) r--; //!if sum == T we can't r-- (but at line44 sum -= val[r];we should add if(sum != T) before),
//! unless line41 ans = max(ans,(r-0)); and line45 r--; will be wrong.
} //!the last time has r++, and this val[r] is not added to sum
ans = max(ans,(r-0)); // r-0+1 -1
//if(sum == T) ans++; this is not enough and r should r++ //!!!!!! this two place
if(ans > n) ans = n;
if(sum != T)sum -= val[r]; //!so if(sum == T) we shouldn't run the sentence sum -= val[r];
r--; //!but here we should do as the normal time
/* above can also be wrote as here
sum += val[0];
while(sum < T && r < n){
r++;sum += val[r];
}
if(sum == T) r++;
//Oh almost the same
ans = max(ans,(r-0));
if(ans > n) ans = n;
if(sum != T)sum -= val[r];
r--;
*/

//cout<<ans<<endl;
//then left , the one short is is the direction the pointer go back
bool first = true; // first time from l to r -> r to l
while(ans != n && ans != 0){
while(sum <= T && -l + r <= n-1 && l > -n){ //!!!!!!l > r-n-1 and at the same time l > -n otherwise maybe TLE
l--; sum += val[(l + n)%n]; //!!!!!!i-- ,first from WA5 to WA7
if(-l < r && first) sum += a;
else {if(first){sum -= a*(-l - 1);sum += a*r; first = false;} } // this will happen only once
//!!!!!! a*(-l - 1) not a*(-l) because the last time sum didn't add a 'a' from WA5 to WA7
}
if(sum < T) break;
ans = max(ans,(r-l)); //! r + -l - 1 + 1
if(first)sum -= val[r];
else sum -= (val[r] + a);
if(r>0) r--;
if(r == 0) break; //!!!!!! r not be r <= 0 from WA7 to WA8

}

//!!!!!! for r == 0 from WA8 to WA13
if(r == 0){
l = 0, sum = 0;
sum += val[0];
while(sum < T && l > -n){
l--; sum += val[(l + n)%n];
}
if(sum == T) l--;
ans = max(ans,(0-l));
if(ans > n) ans = n;
}
printf("%d", ans);
return 0;
}


Thank you all!