"Analog" Books codeforces 279B

Books time limit per test 2 seconds memory limit per test megabytes input standard input output standard output

When Valera had got some free time, he goes to the library to read some books. Today he ' s got t free minutes to read. That's why Valera took N books in the library and for each book he estimated the time he was going to need to read it. Let's number the books by integers from 1 to N. Valera needs AI minutes to read the i-th book.

Valera decided to choose a arbitrary book with number I and read the books one by one, starting from this book. In the other words, he would first read book number I, then book number i + 1, then book number i + 2 and so on. He continues the process until he either runs out of the "free time" or finishes reading the n-th book. Valera reads, the end, that's, he doesn ' t start reading the book if he doesn ' t has enough free time to fi Nish reading it.

Print the maximum number of books Valera can read. Input

The first line contains-integers n and t (1≤n≤105; 1≤t≤109)-the number of books and the number of free minut Es Valera ' s got. The second line contains a sequence of n integers a1, a2, ..., an (1≤ai≤104), where number AI shows the number of Minu TES, the boy, needs to read the i-th book. Output

Print a single integer-the maximum number of books Valera can read. Sample Test (s) input

4 5

3 1 2 1

Output

3

Input

3 3

2 2 3

Output

1

just sweep it back and forth with two pointers.

#include

#include

using namespace std;

Long Long a[110000];

int main()

{

int n;

Long Long T;

scanf ("%d%i64d", &n, &t);

for (int i = 1; I <= n; i++)

scanf ("%i64d", &a[i]);

Long Long now = 0;

int nown = 0;

int max = 0;

int r = N;

for (int l = n; l > 0; l--)

{

Now + = A[l];

nown++;

if (now <= t)

{

if (Nown > Max) max = Nown;

}

Else

while (now > t && r>0)// Note that while statements are to judge the boundary, although they do not think there is a problem, but what wonderful data have ...

{

Now-= A[r];

nown--;

r--;

}

}

printf ("%d\n", Max);

return 0;

}