【模拟】Books  CodeForces 279B

Books

time
limit per test
2 seconds

memory
limit per test
256 megabytes

input
standard input

output
standard output

When Valera has 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 is 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 an arbitrary book with
number i and
read the books one by one, starting from this book. In other words,
he will 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 each book up to the end, that is, he doesn't
start reading the book if he doesn't have enough free time to
finish reading it. 

Print the maximum number of books Valera can read.

Input

The first line contains two integers n and t (1 ≤ n ≤ 105; 1 ≤ t ≤ 109) —
the number of books and the number of free minutes 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 minutes that 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

用两个指针从后往前扫一遍就行了。

#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)//注意while语句都要判断边界，尽管自己觉得不会有问题，但是什么奇葩的数据都有。。。

{

now -= a[r];

nown--;

r--;

}

}

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

return 0;

}