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.
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.
Print a single integer — the maximum number of books Valera can read.
4 5
3 1 2 1
3
3 3
2 2 3
1
最直观的思路是取a[i]，由a[i]开始求和直到sum>t,然后再从a[i+1]开始求和直到sum>t,一直进行到a[n-1].
这样效率很低，可以采取两步优化：
（1）求a[i]+a[i+1]+a[i+2]……和求a[i+1]+a[i+2]+……是有重叠部分的，因此在由a[i]开始累加时，假设累加到a[j],记录j和sum，在由a[i+1]
开始累加时，执行sum-=a[i],那么sum = a[i+1]+……+a[j]就已经求得。
（2）记录已经算得的可读书的最大本数b，如果a[0……(n-1)]中余下的未检测的数的个数小于等于b，可提前结束循环。
下面AC Code只进行了第一步优化。
1 #include <iostream> 2 #include <string> 3 #include <set> 4 #include <map> 5 #include <vector> 6 #include <stack> 7 #include <queue> 8 #include <cmath> 9 #include <cstdio> 10 #include <cstring> 11 #include <algorithm> 12 #include <utility> 13 using namespace std; 14 #define ll long long 15 #define cti const int 16 #define ctll const long long 17 #define dg(i) cout << '*' << i << endl; 18 19 typedef pair<int, int> p; 20 cti N = 100001; 21 int n, t; 22 int a[N]; 23 vector<p> v; 24 25 bool cmp(const p& a, const p& b){return a.second < b.second;} 26 27 int main() 28 { 29 int i, j; 30 while(scanf("%d %d", &n, &t) != EOF) 31 { 32 v.clear(); 33 for(i = 0; i < n; i++) scanf("%d", a + i); 34 if(n == 1) 35 { 36 if(t >= a[0]) puts("1"); 37 else puts("0"); 38 continue; 39 } 40 int k; 41 int sum; 42 for(i = 0; i < n; i++) 43 { 44 if(!i) 45 { 46 k = 0; 47 sum = 0; 48 } 49 else 50 { 51 if(sum) sum -= a[i-1]; 52 } 53 for(j = k; j < n; j++) 54 { 55 if(t >= sum + a[j]) 56 { 57 sum += a[j]; 58 } 59 else 60 { 61 k = j == i ? k + 1 : j; //记录下次累加开始的位置 62 break; 63 } 64 } 65 if(j == n) 66 { 67 v.push_back(make_pair(sum, n - i)); 68 break; 69 } 70 if(j - i) v.push_back(make_pair(sum, j - i)); 71 } 72 if(v.empty()) puts("0"); 73 else 74 { 75 sort(v.begin(), v.end(), cmp); 76 printf("%d\n", (v.end()-1)->second); 77 } 78 } 79 return 0; 80 }