TOJ 3425: Generic Cow Protests -- 树状数组-优快云博客

本文介绍了一道名为GenericCowProtests的算法题，任务是将N头奶牛按正整数组划分，使得每组的和为正数，并计算划分方式的数量。使用树状数组等数据结构进行高效求解。

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3425: Generic Cow Protests

Time Limit(Common/Java):1000MS/3000MS Memory Limit:65536KByte
Total Submit: 46 Accepted:14

Description

Farmer John's N (1 <= N <= 100,000) cows are lined up in a row and numbered 1..N. The cows are conducting another one of their strange protests, so each cow i is holding up a sign with an integer A_i
(-10,000 <= A_i <= 10,000).

FJ knows the mob of cows will behave if they are properly grouped and thus would like to arrange the cows into one or more contiguous groups so that every cow is in exactly one group and that every
group has a nonnegative sum.

Help him count the number of ways he can do this, modulo 1,000,000,009.

By way of example, if N = 4 and the cows' signs are 2, 3, -3, and 1, then the following are the only four valid ways of arranging the cows:

(2 3 -3 1)
(2 3 -3) (1)
(2) (3 -3 1)
(2) (3 -3) (1)

Note that this example demonstrates the rule for counting different orders of the arrangements.

Input

* Line 1: A single integer: N

* Lines 2..N + 1: Line i + 1 contains a single integer: A_i

Output

* Line 1: A single integer, the number of arrangements modulo 1,000,000,009.

Sample Input

Sample Output

Source

USACO Feb 2011

http://www.verydemo.com/demo_c427_i28080.html 有详细的解答

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                           > rpqi; const int MAX_N = 100000 + 2; const int MOD = 1000000009; int sum[MAX_N]; int ordered_sum[MAX_N]; int c[MAX_N]; class Tree_Array { public: Tree_Array(int _n); void update(int pos, int v); int sum(int pos); private: int n; inline int lowbit(int x); }; Tree_Array::Tree_Array(int _n) : n(_n) { } inline int Tree_Array::lowbit(int x) { return x & (-x); } void Tree_Array::update(int pos, int v) { for (int i = pos; i <= n; i += lowbit(i)) { c[i] = (c[i] + v) % MOD; } } int Tree_Array::sum(int pos) { int ret = 0; for (int i = pos; i; i -= lowbit(i)) { ret = (ret + c[i]) % MOD; } return ret; } int main(int argc, char *argv[]) { // freopen("D:\\in.txt", "r", stdin); int n, a, i; cin >> n; Tree_Array ta(n); for (i = 1; i <= n; ++i) { scanf("%d", &a); ordered_sum[i] = sum[i] = sum[i - 1] + a; } sort(ordered_sum + 1, ordered_sum + n + 1); int *end = unique(ordered_sum + 1, ordered_sum + n + 1); for (i = 1; i < n; ++i) { int pos = lower_bound(ordered_sum + 1, end, sum[i]) - ordered_sum; int tmp = ta.sum(pos); if (sum[i] >= 0) { ++tmp; } ta.update(pos, tmp); } cout << (ta.sum(lower_bound(ordered_sum + 1, end, sum[n]) - ordered_sum) + (sum[n] >= 0)) % MOD << endl; return 0; }