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Revenge of Segment Tree (hdu 5086)
Revenge of Segment Tree
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1282 Accepted Submission(s): 324
Problem Description
In computer science, a segment tree is a tree data structure for storing intervals, or segments. It allows querying which of the stored segments contain a given point. It is, in principle, a static structure; that is, its content cannot be modified once the structure is built. A similar data structure is the interval tree. A segment tree for a set I of n intervals uses O(n log n) storage and can be built in O(n log n) time. Segment trees support searching for all the intervals that contain a query point in O(log n + k), k being the number of retrieved intervals or segments. ---Wikipedia Today, Segment Tree takes revenge on you. As Segment Tree can answer the sum query of a interval sequence easily, your task is calculating the sum of the sum of all continuous sub-sequences of a given number sequence.
Input
The first line contains a single integer T, indicating the number of test cases. Each test case begins with an integer N, indicating the length of the sequence. Then N integer Ai follows, indicating the sequence. [Technical Specification] 1. 1 <= T <= 10 2. 1 <= N <= 447 000 3. 0 <= Ai <= 1 000 000 000
Output
For each test case, output the answer mod 1 000 000 007.
Sample Input
2 1 2 3 1 2 3
Sample Output
2 20HintFor the second test case, all continuous sub-sequences are [1], [2], [3], [1, 2], [2, 3] and [1, 2, 3]. So the sum of the sum of the sub-sequences is 1 + 2 + 3 + 3 + 5 + 6 = 20. Huge input, faster I/O method is recommended. And as N is rather big, too straightforward algorithm (for example, O(N^2)) will lead Time Limit Exceeded. And one more little helpful hint, be careful about the overflow of int.代码:#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <string> #include <map> #include <stack> #include <vector> #include <set> #include <queue> #pragma comment (linker,"/STACK:102400000,102400000") #define maxn 1005 #define MAXN 2005 #define mod 1000000007 #define INF 0x3f3f3f3f #define pi acos(-1.0) #define eps 1e-6 typedef long long ll; using namespace std; int main() { int t; __int64 i,N,n,x,s; scanf("%d",&t); while (t--) { s=0; scanf("%I64d",&N); n=(N+1)/2; for (i=1;i<=N;i++) { scanf("%I64d",&x); if (i<=n) s=(s+(x%mod)*((N*i-(i-1)*i)%mod)%mod); else s=(s+(x%mod)*(( N*(N-i+1)-(N-i)*(N-i+1) )%mod))%mod; } printf("%I64d\n",s); } return 0; }
Revenge of Segment Tree (hdu 5086)
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