Revenge of Segment Tree
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1541 Accepted Submission(s): 552
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.
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
1
2
3
1 2 3
20
For 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.
#include <iostream>
#include <stdio.h>
#include <math.h>
#include <stdlib.h>
#include <algorithm>
#include <string.h>
using namespace std;
typedef long long LL;
const int mod = ;
const int N = ;
int n;
LL a[N];
LL dp[N];
int main()
{ int tcase;
scanf("%d",&tcase);
while(tcase--){
scanf("%d",&n);
for(int i=;i<=n;i++){
scanf("%lld",&a[i]);
}
dp[] = a[];
for(int i=;i<=n;i++){
dp[i] = (dp[i-] + (i*a[i])%mod)%mod;
}
LL ans = ;
for(int i=;i<=n;i++){
ans = (ans+dp[i])%mod;
}
printf("%lld\n",ans);
}
return ;
}