You’ve got array a[1], a[2], …, a[n], consisting of n integers. Count the number of ways to split all the elements of the array into three contiguous parts so that the sum of elements in each part is the same.
More formally, you need to find the number of such pairs of indices i, j (2 ≤ i ≤ j ≤ n - 1), that .
Input
The first line contains integer n (1 ≤ n ≤ 5·105), showing how many numbers are in the array. The second line contains n integers a[1], a[2], …, a[n] (|a[i]| ≤ 109) — the elements of array a.
Output
Print a single integer — the number of ways to split the array into three parts with the same sum.
Examples
input
5
1 2 3 0 3
output
2
input
4
0 1 -1 0
output
1
input
2
4 1
output
0
Solution:
#include <cstring> #include <vector> #include <list> #include <map> #include <set> #include <deque> #include <stack> #include <bitset> #include <algorithm> #include <functional> #include <numeric> #include <utility> #include <sstream> #include <iostream> #include <iomanip> #include <cstdio> #include <cmath> #include <cstdlib> #include <ctime> #include <memory.h> #include <cassert> using namespace std; long long s[1234567]; int main() { int n; scanf("%d", &n); s[0] = 0; for (int i = 0; i < n; i++) { int foo; scanf("%d", &foo); s[i + 1] = s[i] + foo; } long long ans = 0; if (s[n] % 3 == 0) { long long u = s[n] / 3, v = 2 * s[n] / 3; long long cnt = 0; for (int i = 1; i < n; i++) { if (s[i] == v) { ans += cnt; } if (s[i] == u) { cnt++; } } } cout << ans << endl; return 0; }
Related posts:
Kate and imperfection
Ancient Prophesy
Tidying Up
One-Way Reform
Haar Features
Lingua Romana
Pearls in a Row
Equalize
Gold Experience
Magic Grid
Chicken or Fish?
Andrey and Problem
Restoring Map
New Year and Hurry
Image Preview
Two Arithmetic Progressions
Rational Resistance
Bash Plays with Functions
Spectator Riots
Dynamic Programming on Broken Profile
Team Rocket Rises Again
Pavel and barbecue
Magnum Opus
0-1 BFS
Sieve of Eratosthenes
Finding a negative cycle in the graph
Dice Tower
K Paths
Tom Riddle's Diary
Chat Order
Tricky Interactor
Merging Two Decks