B - Number of Ways
2015-03-16 23:12
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B - Number of Ways
Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d
& %I64u
Submit Status
Description
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 integersa[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.
Sample Input
Input
Output
Input
Output
Input
Output
小结:
....题目看清楚才发现这是一道彻彻底底的水题,原来以为要用DP做的...
以下是AC代码:
Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d
& %I64u
Submit Status
Description
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 integersa[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.
Sample Input
Input
5 1 2 3 0 3
Output
2
Input
4 0 1 -1 0
Output
1
Input
24 1
Output
0
小结:
....题目看清楚才发现这是一道彻彻底底的水题,原来以为要用DP做的...
以下是AC代码:
#include <iostream> #include <cstdio> typedef long long ll; using namespace std; const int maxn = 500005; ll sum[maxn]; int main() { int n; scanf("%d", &n); sum[0] = 0; for (int i = 0; i < n; i++) { int x; scanf("%d", &x); sum[i+1] = sum[i] + x; } ll ans = 0; if (sum[n] % 3 == 0) { ll one = sum[n] / 3, two = sum[n] / 3 * 2; ll cnt = 0; for (int i = 1; i < n; i++) { if (sum[i] == two) ans += cnt; if (sum[i] == one) cnt++; } } cout << ans << endl; return 0; }
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