hihocoder 2017微软探星夏令营在线技术笔试 1534 Array Partition

#1534 : Array Partition

时间限制:10000ms
单点时限:1000ms
内存限制:256MB

描述

Given an integer array A1, A2 ...
AN, you are asked to split the array into three continuous parts:
A1, A2 ... Ap | Ap+1,
Ap+2, ... Aq | Aq+1,
Aq+2, ... AN.

Let S1, S2 and S3 denote the sums of each part:  

S1 = A1 + A2 + ... +
Ap  

S2 = Ap+1 + Ap+2 + ... +
Aq

S3 = Aq+1 + Aq+2 + ... +
AN

A partition is acceptable if and only if the differences between S1,
S2 and S3 (i.e. |S1 -
S2|, |S2 - S3|, |S3 -
S1|) are no more than 1.  

Can you tell how many different partitions are acceptable?

输入

The first line contains an integer N.  

The second line contains N integers, A1, A2 ...
AN.  

For 30% of the data, N ≤ 100  

For 70% of the data, N ≤ 1000  

For 100% of the data, N ≤ 100000, -1000000 ≤ Ai ≤ 1000000

输出

The number of acceptable partitions.

样例解释

The three acceptable partitions are:  

3 | 2 | 1 0 2  

3 | 2 1 | 0 2  

3 | 2 1 0 | 2

样例输入

5
3 2 1 0 2

这个题不难想到保证满足题目条件的话，s1,s2,s3至少有两个相等，那么把有至少有两个相等的所有情况写出来，就可以看出来可以分总和%3的情况讨论一下：

sum%3 == 0：分成的三个数必须都是相等的sum / 3

sum%3 == 1:必须是有两个是sum/3 + 1，另外一个是sum / 3

所以题目就变成了s1 = a, s2 = b, s3 = c, a + b + c = s 的方案数。

在这里我当时出问题了，没考虑清楚一些0的情况，看了下别人的思路还是挺巧妙地。

map<ll, ll> ma;
ll work(ll x, ll y, ll z) {
ll res = 0;
ma.clear();
ll as = 0;
for (int i = 0; i < n - 1; i++) {
as += a[i];
if(as == x + y) {
res += ma[x];
}
ma[as]++;
}
return res;
}