微软夏令营笔试测验第一题 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
样例输出
3
动态规划问题，按照题目的要求只有可能出现三种情况，x，x，x；x，x，x+1；x，x，x-1；很明显这三种情况是互斥的。
对于每一种情况有O(n)的算法，首先从后向前计算，用一个数组保存当前位置及之后的位置包含多少个和为sum的情况，用一个数组进行统计，之后采用从前到后遍历，遍历确定前半段的和等于固定值之后寻找跳过一个的位置存储的值（保证最少有一个划分元素），相加即可
#include
using namespace std;
int *s, *x, n, c;
int res = 0;
vector<int> cal_pos_sum(int *s,int n,int z){
if (n <= 0)
return vector < int > {};
vector<int> res(n, 0);
if (s[n-1]==z)
res[n - 1] = 1;
int temp = s[n - 1];
for (int i = n-2; i >= 0; --i)
{
temp += s[i];
if (temp == z)
res[i] = res[i + 1] + 1;
else
res[i] = res[i + 1];
}
return res;
}
int count_sum(int x,int y)
{
vector<int> pos_sum;
pos_sum = cal_pos_sum(s,n,y);
int res = 0;
int sum = 0;
for (int i = 0; i < n - 2; ++i)
{
sum += s[i];
if (sum == x){
res += pos_sum[i + 2];
}
}
return res;
}
int main()
{
int all_sum = 0;
int i;
scanf_s("%d", &n);
s = new int[n];
for (i = 0; i"%d", &s[i]);
all_sum += s[i];
}
if ((all_sum + 1) % 3 == 0)
{
int out = 0;
c = (all_sum + 1) / 3;
out = count_sum(c,c - 1) + count_sum(c,c) + count_sum(c - 1,c);
cout << out << endl;
}
else if ((all_sum - 1) % 3 == 0)
{
int out = 0;
c = (all_sum - 1) / 3;
out = count_sum(c,c + 1) + count_sum(c,c) + count_sum(c + 1,c);
cout << out << endl;
}
else if ((all_sum) % 3 == 0)
{
int out = 0;
c = (all_sum) / 3;
out = count_sum(c,c);
cout << out << endl;
}
else
cout << 0 << endl;
return 0;
}