Eugene likes working with arrays. And today he needs your help in solving one challenging task.
An array c is a subarray of an array b if c can be obtained from b by deletion of several (possibly, zero or all) elements from the beginning and several (possibly, zero or all) elements from the end.
Let’s call a nonempty array good if for every nonempty subarray of this array, sum of the elements of this subarray is nonzero. For example, array [−1,2,−3] is good, as all arrays [−1], [−1,2], [−1,2,−3], [2], [2,−3], [−3] have nonzero sums of elements. However, array [−1,2,−1,−3] isn’t good, as his subarray [−1,2,−1] has sum of elements equal to 0.
Help Eugene to calculate the number of nonempty good subarrays of a given array a.
Input
The first line of the input contains a single integer n (1≤n≤2×105) — the length of array a.
The second line of the input contains n integers a1,a2,…,an (−109≤ai≤109) — the elements of a.
Output
Output a single integer — the number of good subarrays of a.
Examples
Input
3
1 2 -3
Output
5
Input
3
41 -41 41
Output
3
Note
In the first sample, the following subarrays are good: [1], [1,2], [2], [2,−3], [−3]. However, the subarray [1,2,−3] isn’t good, as its subarray [1,2,−3] has sum of elements equal to 0.
In the second sample, three subarrays of size 1 are the only good subarrays. At the same time, the subarray [41,−41,41] isn’t good, as its subarray [41,−41] has sum of elements equal to 0.
尺取的题目还是一样的不会，搞了好久。。。
思路：维护一个前缀和应该是挺好想到的。但是关键问题是怎么有序不重复的维护good序列。盲目的去做很容易漏掉一些东西或者重复计算一些东西。最暴力的做法是什么？就是两重循环遍历去找，但是那样的话，时间是不允许的。有一个性质，对于一个区间[l,r]，这个区间里都是good序列的条件是：前缀数组在[l,r]中没有重复的值，例如，如果sum[i]=x,sum[j]也等于x的话，就说明sum[i+1]+sum[i+2]+…sum[j]=0。这样的话问题就转化成，对于一个右端点R，如何找到最小的左端点L，使得[L,R]中没有重复值。这个时候就很容易就想到尺取法了。假如当前右端点的sum值出现过，那么我们找到上一次出现这个值的下一个位置，从这个位置到当前右端点的位置就肯定没有非good序列了，ans加上这一段区间的长度，就代表这一段区间的good序列的数量了。
代码如下：
#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxx=2e5+100;
int a[maxx];
ll sum[maxx];
int n;
int main()
{
scanf("%d",&n);
memset(sum,0,sizeof(sum));
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
sum[i]=sum[i-1]+a[i];
}
map<ll,bool> mp;
mp.clear();
int l=0,r=1;
ll ans=0;
mp[0]=1;
while(r<=n)
{
while(mp[sum[r]])
{
mp[sum[l++]]=0;
}
mp[sum[r]]=1;
ans+=(r-l);
r++;
}
printf("%lld\n",ans);
return 0;
}
努力加油a啊，(o)/~
