HLG 1005 Counting Subsequences

Description

 "47 is the quintessential random number," states the 47 society. And there might be a grain of truth in that.

For example, the first ten digits of the Euler's constant are:

2 7 1 8 2 8 1 8 2 8

And what's their sum? Of course, it is 47.

You are given a sequence S of integers we saw somewhere in the nature. Your task will be to compute how strongly does this sequence support the above claims.

We will call a continuous subsequence of S interesting if the sum of its terms is equal to 47.

E.g., consider the sequence S = (24, 17, 23, 24, 5, 47). Here we have two interesting continuous subsequences: the sequence (23, 24) and the sequence (47).

Given a sequence S, find the count of its interesting subsequences.

Input

The first line of the input file contains an integer T(T <= 10) specifying the number of test cases. Each test case is preceded by a blank line.

The first line of each test case contains the length of a sequence N(N <= 500000). The second line contains N space-separated integers – the elements of the sequence. Sum of any continuous subsequences will fit in 32 bit signed integers.

Output

For each test case output a single line containing a single integer – the count of interesting subsequences of the given sentence.

Sample Input

2

 

13
2 7 1 8 2 8 1 8 2 8 4 5 9

 

7
2 47 10047 47 1047 47 47

Sample Output

3
4

 

总结： hash or map 保存前面数的和

code1： <map>

#include<stdio.h>
#include<map>
using namespace std;
int t,n,tot,x;
map<long long,int> v;
int main()
{
  scanf("%d",&t);
  while(t--)
  {
    scanf("%d",&n);
    v.clear();
    long long res=0,sum=0;
    v[0] = 1;
    while(n--){
      scanf("%d",&x);
      sum+=x;
      tot+=v[sum-47];
      v[sum]++;
    }
    printf("%d\n",tot);
  }
}

 

code:hash

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <set>
using namespace std;
int sum,now,n,ans;
multiset <int> hash;
void solve()
{
    hash.clear();
    scanf("%d",&n);
    sum=ans=0;
    hash.insert(0);
    for(int i=0;i<n;++i)
    {
        scanf("%d",&now);
        sum=now+sum;
        ans+=hash.count(sum-47);
        hash.insert(sum);
    }
    printf("%d\n",ans);
}
int main()
{
    int t;
    for(scanf("%d",&t);t--;)
    solve();
    return 0;
}