HLG 1005 Counting Subsequences [序列定和]

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

code:

#include<stdio.h>
#include<stdlib.h>
#define INFTY 2000000000 
struct node
{
    long z,x;
}q[500010];
int cmp(const void*p1,const void*p2)
{
    struct node*c=(struct node*)p1;
    struct node*d=(struct node*)p2;
    return (c->z)-(d->z);
}
int main()
{
    long n,i,t,p,j,c;
    scanf("%ld",&t);
    while(t--)
    {
    
    
        c=0;q[0].x=0;q[0].z=0;
           scanf("%ld",&n);
        for(i=1;i<=n;i++)
        {
            scanf("%ld",&p);
            q[i].z=q[i-1].z+p;
            q[i].x=i;
        }
        q[n+1].z=INFTY;
        qsort(q,n+1,sizeof(struct node),cmp);
        i=0;j=0;
        while(j<=n)
        {
            if(q[j].z-q[i].z<47)
                j++;
            else if(q[j].z-q[i].z>47)
                i++;
            else
            {
                int ie,je;
                for(ie=i;q[ie+1].z==q[ie].z;ie++);
                ie++;
                for(je=j;q[je+1].z==q[je].z;je++);
                je++;
                while(i<ie&&j<je)
                if(q[i].x>q[j].x)
                        j++;
                    else{
                        c+=(je-j);
                        i++;
                    }
                    i=ie;j=je;
            }
        }
        printf("%ld\n",c);
    }
    return 0;
}