很好的一道思维题目（有坑点的）

最新推荐文章于 2021-10-13 15:33:16 发布

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最新推荐文章于 2021-10-13 15:33:16 发布

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本文介绍了一种算法，用于计算给定整数序列中所有连续子序列的个数，这些子序列的元素之和等于47。通过使用哈希映射来记录已遇到的累积和及其出现次数，该算法能在O(n)时间内解决问题。

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Counting Subsequences

Time Limit: 5000 MS

Memory Limit: 65536 K

Total Submit: 850(264 users)

Total Accepted: 240(208 users)

Rating:

Special Judge: No

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

ac代码：

#include<bits/stdc++.h>
using namespace std;
map<long long  ,int >w;
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {

        int n;
        w.clear();
        scanf("%d",&n);
        long long  sum=0,ans=0;

        w[sum]=1;
        for(int i=0; i<n; i++)
        {
            long long  q;
            scanf("%lld",&q);
            sum+=q;
            w[sum]++;

                ans+= w[sum-47];
        }
        printf("%lld\n",ans);
    }
}

这个题比较坑人的一点就是自然数是包括0的。