题目如下：
Counting Subsequences
Time Limit: 5000 MSMemory Limit: 65536 K
| 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 7 |
| Sample Output |
| 3 4 |
这道题的意思就是给你一个整形的序列，让你计算满足和是47的子序列的个数。序列中的值可能有负数。
所以最直观的方法就是计算所有子序列的和，然后判断是否与47相等。算法的时间复杂度为O(N³)，代码如下：
#include<iostream>
#include<vector>
#include<map>
#define N 47
using namespace std;
template<class type>
int countSubseq(const vector<type> &data)
{
int count = 0;
for(size_t i = 0; i < data.size(); ++i)
{
int sum = 0;
for(int j = 0; j <= i; ++j)
{
int sum = 0;
for(int k = j; k <= i; ++k)
{
sum += data[k];
}
if(sum == N)
{
++count;
}
}
}
return count;
}
int main()
{
int nCase;
cin >> nCase;
for(int i = 0; i < nCase; ++i)
{
int n, d;
cin >> n;
vector<int> data(n, 0);
for(int j = 0; j < n; ++j)
{
cin >> d;
data[j] = d;
}
cout << countSubseq(data) << endl;
}
return 0;
}
通过分析可以看出程序存在许多重复计算，上一个sum可以进行一次加法运算得到下一个sum。经过再次优化使其时间复杂度变为O(N²)，代码如下：
template<class type>
int countSubseq(const vector<type> &data)
{
int count = 0;
for(size_t i = 0; i < data.size(); ++i)
{
int sum = 0;
for(int j = i; j < data.size(); ++j)
{
sum += data[j];
if(sum == N)
{
++count;
}
}
}
return count;
}
但是程序依然超时，所以要设计出复杂度更低的算法才行。
设SUM[i,j] = A[i] + A[i+1] + ... + A[j] (0 <= i <= j < N)，所以SUM[i, j] = SUM[0, j] - SUM[,0, i-1]，也就是说只要求出所有的SUM[0, K] (K∈[0,N))就能计算出任意SUM[i, j].那么我们每次计算K的一种取值得到的SUM时，查找之前计算的SUM是否有与当前SUM差是47的K的个数，可以使用map来降低查找的复杂度，这样时间复杂度降为O(N)，代码如下：
template<class type>
int countSubseq(const vector<type> &data)
{
int count = 0, sum = 0;
map<int, int> sumToSeqCnt;//存储和是sum的序列K的个数
sumToSeqCnt[0] = 1;
for(size_t i = 0; i < data.size(); ++i)
{
sum += data[i];
sumToSeqCnt[sum]++;//计算和是sum的序列K的个数
count += sumToSeqCnt[sum - N];
}
return count;
}
计算子序列和是定值的子序列个数