我遇到一个任务有问题,所以如果你能帮助我一点点。
数字是幸运的&#34;或者&#34;运气不好&#34;。数字是&#34;幸运&#34;就好 数字7 或者每个数字都是4.所以&#34;幸运&#34;数字例如是4,44,7,77。 &#34;不幸&#34;是其他数字 你将获得n元素序列和数字K.你的任务是 计算所有可能的k元素子序列的数量,它们满足一个 条件。条件是在子序列中不能是两个相同的&#34;幸运&#34; 数字。所以例如那里没有77和77 ...... 所有可能的k元素的子序列mod 10 ^ 9 + 7的输出数量 0&lt; N,K < 10 ^ 5
Few examples: Input: 5 2 7 7 3 7 77 Output: 7 Input: 5 3 3 7 77 7 77 Output: 4 Input: 34 17 14 14 14 ... 14 14 14 Output: 333606206
我的代码似乎有用,但是当我尝试计算二项式系数时它太慢了。我正在使用地图。在字符串I中以字符串格式存储数字。在第二个 - 地图 - 部分地图是数字,表示使用的数字(在第一个地图参数中)的次数。所以现在我已经存储了所有不幸的#34;数字存储在一起。也是每一个&#34;幸运&#34;数字在一起。当我像这样存储它时,我只计算所有乘法。例如:
Input
5 2
3 7 7 77 7
Are stored like this: map["other"] = 1 map["7"] = 3 map["77"] = 1
Because k = 2 --> result is: 1*3 + 1*1 + 1*3 = 7.
我认为问题在于计算二项式系数。对于第三个例子,它需要计算(34选择17)并计算很长时间。我找到了this article和this,但我不明白它们是怎样的正在解决这个问题。
我的代码:
#include<iostream>
#include<string>
#include<map>
#include <algorithm>
#include <vector>
using namespace std;
int binomialCoeff(int n, int k)
{
// Base Cases
if (k == 0 || k == n)
return 1;
// Recur
return binomialCoeff(n - 1, k - 1) + binomialCoeff(n - 1, k);
}
int main()
{
int n, k;
cin >> n >> k;
map<string, int> mapa; // create map, string is a number, int represents number of used string-stored numbers ---> so if 7 was used two times, in the map it will be stored like this mapa["7"] == 2 and so on)
for (int i = 0; i < n; i++) // I will load number as string, if this number is "lucky" - digist are all 7 or all 4
{ // every "unlucky" numbers are together, as well as all same "lucky" numbers ---> so 77 and 77 will be stored in one element....
string number;
cin >> number;
char digit = number[0];
bool lucky = false;
if (digit == '7' || digit == '4')
lucky = true;
for (int j = 1; j < number.length(); j++) {
if (digit != '7' && digit != '4')
break;
if (number[j] != digit) {
lucky = false;
break;
}
}
if (lucky)
mapa[number]++;
else
mapa["other"]++;
}
vector<bool> v(mapa.size());
bool lack = k > mapa.size(); //lack of elements in map --> it is when mapa.size() < k; i. e. number of elements in array can't make k-element subsequence.
int rest = lack ? k - mapa.size() + 1 : 1; // how many elements from "unlucky" numbers I must choose, so it makes base for binomial coefficient (n choose rest)
if (lack) //if lack is true, different size of vector
fill(v.begin() + mapa.size(), v.end(), true);
else
fill(v.begin() + k, v.end(), true);
int *array = new int[mapa.size()]; //easier to manipulate with array for me
int sum = 0;
int product = 1;
int index = 0;
for (map<string, int> ::iterator pos = mapa.begin(); pos != mapa.end(); ++pos) // create array from map
{
if (lack && pos->first == "other") { //if lack of elements in map, the number in elemets representing "unlucky" numbers will be binomial coefficient (mapa["other] choose rest)
array[index++] = binomialCoeff(mapa["other"], rest);
continue;
}
array[index++] = pos->second;
}
do { // this will create every posible multiplication for k-elements subsequences
product = 1;
for (int i = 0; i < mapa.size(); ++i) {
if (!v[i]) {
product *= array[i];
}
}
sum += product;
} while (next_permutation(v.begin(), v.end()));
if (mapa["other"] >= k && mapa.size() > 1) { // if number of "unlucky" numbers is bigger than k, we need to compute all possible k-elements subsequences just from "unlucky" number, so binomial coefficient (mapa["other] choose k)
sum += binomialCoeff(mapa["other"], k);
}
cout << sum % 1000000007 << endl;
}