C ++中Kn +（K(n-1)*(K-1)1）+（K(n-2)*(K-1)2）+ ...(K-1)n的和
在问题中，我们是序列K ^ n +（K ^(n-1)*(K-1)^ 1）+（K ^(n-2)*(K-1)的两个数字k和n ）^ 2）+ ...(K-1)^ n。我们的任务是创建一个程序来查找序列的总和。
让我们举个例子来了解这个问题，
Input: n = 3, k = 4Output: 175
Explanation: Sum of the series is
= 4^3 + ( (4^2)*(3^1) ) + ( (4^1)*(3^2) ) + ( (4^0)*(3^3) )
= 64 + 48 + 36 + 27 = 175
解决问题的一种简单方法是使用for循环。找到该系列的每个术语，并将其值加到总和上。
算法
initialise sum = 0;Step 1: for i -> 0 to n.
Step 1.1: update sum: sum += pow(k, n-i) * pow(k, i)
Step 2: return sum.
示例
该程序说明了我们解决方案的工作原理，
#include <iostream>#include <math.h>
using namespace std;
int calcSeriesSum(int k, int n) {
int sum = 0;
for (int i = 0; i <= n; i++) {
int p = pow(k, n-i) * pow((k-1), i);
sum = sum + p;
}
return sum;
}
int main() {
int n = 4;
int K = 2;
cout<<"Sum of the series is "<<calcSeriesSum(K, n);
}
输出结果
Sum of the series is 31
该解决方案效率不高，并且花费的时间约为n。
一个有效的解决方案是找到级数总和的通用公式。
The series K^n + ( K^(n-1) * (K-1)^1 ) + ( K^(n-2) * (K-1)^2 ) + ... (K-1)^nForms a geometric progression. The common ration of this progression is (k-1)/k and the first term is k^n.
sum = K^n + ( K^(n-1) * (K-1)^1 ) + ( K^(n-2) * (K-1)^2 ) + ... (K-1)^n
sum = kn(1 + (k-1)/k + (k-1)2/k2 + … + (k-1)n)
sum = ((kn)(1 - ( (k-1)(n+1))/k(n+1))) / (1 - ((k-1)/k))
sum = kn ( (k(n+1) - (k-1)(n+1))/k(n+1) ) / ( (k - (k-1))/k )
sum = kn ( (k(n+1) - (k-1)(n+1))/k(n+1) ) / (1/k)
sum = kn ( (k(n+1) - (k-1)(n+1))/k(n+1) ) * k
sum = ( k(n+1) - (k-1)(n+1) )
示例
该程序说明了我们解决方案的工作原理，
#include <iostream>#include <math.h>
using namespace std;
int calcSeriesSum(int k, int n) {
return ( pow(k,(n+1)) - pow((k-1),(n+1)) );
;
}
int main() {
int n = 4;
int K = 2;
cout<<"Sum of the series is "<<calcSeriesSum(K, n);
}
输出结果
Sum of the series is 31
以上是 C ++中Kn +（K(n-1)*(K-1)1）+（K(n-2)*(K-1)2）+ ...(K-1)n的和 的全部内容， 来源链接： utcz.com/z/326494.html
