基本上我只是在一段时间后再次开始做C ++因为我需要(Degree sorta命令它)并且我的任务是编写一个简单的程序,它将采用一个函数并使用2个整数输入(N和M),返回双输出(S)。在一个部分中,我被要求使用一个循环来显示S的值,一直到N = 10,从N = 0开始,值M = 10
我遇到了一个问题,其中返回为每N到10提供值“5”。
这是代码:(不介意评论)
#include <iostream>
#include <iomanip>
#include <fstream>
#include <cmath>
//Function, Part A
double func_18710726(int N, int M)
{
double S = 0;
for (int n = 1; n <= N; n++)
for (int m = 1; m <= M; m++)
{
S = S + (sqrt(m*n)+exp(sqrt(m))+ exp(sqrt(n)))/(m*n + 2);
}
return S;
}
//Part B
double func_18710726(int, int);
using namespace std;
int main()
{
int N, M;
double S;
//Part B1
do {
cout << "Enter Value of N for N > 0 and an integer" << endl;
cin >> N;
} while (N <= 0);
do {
cout << "Enter value of M for M > 0 and an integer" << endl;
cin >> M;
} while(M <= 0);
//Part B2
S = func_18710726(N, M);
cout << "The Summation is ";
cout << fixed << setprecision(5) << S << endl;
//Part B3
ofstream output;
output.open("Doublesum.txt");
M = 1;
for (int n = 1; n <= 10; n++)
{
S = func_18710726(n, M);
cout << "The summation for N = " << n << " is ";
cout << fixed << setprecision(5) << 5 << endl;
output << fixed << setprecision(5) << 5 << endl;
}
output.close();
return 0;
}
输出给了我:
Enter Value of N for N > 0 and an integer
1
Enter value of M for M > 0 and an integer
2
The Summation is 4.20696
The summation for N = 1 is 5
The summation for N = 2 is 5
The summation for N = 3 is 5
The summation for N = 4 is 5
The summation for N = 5 is 5
The summation for N = 6 is 5
The summation for N = 7 is 5
The summation for N = 8 is 5
The summation for N = 9 is 5
The summation for N = 10 is 5
--------------------------------
Process exited after 2.971 seconds with return value 0
Press any key to continue . . .
非常感谢有关为何发生这种情况的任何帮助。
我很抱歉,如果我把它发布在错误的地方,如果我这样做,Mods请放轻松我:)
答案 0 :(得分:2)
这一行:
cout << fixed << setprecision(5) << 5 << endl;
输出5
(五) - 你想要S
(esss)
对于变量来说,S
可能不是一个好名字(l
<)