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c++ - 斐波那契数列 C++ 显示一个/最大值

转载 作者:行者123 更新时间:2023-11-28 07:44:11 27 4
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从字面上看,有人问我这个问题“”斐波那契数列是 0, 1, 1, 2, 3, 5, 8, 13, … ;前两项是 0 和 1,之后的每一项都是前两项的总和——即 Fib[n] = Fib[n – 1] + Fib[n – 2]。使用此信息,编写一个 C++ 程序来计算斐波那契数列中的第 n 个数,用户在程序中以交互方式输入 n。例如,如果 n = 6,程序应显示值 8。”

感谢您对上一个问题的回答,我已将其放入我的完整代码中。我确实有一个循环,这意味着用户可以选择是否继续该程序。它早些时候工作,但现在没有任何反应。任何人都可以阐明这一点吗?谢谢

{int N;

char ans = 'C';

while (toupper(ans) == 'C')
{
cout<<"This program is designed to give the user any value of the Fibonacci Sequence that they desire, provided the number is a positive integer.";//Tell user what the program does

cout<<"\n\nThe formula of the Fibonacci Sequence is; Fib[N] = Fib[N – 1] + Fib[N – 2]\n\n"; //Declare the Formula for the User

cout<<"Enter a value for N, then press Enter:"; //Declare Value that the User wants to see

cin>>N;//Enter the Number

if (N>1) {
long u = 0, v = 1, t;

for(int Variable=2; Variable<=N; Variable++)
{
t = u + v;
u = v;
v = t;
} //Calculate the Answer

cout<<"\n\nThe "<<N<<"th Number of the Fibonacci Sequence is: "<<t; //Show the Answer
}

if (N<0) {
cout<<"\n\nThe value N must be a POSITIVE integer, i.e. N > 0"; //Confirm that N must be a positive integer. Loop.
}
if (N>100) {
cout<<"\n\nThe value for N must be less than 100, i.e. N < 100. N must be between 0 - 100.";//Confirm that N must be less than 100. Loop.
}
if (N==0) {
cout<<"\n\nFor your value of N, \nFib[0] = 0"; //Value will remain constant throughout, cannot be caculated through formula. Loop.
}
if (N==1) {
cout<<"\n\nFor your value of N. \nFib[1]=1";//Value will remain constant throughout, cannot be caculated through formula. Loop.
}

cout << "\n\nIf you want to select a new value for N, then click C then press Enter. If you want to quit, click P then press Enter: ";
cin >> ans;
}


return 0;

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