1962-Fibonacci

最新推荐文章于 2025-03-15 22:47:02 发布

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最新推荐文章于 2025-03-15 22:47:02 发布

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原文链接：http://www.cnblogs.com/Rosanna/p/3438659.html

本文介绍了一个类似于斐波那契数列的问题，并提供了一段C++代码实现。该问题要求根据给定的初始值计算特定位置的序列值。代码通过迭代的方式解决了这一问题。

描述

This is an easy problem.I think Fibonacci sequence is familiar to you.Now there is another one.

Here a and b are constants.Gice you a,b,and n,your task is to calculate the f[n].

输入

First line of input comes a positive integer T（T<=10) indicating the number of test cases.Each test case contains three positive integer a,b and n(a<=10,b<=10,n<=30)

输出

Print one line containing an integer,i.e.f[n],for each test case.

样例输入

1 2 3

1 3 6

样例输出

#include<iostream>
using namespace std;

int main()
{
    int m,a,b,n,t;
    int sum=0;
    cin>>m;
    while(m--)
    {
        cin>>a>>b>>n;
        if(n==1) cout<<a<<endl;
        else if(n==2) cout<<b<<endl;
        else{
        for(int i=3;i<=n;i++)
        {
            if(i%2==1) t=a+b;            
            else
            {
                t=a+a+b+b;
                a=b+a;
                b=t;
            }
        }
        cout<<t<<endl;
        }
    }
    return 0;
}

转载于:https://www.cnblogs.com/Rosanna/p/3438659.html