9 interview questions to prove safety Offer --- --- Fibonacci number

To prove safety Offer- face questions 9 --- Fibonacci number

1, Question 1: Output Fibonacci number

Input n, output Fibonacci number is located in bit n value.

Recursive manner efficiency will be low, because there are a large number of repeat calculations. So at this time in order to push is a good idea.
My solution: front to back calculation, the calculated values are stored in the array.

#include <iostream>
#include <vector>
using namespace std;

long long CalculateNums(vector<long long> nums,unsigned int n)
{
    nums[0] = 0;
    nums[1] = 1;
    //每次计算时先检测之前是否已经计算过
    if(n<nums.size()){
        return nums[n];
    }

    //接着从未计算过的位置开始计算
    for(int i=nums.size();i<=n;i++){
        nums.push_back(nums[i-1] + nums[i-2]);
    }
    return nums[n];
}

int main()
{
    int n;
    vector<long long> nums;
    //给初始的两个数预留空间
    nums.resize(2);
    while(cin>>n){
        if(n<0){
            cout<<"输入范围错误"<<endl;
        }
        else{
            cout<<CalculateNums(nums,n)<<endl;
        }
    }

    return 0;
}

2, Topic 2: frog jump step 1

Frog jump step 1: one can choose a frog or a two-step. N-layer jump request jump method has several steps. (N <= 30)

In fact, on stairs and jump Fibonacci number is basically the same. Not repeat wrote.

3, Question 3: frog jump step 2

Frog jump step 2: one can choose frog 1,2,, n-1, n steps. N-layer jump request jump method has several steps. (N <= 30)

This question is considered a good rule: f (n) = 2 ^ (n-1)

My solution: I use the power of fast count

long long CalculateWays(unsigned int n)
{
    if(n==0)
        return 0;
    if(n==1)
        return 1;
    n--;

    long long result = 1;
    long long base = 2;
    while(n != 0){
        if(n%2 !=0){
            result *= base;
        }
        base *= base;
        n /= 2;
    }

    return result;
}

Gangster Solution: bit operation, a line of code to get!

long long CalculateWays(int n)
{
    if(n==0)
        return 0;
    return 1<<(n-1);
}

4. Question 4: 3 steps frog jump

Frog jump step 3: 1 frogs can pick one or two-step. N-layer jump request jump method has several steps. (500> = n> = 100)

This question is clearly defined scope beyond the scope of long long, so use a large integer add operation.

My solution: long-winded large integer +

#include <iostream>
#include <string>
using namespace std;

string CalculateWays(unsigned int n);

string BigNumAdd(string s1,string s2);

int main()
{

    int n;
    cin>>n;
    if(n<=0){
        cout<<"输入范围错误";
    }
    else{
        cout<<CalculateWays(n);
    }

    return 0;
}

string CalculateWays(unsigned int n)
{
    if(n==1)
        return "1";
    if(n==2)
        return "2";
    string lastlast = "1";
    string last = "2";
    string result = "0";
    for(int i=3;i<=n;i++){
        result = BigNumAdd(lastlast,last);
        lastlast = last;
        last = result;
    }
    return result;
}

string BigNumAdd(string s1,string s2)
{
    int len1 = s1.size()-1;
    int len2 = s2.size()-1;
    string result = "";
    int carry = 0;

    //先算两个数都有的位
    while(len1>=0 && len2>=0){
        int add = s1[len1] - 48 + s2[len2] - 48 + carry;
        carry = add/10;
        add%=10;

        char addChar = add+48;
        result = addChar + result;
        len1--;
        len2--;
    }

    //再算位数多的
    while(len1>=0){
        int add = s1[len1] - 48 + carry;
        carry = add/10;
        add%=10;

        char addChar = add+48;
        result = addChar + result;
        len1--;
    }
    while(len2>=0){
        int add = s2[len2] - 48 + carry;
        carry = add/10;
        add%=10;

        char addChar = add+48;
        result = addChar + result;
        len2--;
    }
    //最后是否进位
    if(carry > 0){
        char addChar = carry+48;
        result = addChar + result;
    }
    return result;
}

Gangster's solution: high-quality large integer +

#include <iostream>
#include <string>
#include <vector>
using namespace std;

string BigNumAddString(string s1,string s2)
{
    string& longStr = s1.size()>=s2.size()?s1:s2;
    string& shortStr = s1.size()<s2.size()?s1:s2;
    //给短的字符串前面填充0
    shortStr.insert(0,longStr.size()-shortStr.size(),'0');

    string result;
    //给result预留足够的空间
    result.resize(s1.size()+s2.size()+1);

    //取各自最后一个字符的下标（要从后向前加）
    int longIndex = longStr.size()-1;
    int shortIndex = shortStr.size()-1;
    int resultIndex = result.size()-1;
    //进位标志
    int carry = 0;

    //加运算
    while(shortIndex>=0){
        int add = longStr[longIndex--] - '0' + shortStr[shortIndex--] - '0' + carry;
        carry = add/10;
        result[resultIndex--] = add%10 + '0';
    }
    //最后是否进位
    if(carry > 0){
        result[resultIndex--] = carry + '0';
    }
    //截取从这个参数下标开始一直到结尾（把前面多余的0截掉）
    result = result.substr(resultIndex+1);
    return result;
}

string CalculateWays(unsigned int n)
{
    //也可以使用一个数组，把每次计算结果存进去
    if(n==1)
        return "1";
    if(n==2)
        return "2";
    string lastlast = "1";
    string last = "2";
    string result = "0";
    for(int i=3;i<=n;i++){
        result = BigNumAddString(lastlast,last);
        lastlast = last;
        last = result;
    }
    return result;
}

int main()
{
    int n;
    cin>>n;
    if(n<=0){
        cout<<"输入范围错误"<<endl;
    }
    else{
        cout<<CalculateWays(n);
    }

    return 0;
}

5 new knowledge

Some containers may be used string and resize pre-allocated space, so that the object can be used directly + [] in the form of assignment.