腾讯笔试题——用1，1，2，2，4，4，8，8...2^i,2^i拼凑成一个整数n，求问多少种拼凑方法

用1，1，2，2，4，4，8，8…2^i,2^i拼凑成一个整数n，求问多少种拼凑方法

话不多说，上代码：

#include<iostream>#include<math.h>#include<time.h>#include<set>using namespace std;int nums[63];//num[i]表示2^i的个数，只有0,1,2三个取值//回溯法int IsOk(long long n,int *nums,int len){    long long sum = 0;    for (int i = 0; i < len; i++)        sum += nums[i] * pow(2, i);    if (sum==n)        return 0;    else if (sum > n)        return -1;    else        return 1;}void solve(long long n, int index, int *nums, int len,int &count){    if (index >= len)        return;    for (int i = 0; i <= 2; i++)    {        nums[index] = i;        if (IsOk(n, nums, len) == 0)            count++;        else if (IsOk(n, nums, len) == 1)            solve(n, index + 1, nums, len, count);    }    nums[index] = 0;//回溯法，要撤回上一步的假设}long long DP(long long n)//使用动态规划方法{    int len = log(n) / log(2) + 1;    long long **dp = new long long*[n + 1];    for (long long i = 0; i <= n; i++)    {        dp[i] = new long long[len];    }    for (int i = 0; i < len; i++)        for (long long j = 0; j <= n; j++)            dp[j][i] = 0;    //dp[n][i]表示使用1，1，2，2，4，4，...，2^i可以组合出n的方案数    for (int i = 0; i < len; i++)        dp[0][i] = 1;    if (n == 1||n==2)        return n;    dp[1][0] = 1;    dp[2][0] = 1;    for (int i = 3; i <= n; i++)        dp[i][0] = 0;    //dp[n][i]=//  cout << "len=" << len << endl;    for (int i = 1; i < len; i++)        for (long long j = 1; j <= n; j++)        {            for (int m = 0; m <= 2; m++)                if (j - pow(2, i)*m >= 0)                {                    dp[j][i] = dp[j][i] + dp[(long long)(j - pow(2, i)*m)][i - 1];                    //cout <<"j="<< j << " " << "i=" << i << " " << "m=" << m<<" "<< dp[j][i]<<endl;                }        }    return dp[n][len - 1];}int solve3(long long n){    long long stop = n / 2;    long long res = 0;    set<int> myset;    /*    将硬币分为两份：1，2，4，8，16，.....和1，2，4，8，16....    组成两个数值为a,b的两个数字，他们的和是a+b=n;     a在每一份中只可能有一种组合方式（二进制的思想）    */    for (int i = 1; i <= stop; i++)    {        res = i ^ (n - i);        myset.insert(res);    }    //对于1,2,4,8结果再加1.    int len = log(n) / log(2) + 1;    if (pow(2, len-1) == n)        return myset.size() + 1;    return myset.size();}int main(){    for (int i = 0; i < 63; i++)        nums[i] = 0;    long long n;    clock_t start, finish;    while (true)    {        cin >> n;        int len = log(n) / log(2) + 1;        int count = 0;        start = clock();        solve(n, 0, nums, len, count);        cout << count << endl;        finish = clock();        cout << "回溯法耗费时间为" << (double)(finish - start) / CLOCKS_PER_SEC<<"秒"<<endl;        //        //        start = clock();        long long res = DP(n);        cout << res << endl;        finish = clock();        cout << "动态规划方法耗费时间为" << (double)(finish - start) / CLOCKS_PER_SEC << "秒" << endl;        start = clock();        res = solve3(n);        cout << res << endl;        finish = clock();        cout << "第三种方法耗费时间为" << (double)(finish - start) / CLOCKS_PER_SEC << "秒" << endl;    }    return 0;   }

输入a,b，A，B，求问：能否经过同时+1或者同时*2变成A，Bint method1(int a,int b,int A,int B){    if(a>A || b>B) return -1;    int c = a-b,        d = A-B;    if(c*d < 0) return -1;    if(c==0 && d==0){        // ok    }else if(c!=0 && d!=0){        if(d%c!=0) return -1;        int e = d/c;        if(e&(e-1)!=0) return -1;    }else return -1;    int res = 0;    while(true){        if(A==a) return res;        if(A%2==0){            if(A/2>=a){                A>>=1;                res++;            }else{                res += A-a;                return res;            }        }else{            A-=1;            res++;        }    }}

我的方法：

#include<iostream>using namespace std;bool fun(int a,int A,int b,int B,int &count){    if (a == A&&b==B)        return true;    if (a > A||b>B)        return false;    int count1 = 0;    int count2 = 0;    bool res1 = fun(a + 1, A, b + 1, B, count1);    bool res2 = fun(2 * a, A, 2 * b, B, count2);    if (res1 || res2)    {        if (res1&&res2)        {            count = count1 > count2 ? count2 + 1 : count1 + 1;            return true;        }        if (res1)        {            count = count1 + 1;            return true;        }        if (res2)        {            count = count2 + 1;            return true;        }    }    else    {        count = -1;        return false;    }}int main(){    int a;    int A;    //cin >> a;    //cin >> A;    a = 100;    A = 408;    int b = 1000;    int B = 4009;    int count = 0;    fun(a, A,b,B,count);    cout << count << endl;return 0;}