2018快手实习笔试题

// 第二题O(n)空间
#if 0

typedef long long ll;


int main()
{
    ll n;

    while (cin >> n)
    {
        if (n < 5)
        {
            cout << 1 << endl;
            continue;
        }
        vector<ll> dp(n + 1);
        dp[0] = dp[1] = dp[2] = dp[3] = dp[4] = 1;


        for (ll i = 5; i <= n; i++)
        {
            dp[i] = 2018 * dp[i - 1] % 1000000003 + 2017 * dp[i - 2] % 1000000003 + 2016 * dp[i - 3] % 1000000003 + 2015 * dp[i - 4] % 1000000003 + 2014 * dp[i - 5] % 1000000003;
        }

        cout << dp[n] % 1000000003 << endl;
    }

    return 0;
}

#endif

// 第二题 时间复杂度超时
#if 0

typedef long long ll;

inline ll ksc(ll x, ll y, ll mod)
{
    return (x*y - (ll)((long double)x / mod*y)*mod + mod) % mod;
}


int main()
{
    ll n;

    while (scanf("%lld",&n)!=EOF)
        //while (cin >> n)
    {
        if (n < 5)
        {
            //cout << 1 << endl;
            printf("1
");
            continue;
        }
        vector<ll> dp(6);
        dp[0] = dp[1] = dp[2] = dp[3] = dp[4] = 1;

        // 使用6个空间，然后每次都需要移位
        for (ll i = 5; i <= n; i++)
        {
            dp[5] = 2018 * (dp[4] % 1000000003) + 2017 * (dp[3] % 1000000003) + 2016 * (dp[2] % 1000000003) + 2015 * (dp[1] % 1000000003) + 2014 * (dp[0] % 1000000003);
            for (int j = 0; j < 5; j++)
            {
                dp[j] = dp[j + 1];
            }
        }
        printf("%lld
", dp[5] % 1000000003);

        // 不用移位
        //for (ll i = 5; i <= n; i++)
        //{
        //    dp[i % 6] = ksc(2018, dp[(i - 1) % 6], 1000000003) + ksc(2017, dp[(i - 2) % 6], 1000000003) + ksc(2016, dp[(i - 3) % 6], 1000000003) + ksc(2015, dp[(i - 4) % 6], 1000000003) + ksc(2014, dp[(i - 5) % 6], 1000000003);
        //    //dp[i % 6] = 2018 * (dp[(i - 1) % 6] % 1000000003) + 2017 * (dp[(i - 2) % 6] % 1000000003) + 2016 * (dp[(i - 3) % 6] % 1000000003) + 2015 * (dp[(i - 4) % 6] % 1000000003) + 2014 * (dp[(i - 5) % 6] % 1000000003);
        //}
        ////cout << dp[n%6] % 1000000003 << endl;
        //printf("%lld
", dp[n % 6] % 1000000003);
    }

    return 0;
}

#endif

// 第二题 时间复杂度超时
#if 0

typedef long long ll;

long long quick_multiply(long long a, long long b, long long mod) {
    long long result = 0;
    while (b) {
        result = (result + (b % 2 * a) % mod) % mod;
        a = a * 2 % mod;
        b = b / 2;
    }
    return result;
}

inline ll ksc(ll x, ll y, ll mod)
{
    return (x*y - (ll)((long double)x / mod*y)*mod + mod) % mod;
}

int main()
{
    ll n;

    while (scanf("%lld", &n) != EOF)
        //while (cin >> n)
    {
        if (n < 5)
        {
            //cout << 1 << endl;
            printf("1
");
            continue;
        }
        vector<ll> dp(6);
        dp[0] = dp[1] = dp[2] = dp[3] = dp[4] = 1;


        for (ll i = 5; i <= n; i++)
        {
            dp[i % 6] = quick_multiply(2018, dp[(i - 1) % 6], 1000000003) + quick_multiply(2017, dp[(i - 2) % 6], 1000000003) + quick_multiply(2016, dp[(i - 3) % 6], 1000000003) + quick_multiply(2015, dp[(i - 4) % 6], 1000000003) + quick_multiply(2014, dp[(i - 5) % 6], 1000000003);
            //dp[i % 6] = 2018 * (dp[(i - 1) % 6] % 1000000003) + 2017 * (dp[(i - 2) % 6] % 1000000003) + 2016 * (dp[(i - 3) % 6] % 1000000003) + 2015 * (dp[(i - 4) % 6] % 1000000003) + 2014 * (dp[(i - 5) % 6] % 1000000003);
        }

        //cout << dp[n%6] % 1000000003 << endl;
        printf("%lld
", dp[n % 6] % 1000000003);
    }

    return 0;

}

#endif


#if 0

// 快速幂运算求解斐波拉契数列
#include <cstdio>
#include <iostream>

using namespace std;

const int MOD = 10000;

struct matrix
{
    int m[2][2];
}ans, base;

matrix multi(matrix a, matrix b)
{
    matrix tmp;
    for(int i = 0; i < 2; ++i)
    {
        for(int j = 0; j < 2; ++j)
        {
            tmp.m[i][j] = 0;
            for(int k = 0; k < 2; ++k)
                tmp.m[i][j] = (tmp.m[i][j] + a.m[i][k] * b.m[k][j]) % MOD;
        }
    }
    return tmp;
}
int fast_mod(int n)  // 求矩阵 base 的  n 次幂 
{
    base.m[0][0] = base.m[0][1] = base.m[1][0] = 1;
    base.m[1][1] = 0;
    ans.m[0][0] = ans.m[1][1] = 1;  // ans 初始化为单位矩阵 
    ans.m[0][1] = ans.m[1][0] = 0;
    while (n)
    {
        if (n & 1)  //实现 ans *= t; 其中要先把 ans赋值给 tmp，然后用 ans = tmp * t 
        {
            ans = multi(ans, base);
        }
        base = multi(base, base);
        n >>= 1;
    }
    return ans.m[0][1];
}

int main()
{
    int n;
    while (scanf("%d", &n) && n != -1)
    {
        printf("%d
", fast_mod(n));
    }
    return 0;
}

#endif

//计算(x^y) % N; 注：(x^y)表示x的y次方
#if 0

int main()
{
    int x, y, N;

    cin >> x >> y >> N;

    long res = 1;
    x = x % N; //开始

    while (y > 0) {
        if (y % 2 == 1) //等价于 if(y&1)
            res = (res * x) % N;
        y /= 2; //y>>1; 分解y为二进制编码
        x = (x * x) % N;
    }

    cout << res << endl;

    return 0;
}

#endif

• 快手2018春季校园招聘笔试试卷--算法A试卷编程题题解

// 第一题
#if 0

int main()
{
    int N, M;
    cin >> N >> M;

    unordered_set<string> st;

    for (int i = 0; i < N;i++)
    {
        string temp;
        cin >> temp;
        st.insert(temp);
    }

    for (int i = 0; i < M;i++)
    {
        string temp;
        cin >> temp;
        if (st.find(temp)!=st.end()) //找到了
        {
            cout << "NO" << endl;
        }
        else
        {
            st.insert(temp);
            cout << "YES" << endl;
        }
    }


    return 0;
}
#endif