“字节跳动-文远知行杯”广东工业大学第十四届程序设计竞赛

a、hzy 和zsl 的生存挑战

思路：看见“假设他们都足够机智”就够了，全输出1.00就ac

#include
#define INF 0x3f3f3f3f
#define ll long long
#define mem(ar,num) memset(ar,num,sizeof(ar))
#define me(ar) memset(ar,0,sizeof(ar))
#define lowbit(x) (x&(-x))
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define lcm(a,b) ((a)*(b)/(__gcd((a),(b))))
#define Max 3e5+10
#define mod 192600817
using namespace std;

int main() {
    printf("%.2f\n",4/4.0);
    printf("%.2f\n",4/4.0);
    printf("%.2f\n",4/4.0);
    printf("%.2f\n",4/4.0);
    return 0;
}

b、人类史上最大最好的希望事件

思路：暴力算。。。

#include
#define INF 0x3f3f3f3f
#define ll long long
#define mem(ar,num) memset(ar,num,sizeof(ar))
#define me(ar) memset(ar,0,sizeof(ar))
#define lowbit(x) (x&(-x))
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define lcm(a,b) ((a)*(b)/(__gcd((a),(b))))
#define Max 3e5+10
#define mod 192600817
using namespace std;
ll t, a, b, c, d, fb[50010], m[50010];
void init() {
    fb[1] = 1;
    m[1] = 1;
    for(int i = 2; i < 50010; i++) {
        fb[i] = (fb[i - 1] + fb[i - 2]) % mod;
        m[i] = (fb[i] * fb[i]) % mod + m[i - 1];
    }
}
int main() {
    while(cin >> t) {
        init();
        while(t--) {
            cin >> a >> b >> c >> d;
            a = a * 4 + b + 1;
            c = c * 4 + d + 1;
            if(a > c)
                swap(a, c);
            ll ans = 0;
            ans = (m[c] - m[a - 1]) % mod;
            cout << ans % mod << endl;
        }
    }
    return 0;
}

c、超级无敌简单题

思路：数据没那么大，预处理一下，遇见不行的及时return。

#include
#define INF 0x3f3f3f3f
#define ll long long
#define mem(ar,num) memset(ar,num,sizeof(ar))
#define me(ar) memset(ar,0,sizeof(ar))
#define lowbit(x) (x&(-x))
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define lcm(a,b) ((a)*(b)/(__gcd((a),(b))))
#define Max 3e5+10
#define mod 192600817
using namespace std;
int ge[150010], g;
int gts(int y) {
    int ans = 0, x = y;
    while(x != 0) {
        int k = x % 10;
        x /= 10;
        ans += k * k;
    }
    return ans;
}
int isge(int k) {
    int s=k;
    while(1){
        s=gts(s);
        if(s==1)return 1;
        if(s==2||s==3||s==4||s==5||s==6||s==8||s==9||s==14||s==15||s==16||s==17||s==18||s==25||s==30)return 0;
    }
}
int main() {
    for(int i = 1; i <= 1000000; i++) {
        if(isge(i)) {
            ge[++g] = i;
        }
    }
    int t;
    cin >> t;
    while(t--) {
        int k;
        cin >> k;
        cout << ge[k] << endl;
    }
    return 0;
}

g、简单数学题

思路：在纸上分开写后发现可以套用这个公式。

然后利用等比数列求和来简化计算

下面贴一张常用的组合数公式：

#include
#define INF 0x3f3f3f3f
#define ll long long
#define mem(ar,num) memset(ar,num,sizeof(ar))
#define me(ar) memset(ar,0,sizeof(ar))
#define lowbit(x) (x&(-x))
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define lcm(a,b) ((a)*(b)/(__gcd((a),(b))))
#define Max 3e5+10
#define mod 1000000007
using namespace std;
ll n;
ll poww(ll a, ll b) {
    ll ans = 1, push = a;
    while(b > 0) {
        if(b & 1) {
            ans *= push;
            ans %= mod;
        }
        push *= push;
        push %= mod;
        b >>= 1;
    }
    return ans % mod;
}
int main() {
    while(~scanf("%lld", &n)) {
        cout << (n - 1) % mod * poww(2, n) % mod + 1 << endl;
    }
    return 0;
}

j、Count

思路：矩阵快速幂。

将转换成

f(n+1)=f(n)+2*f(n-1)+     根据这个构造矩阵。

#include
#define INF 0x3f3f3f3f
#define ll long long
#define mem(ar,num) memset(ar,num,sizeof(ar))
#define me(ar) memset(ar,0,sizeof(ar))
#define lowbit(x) (x&(-x))
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define lcm(a,b) ((a)*(b)/(__gcd((a),(b))))
#define Max 3e5+10
#define mod 123456789
using namespace std;
ll t, n, k;
struct mtx {
    ll x[6 + 1][6 + 1];
    mtx() {
        memset(x, 0, sizeof x);
    }
} ans;
mtx operator *(const mtx &a, const mtx &b) {
    mtx c;
    for(int i = 0; i < 6; i++)
        for(int j = 0; j < 6; j++)
            for(int k = 0; k < 6; k++)
                c.x[i][j] = (c.x[i][j] + a.x[i][k] * b.x[k][j] % mod) % mod;
    return c;
}
mtx operator ^(mtx a, ll k) {
    mtx ret;
    for(int i = 0; i < 6; ++i)
        ret.x[i][i] = 1;
    while(k) {
        if(k & 1)
            ret = ret * a;
        a = a * a;
        k >>= 1;
    }
    return ret;
}
int main() {
    cin >> t;
    while(t--) {
        mtx ak;
        cin >> n;
        if(n == 1) {
            cout << 1 << endl;
            continue;
        } else if(n == 2) {
            cout << 2 << endl;
            continue;
        }
        ak.x[0][0] = 1;
        ak.x[0][1] = 2;
        ak.x[0][2] = 1;
        ak.x[0][3] = 3;
        ak.x[0][4] = 3;
        ak.x[0][5] = 1;
        ak.x[1][0] = 1;
        ak.x[2][2] = 1;
        ak.x[2][3] = 3;
        ak.x[2][4] = 3;
        ak.x[2][5] = 1;
        ak.x[3][3] = 1;
        ak.x[3][4] = 2;
        ak.x[3][5] = 1;
        ak.x[4][4] = 1;
        ak.x[4][5] = 1;
        ak.x[5][5] = 1;
        mtx ans = ak ^ (n - 2);
        ll as = (2 * ans.x[0][0] % mod + ans.x[0][1] + 8 * ans.x[0][2] % mod + 4 * ans.x[0][3] % mod + 2 * ans.x[0][4] % mod + ans.x[0][5]) % mod;
        cout << as << endl;
    }
    return 0;
}