hdoj 1568 && hdoj 5344 && hdoj 5444

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这里记录三道水题~~~

Fibonacci

$F[n] = \frac 1{\sqrt5} * [(\frac{1 + \sqrt5}{2}) ^ n - (\frac{1 - \sqrt5}{2}) ^ n]$

发现$(\frac{1 - \sqrt5}{2}) ^ n$是可以忽略的，然后就很随意了。
$(\frac{1 + \sqrt5}{2}) ^ n$取对数$n * log10(\frac{1 + \sqrt5}{2})$

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
using namespace std;
const int MAXN = 3e6 + 1;
const int MOD = 1e9 + 7;
typedef long long LL;
void add(LL &x, LL y) { x += y; x %= MOD; }
typedef pair<double, double> pii;
int f[30];
int main()
{
    // fn = 1/sqrt(5) * (((1 + sqrt(5)) / 2) ^ n - (1 - sqrt(5)) / 2) ^ n)
    double a = (1 + sqrt(5)) / 2;
    double b = (1 - sqrt(5)) / 2;
    double c = 1.0 / sqrt(5);
    // printf("%lf %lf %lf\n", a, b, c);
    f[0] = 0, f[1] = 1;
    for(int i = 2; i <= 21; i++) {
        f[i] = f[i - 1] + f[i - 2];
    }
    // cout << f[21] << endl;
    // printf("%.6lf\n", pow(b, 21));
    int n;
    while(scanf("%d", &n) != EOF) {
        if(n <= 20) {
            printf("%d\n", f[n]);
            continue;
        }
        double T = log10(c) + n * log10(a);
        // printf("%lf\n", T);
        double X = T - (LL)T;
        printf("%d\n", (int)(pow(10, X) * 1000));
    }
    return 0;
}

MZL’s xor

不同数相加异或完之后就没了。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
using namespace std;
const int MAXN = 5e5 + 1;
const int MOD = 1e9 + 7;
typedef long long LL;
void add(LL &x, LL y) { x += y; x %= MOD; }
typedef pair<double, double> pii;
int main()
{
    int t; scanf("%d", &t);
    while(t--) {
        int n, m, z, l;
        scanf("%d%d%d%d", &n, &m, &z, &l);
        LL a = 0; LL ans = 0;
        for(int i = 2; i <= n; i++) {
            a = (a * m + z) % l;
            ans ^= (a * 2);
        }
        printf("%lld\n", ans);
    }
    return 0;
}

Elven Postman

模拟二叉树，小左大右。阅读理解题~~~

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
using namespace std;
const int MAXN = 2e3 + 1;
const int MOD = 1e9 + 7;
typedef long long LL;
void add(LL &x, LL y) { x += y; x %= MOD; }
typedef pair<double, double> pii;
int Next[MAXN][2], val[MAXN], L, root;
int newnode() {
    Next[L][0] = Next[L][1] = -1;
    val[L++] = 0;
    return L - 1;
}
void Init() { L = 0; root = newnode(); }
void Insert(int x, int u) {
    if(x < val[u]) {
        if(Next[u][0] == -1) {
            Next[u][0] = newnode();
            val[Next[u][0]] = x;
            return;
        }
        Insert(x, Next[u][0]);
    }
    else {
        if(Next[u][1] == -1) {
            Next[u][1] = newnode();
            val[Next[u][1]] = x;
            return;
        }
        Insert(x, Next[u][1]);
    }
}
void Work(int x, int u) {
    if(val[u] == x) {
        return;
    }
    else if(x < val[u]) {
        printf("E");
        Work(x, Next[u][0]);
    }
    else {
        printf("W");
        Work(x, Next[u][1]);
    }
}
int main()
{
    int t; scanf("%d", &t);
    while(t--) {
        int n; scanf("%d", &n); Init();
        for(int i = 1; i <= n; i++) {
            int a; scanf("%d", &a);
            if(i == 1) {
                val[root] = a;
            }
            else {
                Insert(a, root);
            }
        }
        int q; scanf("%d", &q);
        while(q--) {
            int x; scanf("%d", &x);
            if(val[root] != x) {
                Work(x, root);
            }
            printf("\n");
        }
    }
    return 0;
}