hdoj 1568 && hdoj 5344 && hdoj 5444

来源：互联网发布：淘宝的阿玛尼这么便宜编辑：程序博客网时间：2024/04/28 00:04

这里记录三道水题~~~

Fibonacci

F[n]=15√∗[(1+5√2)n−(1−5√2)n]

发现(1−5√2)n是可以忽略的，然后就很随意了。
(1+5√2)n取对数n∗log10(1+5√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;}

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