2019.11.09考试解题报告

Posted 2020-11-29 loceaner

总结

期望得分：(100+ 30 + 40 = 170)
实际得分：(100 + 50 + 40 = 190)

简单记录一下

---

思路&&代码

T1

(n^2算法)

#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

const int A = 1e5 + 11;
const int B = 1e6 + 11;

inline int read() {
    char c = getchar(); int x = 0, f = 1;
    for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;
    for( ; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + (c ^ 48);
    return x * f;
}

int n, m, all, cnt;
int d[A], w[A], a[A], la[A];

inline bool check(int x) {
    memset(la, 0, sizeof(la));
    for(int i = 1; i <= n; i++) a[i] = d[i];
    for(int i = 1; i <= x; i++) if(a[i]) a[la[a[i]]] = 0, la[a[i]] = i; 
    int tl = 0, cnt = 0;
    for(int i = 1; i <= x; i++) {
        if(a[i]) { tl -= w[a[i]]; if(tl < 0) return 0; else cnt++; }
        else tl++;
    }
    return cnt == m; 
}

int main() {
    freopen("generals.in", "r", stdin);
    freopen("generals.out", "w", stdout);
    n = read(), m = read();
    if(n < m) return puts("-1"), 0;
    for(int i = 1; i <= n; i++) d[i] = read();
    for(int i = 1; i <= m; i++) w[i] = read(), all += w[i];
    if(all > n) return puts("-1"), 0;
    int l = 0, r = n, ans = -1;
    while(l <= r) {
        int mid = (l + r) >> 1;
        if(check(mid)) ans = mid, r = mid - 1;
        else l = mid + 1;
    }
    return cout << ans << '
', 0;
}

*/

T2

找规律

#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define int long long
using namespace std;

const int A = 1e5 + 11;
const int B = 1e6 + 11;
const int mod = 1e9 + 7;

inline int read() {
    char c = getchar(); int x = 0, f = 1;
    for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;
    for( ; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + (c ^ 48);
    return x * f;
}

int n, m, a[A], k, ans = 0, pd[A];

int power(int a, int b, int res = 1) {
    if(a == 0 && b == 0) return 1;
    while(b) {
        if(b & 1) res = res * a % mod;
        a = a * a % mod; b >>= 1;
    } return res;
}

int an[100][100];


signed main() {
    freopen("transfer.in", "r", stdin);
    freopen("transfer.out", "w", stdout);
    an[1][1] = 1;
    an[2][1] = 1,an[2][2] = 2;
    an[3][1] = 4,an[3][2] = 2,an[3][3] = 9;
    an[4][1] = 27,an[4][2] = 8,an[4][3] = 9,an[4][4] = 64;
    an[5][1] = 256,an[5][2] = 54,an[5][3] = 36,an[5][4] = 64,an[5][5] = 625;
    an[6][1] = 3125,an[6][2] = 512,an[6][3] = 243,an[6][4] = 256,an[6][5] = 625,an[6][6] = 7776;
    an[7][1] = 46656,an[7][2] = 625,an[7][3] = 2304,an[7][4] = 1728,an[7][5] = 2500,an[7][6] = 7776,an[7][7] = 117649;
    int T = read();
    while(T--) {
        n = read(), k = read();
        if(k == n) {
            int now = power(n, n - 1) % mod;
            now = (now % mod + mod) % mod;
            cout << now << '
';
            continue;
        } else if(k == 1) {
            int now = power(n - 1, n - 1) % mod;
            now = (now % mod + mod) % mod;
            cout << now << '
';
            continue;
        } else if(n <= 7) {
            cout << an[n][k] % mod << '
';
            continue;
        } else {
            int oo = power(n - k, n - k) % mod * power(k, k - 1);
            oo = (oo % mod + mod) % mod;
            cout << oo << '
';
            continue;
        }
    }
    return 0;
}

T3

树形(DP)预处理，然后伪(O(1))查询（快速幂呀~~）

#include <bits/stdc++.h>
#define int long long
using namespace std;

const int mod = 1e9 + 7;
const int A = 2e5 + 11;
const int B = 2e3 + 11;

inline int read() {
    char c = getchar();
    int x = 0, f = 1;
    for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;
    for( ; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + (c ^ 48);
    return x * f;
}

int n, m, ans;
struct node { int to, nxt, val; } e[A << 1];
int head[A], cnt, dis[B][B], o[A][2];

inline void add(int from, int to, int val) {
    e[++cnt].to = to;
    e[cnt].nxt = head[from];
    e[cnt].val = val;
    head[from] = cnt;
}

int power(int a, int b, int res = 1) {
    while(b) {
        if(b & 1) res = res * a % mod;
        a = a * a % mod; b >>= 1;
    } return res;
}

int tot[A][2], siz[A][2], f[A][2];

void dfs1(int now, int fa, int ww) {
    f[1][ww & 1] += ww, f[1][ww & 1] %= mod;
    tot[1][ww & 1]++;
    siz[now][0] = 1;
    for(int i = head[now]; i; i = e[i].nxt) {
        int to = e[i].to;
        if(to == fa) continue;
        dfs1(to, now, ww + e[i].val);
        if(e[i].val & 1) siz[now][0] += siz[to][1], siz[now][1] += siz[to][0];
        else siz[now][0] += siz[to][0], siz[now][1] += siz[to][1];
    }
}

void dfs2(int now, int fa) {
    for(int i = head[now]; i; i = e[i].nxt) {
        int to = e[i].to;
        if(to == fa) continue;
        if(e[i].val & 1) {
            tot[to][0] = tot[now][1], tot[to][1] = tot[now][0];
            f[to][1] = (((f[now][0] % mod + e[i].val * (tot[to][1] - 2 * siz[to][1]) % mod) % mod) % mod) % mod;
            f[to][0] = (((f[now][1] % mod + e[i].val * (tot[to][0] - 2 * siz[to][0]) % mod) % mod) % mod) % mod;
        } else {
            tot[to][0] = tot[now][0], tot[to][1] = tot[now][1];
            f[to][1] = (((f[now][1] % mod + e[i].val * (tot[to][1] - 2 * siz[to][1]) % mod) % mod) % mod) % mod;
            f[to][0] = (((f[now][0] % mod + e[i].val * (tot[to][0] - 2 * siz[to][0]) % mod) % mod) % mod) % mod;
        }
        dfs2(to, now);
    }
}

signed main() {
    freopen("meet2.in", "r", stdin);
    freopen("meet.out", "w", stdout);
    n = read(), m = read();
    for(int i = 1, u, v, w; i < n; i++) {
        u = read(), v = read(), w = read();
        add(u, v, w), add(v, u, w);
    }
    dfs1(1, 0, 0), dfs2(1, 0);
    while(m--) {
        int x = read(), y = read();
        int qwq = f[x][y] % mod * power(tot[x][y], mod - 2) % mod;
        qwq = (qwq % mod + mod) % mod; cout << qwq << '
';
    }
    return 0;
}