Codeforces Round #259 (Div. 1) (A,B,C)

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Codeforces Round #259 (Div. 1)

A題：最大值為i的期望為(in?(i?1)n)?i/mn，那麼總期望為∑m1(in?(i?1)n)?i/mn,然後化簡一下公式，答案為m?∑m?11i/mn

B題：狀壓DP，只需要用到小於59的素數，一共有16個，dp[n][s]表示當前放到第n個數字，素數使用的集合為s的最小值，S[k]表示k數字對應會用掉哪幾個素數，然後進行狀態轉移
dp(n, s) = dp(n - 1, s^S[k]) + abs(k - num[n])

C題：dfs構造，題目中其實給了提示，最多不超過4n步，那麼其實如果給定的是一棵樹，在這個步驟內是絕對有辦法構造出來的，可以選一個根結點出來，然後當成一棵樹，對於一條路徑，如果走兩遍，其實就等於沒走，那麼對於每條路徑，只需要走一遍走到底，再走回根結點，如果當前一個位置不滿足，就退一步在進一步，然後注意根結點，其實根結點是無法回退的，但是其實根結點如果不滿足，那麼我只要讓最後回到根結點的一步不走就可以了，這樣搜完，在判斷一下還有沒結點不滿足，如果有，肯定是多棵樹構造不出來的情況

代碼：

A:

#include <cstdio>#include <cstring>#include <cmath>int m, n;int main() {    scanf("%d%d", &m, &n);    double ans = 0;    for (int i = 1; i < m; i++)ans += pow((m - i) * 1.0 / m, n);    printf("%.12lf\n", m * 1.0 - ans);    return 0;}

B:

#include <cstdio>#include <cstring>#include <cstdlib>const int N = 105;const int MAXN = (1<<16) + 5;const int INF = 0x3f3f3f3f;int n, num[N], prime[N], pn = 0, vis[N], to[N];void get_prime() {    for (int i = 2; i < 59; i++) {if (vis[i]) continue;prime[pn++] = i;for (int j = i * i; j < 60; j += i) {    vis[j] = 1;}    }}int tra(int num) {    int ans = 0;    for (int i = 0; i < pn; i++) {if (num % prime[i] == 0) ans |= (1<<i);while (num % prime[i] == 0) num /= prime[i];    }    return ans;}int dp[N][MAXN], zh[N][MAXN][2];void print(int now, int s) {    if (now == 0) return;    print(now - 1, zh[now][s][0]);    printf("%d%c", zh[now][s][1], now == n? '\n' : ' ');}int main() {    get_prime();    scanf("%d", &n);    for (int i = 1; i <= n; i++)scanf("%d", &num[i]);    for (int i = 1; i < 59; i++)to[i] = tra(i);    int maxs = (1<<pn);    memset(dp, INF, sizeof(dp));    memset(dp[0], 0, sizeof(dp[0]));    for (int i = 1; i <= n; i++) {for (int j = 0; j < maxs; j++) {    for (int k = 1; k < 60; k++) {if ((j&to[k]) != to[k]) continue;int tmp = dp[i - 1][j^to[k]] + abs(k - num[i]);if (dp[i][j] > tmp) {    dp[i][j] = tmp;    zh[i][j][0] = (j^to[k]);    zh[i][j][1] = k;}    }}    }    int Min = INF, Min_v;    for (int i = 0; i < maxs; i++) {if (dp[n][i] < Min) {    dp[n][i] = Min;    Min_v = i;}    }    print(n, Min_v);    return 0;}

C:

#include <cstdio>#include <cstring>#include <vector>using namespace std;const int N = 100005;int n, m, x[N], now[N], vis[N], nv;vector<int> g[N], ans;void dfs(int u, int p) {    vis[u] = 1;    ans.push_back(u);    now[u] ^= 1;    for (int i = 0; i < g[u].size(); i++) {int v = g[u][i];if (vis[v]) continue;dfs(v, u);    }    if (x[u]^now[u]) {if (p != 0) {    ans.push_back(p);    ans.push_back(u);    now[p] ^= 1;    now[u] ^= 1;}    }    if (p != 0) {ans.push_back(p);now[p] ^= 1;    }}bool judge() {    if (x[nv]^now[nv]) {if (ans.size() == 0) return false;now[nv] ^= 1;ans.pop_back();    }    for (int i = 1; i <= n; i++)if (x[i]^now[i]) return false;    return true;}int main() {    scanf("%d%d", &n, &m);    int u, v;    while (m--) {scanf("%d%d", &u, &v);g[u].push_back(v);g[v].push_back(u);    }    for (int i = 1; i <= n; i++)scanf("%d", &x[i]);    memset(now, 0, sizeof(now));    for (int i = 1; i <= n; i++) {if (x[i]^now[i]) {    nv = i;    dfs(i, 0);    break;}    }    if (judge()) {int tmp = ans.size();printf("%d\n", tmp);for (int i = 0; i < tmp; i++) {    printf("%d%c", ans[i], i == ans.size() - 1 ? '\n' : ' ');}    }    else printf("-1\n");    return 0;}