Codeforces Round #263 (Div. 1)

A:贪心，排个序，然后从后往前扫一遍，计算后缀和，之后在从左往右扫一遍计算答案

B：树形DP，0表示没有1，1表示有1，0遇到0必然合并，0遇到1也必然合并，1遇到0必然合并，1遇到1，必然切断，按照这样去转移即可

C：树状数组，再利用启发式合并，开一个l，r记录当前被子左右下标，和一个flip表示是否翻转

代码：

A：

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

typedef long long ll;
const int N = 300005;

int n;
ll a[N], sum[N];

int main() {
	scanf("%d", &n);ll ans = 0;
 	for (int i = 0; i < n; i++) {
    	scanf("%lld", &a[i]);
     ans += a[i];
	}
	sort(a, a + n);
	for (int i = n - 1; i >= 0; i--)
		sum[i] = a[i] + sum[i + 1];

	for (int i = 0; i < n - 1; i++) {
		ans += sum[i];
	}
	printf("%lld\n", ans);
	//system("pause");
	return 0;
}

B：

#include <cstdio>
#include <cstring>
#include <vector>
#include <cstdlib>
using namespace std;

const int N = 100005;
typedef long long ll;
const ll MOD = 1000000007;
int n, node[N];
vector<int> g[N];
ll dp[N][2];

ll pow_mod(ll x, ll k) {
    ll ans = 1;
    while (k) {
        if (k&1) ans = ans * x % MOD;
        x = x * x % MOD;
        k >>= 1;
    }
    return ans;
}

ll inv(ll x) {
    return pow_mod(x, MOD - 2);
}

void init() {
    scanf("%d", &n);
    int u;
    for (int i = 1; i < n; i++) {
        scanf("%d", &u);
        g[u].push_back(i);
    }
    for (int i = 0; i < n; i++)
        scanf("%d", &node[i]);
}

void dfs(int u) {
    if (g[u].size() == 0) {
        dp[u][node[u]] = 1;
        return;
    }
    for (int i = 0; i < g[u].size(); i++)
        dfs(g[u][i]);
    dp[u][0] = dp[u][1] = 1;
    if (node[u]) {
        dp[u][0] = 0;
        for (int i = 0; i < g[u].size(); i++) {
            int v = g[u][i];
            dp[u][1] = dp[u][1] * (dp[v][0] + dp[v][1]) % MOD;
        }
    }
    else {
        ll cnt = 0;
        ll mul = 1;
        for (int i = 0; i < g[u].size(); i++) {
            int v = g[u][i];
            dp[u][0] = dp[u][0] * (dp[v][0] + dp[v][1]) % MOD;
            mul = mul * (dp[v][0] + dp[v][1]) % MOD;
        }
        dp[u][1] = 0;
        for (int i = 0; i < g[u].size(); i++){
            int v = g[u][i];
            dp[u][1] = (dp[u][1] + mul * inv((dp[v][0] + dp[v][1]) % MOD) % MOD * dp[v][1]) % MOD;
        }
    }
}

int main() {
    init();
    dfs(0);
    printf("%lld\n", dp[0][1] % MOD);
    //system("pause");
    return 0;
}

C：

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

#define lowbit(x) (x&(-x))

const int N = 100005;

int n, q, bit[N];

void add(int x, int v) {
    while (x < N) {
        bit[x] += v;
        x += lowbit(x);
    }
}

int get(int x) {
    int ans = 0;
    while (x) {
        ans += bit[x];
        x -= lowbit(x);
    }
    return ans;
}

int main() {
    scanf("%d%d", &n, &q);
    for (int i = 1; i <= n; i++)
        add(i, 1);
    int tp, a, b;
    int l = 1, r = n, flip = 0;
    while (q--) {
        scanf("%d%d", &tp, &a);
        if (tp == 1) {
            int tl = l, tr = r;
            if (a <= (r - l + 1) / 2) {
                if (flip) {
                    for (int i = a; i >= 1; i--) {
                        add(r - 2 * i + 1, get(tr - a + i) - get(tr - a + i - 1));
                        r--;
                    }
                }
                else {
                    for (int i = a; i >= 1; i--) {
                        add(l + 2 * i - 1, get(tl + a - i) - get(tl + a - i - 1));
                        l++;
                    }
                }
            } else {
                a = r - a - l + 1;
                if (!flip) {
                    for (int i = a; i >= 1; i--) {
                        add(r - 2 * i + 1, get(tr - a + i) - get(tr - a + i - 1));
                        r--;
                    }
                }
                else {
                    for (int i = a; i >= 1; i--) {
                        add(l + 2 * i - 1, get(tl + a - i) - get(tl + a - i - 1));
                        l++;
                    }
                }
                flip ^= 1;
            }
        }
        else {
            scanf("%d", &b);
            if (flip) {
                a = r - a;
                b = r - b;
                swap(a, b);
            } else {
                a += l - 1;
                b += l - 1;
            }
            printf("%d\n", get(b) - get(a));
        }
    }
    return 0;
}