线段树选做

1|0[NOIP2012]借教室

简单的区间加和区间查询

#include <bits/stdc++.h>

using namespace std;

using i32 = int32_t;
using i64 = long long;
using ldb = long double;

//#define int i64

using vi = vector<int>;
using pii = pair<int, int>;

const int mod = 1e9 + 7;

struct Node {
    int l, r, value, add;
    Node *left, *right;

    Node(int l, int r, int value, Node *left, Node *right)
            : l(l), r(r), value(value), left(left), right(right) {
        add = 0;
    }
} *root;

Node *build(int l, int r, const vector<int> &v) {
    if (l == r) {
        return new Node(l, r, v[l], nullptr, nullptr);
    }
    int mid = (l + r) / 2;
    Node *left = build(l, mid, v);
    Node *right = build(mid + 1, r, v);
    return new Node(l, r, min(left->value, right->value), left, right);
}

void mark(int v, Node *cur) {
    cur->add += v;
    cur->value += v;
    return;
}

void pushdown(Node *cur) {
    if (cur->add == 0) return;
    mark(cur->add, cur->left);
    mark(cur->add, cur->right);
    cur->add = 0;
    return;
}

void modify(int l, int r, int v, Node *cur) {
    if (l > cur->r or r < cur->l) return;
    if (l <= cur->l and cur->r <= r) {
        mark(v, cur);
        return;
    }
    pushdown(cur);
    int mid = (cur->l + cur->r) / 2;
    if (l <= mid) modify(l, r, v, cur->left);
    if (r > mid) modify(l, r, v, cur->right);
    cur->value = min(cur->left->value, cur->right->value);
    return;
}

int query(int l, int r, Node *cur) {
    if (l > cur->r or r < cur->l) return INT_MAX;
    if (l <= cur->l and cur->r <= r) return cur->value;
    pushdown(cur);
    int mid = (cur->l + cur->r) / 2, res = INT_MAX;
    if (l <= mid) res = min(res, query(l, r, cur->left));
    if (r > mid) res = min(res, query(l, r, cur->right));
    return res;
}

i32 main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);
    int n, m;
    cin >> n >> m;

    vi a(n + 1);
    for (int i = 1; i <= n; i++)
        cin >> a[i];

    root = build(1, n, a);
    vi res;
    for (int i = 1, l, r, v; i <= m; i++) {
        cin >> v >> l >> r;
        if (query(l, r, root) < v) {
            cout << "-1\n" << i << "\n";
            return 0;
        }
        modify(l, r, -v, root);
    }

    cout << "0\n";
    return 0;
}

2|0数据结构

https://ac.nowcoder.com/acm/problem/19246

维护复杂了一点

#include <bits/stdc++.h>

using namespace std;

using i32 = int32_t;
using i64 = long long;
using ldb = long double;

#define int i64

using vi = vector<int>;

struct Node {
    int l, r, value, square, add, mul;
    Node *left, *right;

    Node(int l, int r, int value, int square, Node *left, Node *right)
            : l(l), r(r), value(value), square(square), left(left), right(right) {
        add = 0, mul = 1;
    }
} *root;

Node *build(int l, int r, const vector<int> &v) {
    if (l == r) {
        return new Node(l, r, v[l], v[l] * v[l], nullptr, nullptr);
    }
    int mid = (l + r) / 2;
    Node *left = build(l, mid, v);
    Node *right = build(mid + 1, r, v);
    return new Node(l, r, left->value + right->value, left->square + right->square, left, right);
}

void markMul(int v, Node *cur) {
    cur->mul *= v;
    cur->add *= v;
    cur->value *= v;
    cur->square *= v * v;
    return;
}

void markAdd(int v, Node *cur) {
    cur->square += (cur->r - cur->l + 1) * v * v + 2 * v * cur->value;
    cur->value += v * (cur->r - cur->l + 1);
    cur->add += v;
    return;
}

void pushdown(Node *cur) {
    if (cur->mul != 1) {
        markMul(cur->mul, cur->left);
        markMul(cur->mul, cur->right);
        cur->mul = 1;
    }
    if (cur->add != 0) {
        markAdd(cur->add, cur->left);
        markAdd(cur->add, cur->right);
        cur->add = 0;
    }
    return;
}

int calcValue(int l, int r, Node *cur) {
    if (l > cur->r or r < cur->l) return 0;
    if (l <= cur->l and cur->r <= r) return cur->value;
    pushdown(cur);
    int mid = (cur->l + cur->r) / 2, sum = 0;
    if (l <= mid) sum += calcValue(l, r, cur->left);
    if (r > mid) sum += calcValue(l, r, cur->right);
    return sum;
}

int calcSquare(int l, int r, Node *cur) {
    if (l > cur->r or r < cur->l) return 0;
    if (l <= cur->l and cur->r <= r) return cur->square;
    pushdown(cur);
    int mid = (cur->l + cur->r) / 2, sum = 0;
    if (l <= mid) sum += calcSquare(l, r, cur->left);
    if (r > mid) sum += calcSquare(l, r, cur->right);
    return sum;
}

void modifyMul(int l, int r, int v, Node *cur) {
    if (l > cur->r or r < cur->l) return;
    if (l <= cur->l and cur->r <= r) {
        markMul(v, cur);
        return;
    }
    pushdown(cur);
    int mid = (cur->l + cur->r) / 2;
    if (l <= mid) modifyMul(l, r, v, cur->left);
    if (r > mid) modifyMul(l, r, v, cur->right);
    cur->value = cur->left->value + cur->right->value;
    cur->square = cur->left->square + cur->right->square;
    return;
}

void modifyAdd(int l, int r, int v, Node *cur) {
    if (l > cur->r or r < cur->l) return;
    if (l <= cur->l and cur->r <= r) {
        markAdd(v, cur);
        return;
    }
    pushdown(cur);
    int mid = (cur->l + cur->r) / 2;
    if (l <= mid) modifyAdd(l, r, v, cur->left);
    if (r > mid) modifyAdd(l, r, v, cur->right);
    cur->value = cur->left->value + cur->right->value;
    cur->square = cur->left->square + cur->right->square;
    return;
}

i32 main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);

    int n, m;
    cin >> n >> m;

    vector<int> a(n + 1);
    for (int i = 1; i <= n; i++)
        cin >> a[i];

    root = build(1, n, a);

    for (int opt, l, r, v; m; m--) {
        cin >> opt >> l >> r;
        if (opt == 1) {
            cout << calcValue(l, r, root) << "\n";
        } else if (opt == 2) {
            cout << calcSquare(l, r, root) << "\n";
        } else if (opt == 3) {
            cin >> v;
            modifyMul(l, r, v, root);
        } else {
            cin >> v;
            modifyAdd(l, r, v, root);
        }
    }
    return 0;
}

3|0线段树

https://ac.nowcoder.com/acm/problem/212880

两两乘积和不好维护，可以用区间和、区间平方和导出。

#include <bits/stdc++.h>

using namespace std;

using i32 = int32_t;
using i64 = long long;
using ldb = long double;

#define int i64

int mod;

int power(int x, int y) {
    int ans = 1;
    while (y) {
        if (y & 1) ans = ans * x % mod;
        x = x * x % mod, y /= 2;
    }
    return ans;
}

int inv(int x) {
    return power(x, mod - 2);
}


using vi = vector<int>;

struct Node {
    int l, r, value, square, add, mul;
    Node *left, *right;

    Node(int l, int r, int value, int square, Node *left, Node *right)
            : l(l), r(r), value(value), square(square), left(left), right(right) {
        add = 0, mul = 1;
    }
} *root;

Node *build(int l, int r, const vector<int> &v) {
    if (l == r) {
        return new Node(l, r, v[l] % mod, v[l] * v[l] % mod, nullptr, nullptr);
    }
    int mid = (l + r) / 2;
    Node *left = build(l, mid, v);
    Node *right = build(mid + 1, r, v);
    return new Node(l, r, (left->value + right->value) % mod, (left->square + right->square) % mod, left, right);
}

void markMul(int v, Node *cur) {
    (cur->mul *= v) %= mod;
    (cur->add *= v) %= mod;
    (cur->value *= v) %= mod;
    (cur->square *= v * v % mod) %= mod;
    return;
}

void markAdd(int v, Node *cur) {
    (cur->square += (cur->r - cur->l + 1) * v % mod * v % mod + 2 * v % mod * cur->value % mod) %= mod;
    (cur->value += v * (cur->r - cur->l + 1) % mod) %= mod;
    (cur->add += v) %= mod;
    return;
}

void pushdown(Node *cur) {
    if (cur->mul != 1) {
        markMul(cur->mul, cur->left);
        markMul(cur->mul, cur->right);
        cur->mul = 1;
    }
    if (cur->add != 0) {
        markAdd(cur->add, cur->left);
        markAdd(cur->add, cur->right);
        cur->add = 0;
    }
    return;
}

int calcValue(int l, int r, Node *cur) {
    if (l > cur->r or r < cur->l) return 0;
    if (l <= cur->l and cur->r <= r) return cur->value;
    pushdown(cur);
    int mid = (cur->l + cur->r) / 2, sum = 0;
    if (l <= mid) (sum += calcValue(l, r, cur->left)) %= mod;
    if (r > mid) (sum += calcValue(l, r, cur->right)) %= mod;
    return sum;
}

int calcSquare(int l, int r, Node *cur) {
    if (l > cur->r or r < cur->l) return 0;
    if (l <= cur->l and cur->r <= r) return cur->square;
    pushdown(cur);
    int mid = (cur->l + cur->r) / 2, sum = 0;
    if (l <= mid) (sum += calcSquare(l, r, cur->left)) %= mod;
    if (r > mid) (sum += calcSquare(l, r, cur->right)) %= mod;
    return sum;
}

void modifyMul(int l, int r, int v, Node *cur) {
    if (l > cur->r or r < cur->l) return;
    if (l <= cur->l and cur->r <= r) {
        markMul(v, cur);
        return;
    }
    pushdown(cur);
    int mid = (cur->l + cur->r) / 2;
    if (l <= mid) modifyMul(l, r, v, cur->left);
    if (r > mid) modifyMul(l, r, v, cur->right);
    cur->value = (cur->left->value + cur->right->value) % mod;
    cur->square = (cur->left->square + cur->right->square) % mod;
    return;
}

void modifyAdd(int l, int r, int v, Node *cur) {
    if (l > cur->r or r < cur->l) return;
    if (l <= cur->l and cur->r <= r) {
        markAdd(v, cur);
        return;
    }
    pushdown(cur);
    int mid = (cur->l + cur->r) / 2;
    if (l <= mid) modifyAdd(l, r, v, cur->left);
    if (r > mid) modifyAdd(l, r, v, cur->right);
    cur->value = (cur->left->value + cur->right->value) % mod;
    cur->square = (cur->left->square + cur->right->square) % mod;
    return;
}

void solve(){
    int n, m;
    cin >> n >> m >> mod;
    vector<int> a(n + 1);
    for (int i = 1; i <= n; i++)
        cin >> a[i];

    root = build(1, n, a);

    for (int opt, l, r, v; m; m--) {
        cin >> opt >> l >> r;
        if (opt == 1) {
            cin >> v;
            modifyAdd(l, r, v, root);
        } else if (opt == 2) {
            cin >> v;
            modifyMul(l, r, v, root);
        } else {
            int x = calcValue(l, r, root), y = calcSquare(l, r, root);
            cout << (x * x % mod - y % mod + mod) % mod * inv(2) % mod << "\n";
        }
    }
}

i32 main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);

    int TC;
    for( cin >> TC; TC; TC --)
        solve();

    return 0;
}

4|0仓鼠的鸡蛋

https://ac.nowcoder.com/acm/problem/226170

在线段树上二分查找，并返回坐标。

#include <bits/stdc++.h>

using namespace std;

using i32 = int32_t;
using i64 = long long;

const int inf = INT_MAX / 2;

using vi = vector<int>;
using pii = pair<int, int>;

struct Node {
    int l, r, val, cnt;
    Node *left, *right;

    Node(int l, int r, int val, int cnt, Node *left = nullptr, Node *right = nullptr)
            : l(l), r(r), val(val), cnt(cnt), left(left), right(right) {};
} *root;

Node *build(int l, int r, int m, int k) {
    if (l == r) return new Node(l, r, m, k);
    int mid = (l + r) / 2;
    Node *left = build(l, mid, m, k);
    Node *right = build(mid + 1, r, m, k);
    return new Node(l, r, m, 0, left, right);
}

void modify(int p, int delta, Node *cur) {
    if (p < cur->l or cur->r < p) return;
    if (cur->l == cur->r) {
        cur->val += delta;
        cur->cnt--;
        if (cur->cnt == 0) cur->val = -inf;
        return;
    }
    int mid = (cur->l + cur->r) / 2;
    if (p <= mid) modify(p, delta, cur->left);
    if (p > mid) modify(p, delta, cur->right);
    cur->val = max(cur->left->val, cur->right->val);
    return;
}

int query(int x, Node *cur) {
    if (cur->l == cur->r) return cur->l;
    if (cur->left->val >= x) return query(x, cur->left);
    if (cur->right->val >= x) return query(x, cur->right);
    return -1;
}


void solve() {
    int n, m, k;
    cin >> n >> m >> k;
    root = build(1, n, m, k);
    for (int x, l, r, ans, i = 1; i <= n; i++) {
        cin >> x;
        ans = query(x, root);
        cout << ans << "\n";
        if (ans != -1) modify(ans, -x, root);
    }
    return;
}

int main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);
    int TC;
    for (cin >> TC; TC; TC--)
        solve();
    return 0;
}

5|0P5490 【模板】扫描线

https://www.luogu.com.cn/problem/P5490

扫描线模板题

// luogu P5490
#include<bits/stdc++.h>

using namespace std;


using i32 = int32_t;
using i64 = long long;

#define int i64

const int inf = INT_MAX / 2;

using seg = array<int, 4>; // {x, y1, y2, v}
using vi = vector<int>;

struct Node {
    int l, r, value, cnt;// value 被覆盖子区间长度，cnt 当前区间被完整覆盖的次数。
    Node *left, *right;

    Node(int l, int r, int value, int cnt, Node *left, Node *right)
            : l(l), r(r), value(value), cnt(cnt), left(left), right(right) {};
} *root;

vi raw;// 原始坐标

int val(int x) { // // x 是原始值，返回哈希值
    int i = lower_bound(raw.begin(), raw.end(), x) - raw.begin();
    if (raw[i] != x) return -1;
    return i;
}

Node *build(int l, int r) {
    if (l == r) return new Node(l, r, 0, 0, nullptr, nullptr);
    int mid = (l + r) / 2;
    auto left = build(l, mid), right = build(mid + 1, r);
    return new Node(l, r, 0, 0, left, right);
}

void maintain(Node *cur) {
    if (cur->cnt > 0) cur->value = raw[cur->r + 1] - raw[cur->l];
    else if (cur->left == nullptr) cur->value = 0;
    else cur->value = cur->left->value + cur->right->value;
    return;
}

void modify(int l, int r, int v, Node *cur) {
    if (l > cur->r or r < cur->l) return;
    if (l <= cur->l and cur->r <= r) {
        cur->cnt += v;
        maintain(cur);
        return;
    }
    int mid = (cur->l + cur->r) / 2;
    if (l <= mid) modify(l, r, v, cur->left);
    if (r > mid) modify(l, r, v, cur->right);
    maintain(cur);
    return;
}

i32 main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);
    int n;
    cin >> n;

    vector<seg> p;
    for (int i = 1, xa, xb, ya, yb; i <= n; i++) {
        cin >> xa >> ya >> xb >> yb;
        p.push_back(seg{xa, ya, yb, 1});
        p.push_back(seg{xb, ya, yb, -1});
        raw.push_back(ya), raw.push_back(yb);
    }
    sort(p.begin(), p.end());
    sort(raw.begin(), raw.end());
    raw.resize(unique(raw.begin(), raw.end()) - raw.begin());
    int tot = raw.size() - 1;
    root = build(0, tot);
    modify(val(p[0][1]), val(p[0][2]) - 1, p[0][3], root);
    int res = 0;
    for (int i = 1; i < p.size(); i++) {
        res += (p[i][0] - p[i - 1][0]) * root->value;
        modify(val(p[i][1]), val(p[i][2]) - 1, p[i][3], root);
    }
    cout << res;
    return 0;
}

6|0P4588 [TJOI2018] 数学计算

做法很多，这里提供一个用线段树解决的思路。用一个大小为$Q$的数组，初始全为$1$，操作 $1$ 就是第$i$为修改为$m$，操作$2$就是第$pos$位修改为$1$，然后求一个区间乘积就好。

#include<bits/stdc++.h>

using namespace std;


using i32 = int32_t;
using i64 = long long;

#define int i64

const int inf = INT_MAX / 2;

int mod;

struct Node {
    int l, r, val;
    Node *left, *right;

    Node(int l, int r, int val, Node *left = nullptr, Node *right = nullptr)
            : l(l), r(r), val(val), left(left), right(right) {};
} *root;

Node *build(int l, int r) {
    if (l == r) return new Node(l, r, 1);
    int mid = (l + r) / 2;
    auto left = build(l, mid), right = build(mid + 1, r);
    return new Node(l, r, 1, left, right);
}

void modify(int p, int val, Node *cur) {
    if (p < cur->l or p > cur->r) return;
    if (cur->l == cur->r) {
        cur->val = val % mod;
        return;
    }
    int mid = (cur->l + cur->r) / 2;
    if (p <= mid) modify(p, val, cur->left);
    if (p > mid) modify(p, val, cur->right);
    cur->val = (cur->left->val * cur->right->val) % mod;
    return;
}

using vi = vector<int>;

void debug(Node *cur) {
    cerr << cur->l << " " << cur->r << " " << cur->val << "\n";
    if (cur->left != nullptr)debug(cur->left), debug(cur->right);
}

void solve() {
    int n;
    cin >> n >> mod;
    root = build(1, n);
    for (int i = 1, opt, x; i <= n; i++) {
        cin >> opt >> x;
        if (opt == 1) modify(i, x, root);
        else modify(x, 1, root);
        cout << root->val << "\n";
    }
    return;
}

i32 main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);
    int TC;
    for (cin >> TC; TC; TC--)
        solve();
    return 0;
}

7|0D. Traffic Jams in the Land

https://codeforces.com/problemset/problem/498/D

一个很有用的性质是$lcm (2,3,4,5,6) = 60$，因此对于到达时间我们模$60$就好了。

然后对于线段树节点的信息$\{l,r,val[60]\}$就可以用$val[i]$表示$i$的时间到达$l$并走到$r$所需要的代价。这样线段是维护就变成了$O(60\log n)$

#include <bits/stdc++.h>

using namespace std;

using vi = vector<int>;

struct Node {
    int l, r;
    vector<int> val;
    Node *left, *right;

    Node(int l, int r, int x) : l(l), r(r), val(60) {
        left = right = nullptr;
        for (int i = 0; i < 60; i++)
            val[i] = 1 + (i % x == 0);
    }

    Node(int l, int r, Node *left, Node *right)
            : l(l), r(r), left(left), right(right), val(60) {}
} *root;

void maintain(Node *cur) {
    for (int i = 0, t; i < 60; i++) {
        t = cur->left->val[i];
        cur->val[i] = t + cur->right->val[(i + t) % 60];
    }
    return;
}

Node *build(int l, int r, const vi &v) {
    if (l == r) return new Node(l, r, v[l]);
    int mid = (l + r) / 2;
    Node *left = build(l, mid, v);
    Node *right = build(mid + 1, r, v);
    Node *cur = new Node(l, r, left, right);
    maintain(cur);
    return cur;
}

void modify(int p, int delta, Node *cur) {
    if (p < cur->l or cur->r < p) return;
    if (cur->l == cur->r) {
        for (int i = 0; i < 60; i++)
            cur->val[i] = 1 + (i % delta == 0);
        return;
    }
    int mid = (cur->l + cur->r) / 2;
    if (p <= mid) modify(p, delta, cur->left);
    if (p > mid) modify(p, delta, cur->right);
    maintain(cur);
    return;
}

int query(int l, int r, int start, Node *cur) {
    if (l > cur->r or r < cur->l) return 0;
    if (l <= cur->l and cur->r <= r) return cur->val[start];
    int mid = (cur->l + cur->r) / 2, ans = 0;
    if (l <= mid) ans += query(l, r, (start + ans) % 60, cur->left);
    if (r > mid) ans += query(l, r, (start + ans) % 60, cur->right);
    return ans;
}


int main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);

    int n;
    cin >> n;

    vi a(n + 1);
    for (int i = 1; i <= n; i++) cin >> a[i];

    root = build(1, n, a);

    int m;
    cin >> m;
    for (char opt; m; m--) {
        cin >> opt;
        if (opt == 'C') {
            int x, d;
            cin >> x >> d;
            modify(x, d, root);
        } else {
            int x, y;
            cin >> x >> y, y--;
            cout << query(x, y, 0, root) << "\n";
        }
    }
    return 0;
}

8|0F - Vacation Query

要用线段树实现区间翻转和区间最长连续子段。

其实可以直接维护出最长的连续$1$的同时维护出最长连续$0$，当需要翻转的时候直接交换就好了。

#include<bits/stdc++.h>

using namespace std;

using i32 = int32_t;
using i64 = long long;

#define int i64

const int inf = INT_MAX / 2;

using vi = vector<int>;

struct Info {
    array<int, 2> pre, suf, val, len;

    Info() : pre{0, 0}, suf{0, 0}, val{0, 0}, len{0, 0} {}

    Info(int x) {
        pre[x] = suf[x] = val[x] = 1;
        pre[x ^ 1] = suf[x ^ 1] = val[x ^ 1] = 0;
        len[x] = 1, len[x ^ 1] = -inf;
    }

    void swap() {
        std::swap(pre[0], pre[1]);
        std::swap(suf[0], suf[1]);
        std::swap(val[0], val[1]);
        std::swap(len[0], len[1]);
    }

    friend Info operator+(Info x, Info y) {
        Info ans;
        for (int i = 0; i < 2; i++) {
            ans.pre[i] = std::max(x.pre[i], x.len[i] + y.pre[i]);
            ans.suf[i] = std::max(y.suf[i], x.suf[i] + y.len[i]);
            ans.val[i] = std::max({x.val[i], y.val[i], x.suf[i] + y.pre[i]});
            ans.len[i] = x.len[i] + y.len[i];
        }
        return ans;
    }
};

struct Node {
    i32 l, r, tag;
    Info info;

    Node *left, *right;

    Node() {};

    Node(int p, int v) : info(v) {
        l = r = p, tag = 0;
        left = right = nullptr;
    }

    Node(int l, int r, Node *left, Node *right)
            : l(l), r(r), left(left), right(right) {
        tag = 0;
        info = left->info + right->info;
    }
};

void maintain(Node *cur) {
    if (cur->left == nullptr) return;
    cur->info = cur->left->info + cur->right->info;
}

void mark(Node *cur) {
    cur->tag ^= 1;
    cur->info.swap();
    return;
}

void pushdown(Node *cur) {
    if (cur->tag == 0) return;
    if (cur->left != nullptr) {
        mark(cur->left);
        mark(cur->right);
    }
    cur->tag = 0;
    return;
}

Node *build(i32 l, i32 r, const string &s) {
    if (l == r) return new Node(l, s[l - 1] - '0');
    i32 mid = (l + r) / 2;
    auto left = build(l, mid, s), right = build(mid + 1, r, s);
    return new Node(l, r, left, right);
}

Info query(i32 l, i32 r, Node *cur) {
    if (cur == nullptr) return Info();
    if (l > cur->r or r < cur->l) return Info();
    if (l <= cur->l and cur->r <= r) return cur->info;
    pushdown(cur);
    return query(l, r, cur->left) + query(l, r, cur->right);
}

void modify(i32 l, i32 r, Node *cur) {
    if (cur == nullptr) return;
    if (l > cur->r or r < cur->l) return;
    if (l <= cur->l and cur->r <= r) {
        mark(cur);
        return;
    }
    pushdown(cur);
    modify(l, r, cur->left);
    modify(l, r, cur->right);
    maintain(cur);
}

i32 main() {
    ios::sync_with_stdio(false), cin.tie(nullptr);
    i32 n, q;
    string s;

    cin >> n >> q >> s;
    Node *root = build(1, n, s);
    for (i32 op, l, r; q; q--) {
        cin >> op >> l >> r;
        if (op == 1) modify(l, r, root);
        else cout << query(l, r, root).val[1] << "\n";
    }
    return 0;
}

__EOF__

本文作者：PHarr
本文链接：https://www.cnblogs.com/PHarr/p/18422551.html
关于博主：前OIer，SMUer
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