1 条题解

  • 0
    @ 2026-10-27 4:21:40

    思路与算法

    读到行尾/EOF,流式维护最小值、最大值、总和和个数。平均值按总和的绝对值乘1000再除个数,保留商和余数,按最近值、恰好一半时末位取偶的标准浮点输出舍入规则处理,最后补足3位小数及符号。

    正确性说明

    每次读数更新区间极值与累计和,归纳后即为所有输入的统计量。商是平均值绝对值放大1000后的整数部分,余数与除数的一半比较决定最接近的千分位;正负对称处理保证负数准确。大整数避免输入只声明不超过1000却未声明负数下界时的截断。

    复杂度

    O(qD)时间、O(D)额外空间,q≤100,D为输入与总和位数;含极值、累计值和十进制格式化初始化。

    易错点

    输入没有预先给个数,不能把第一个数当n。平均值保留3位,最小最大值仍输出整数;不在每次读入时提前舍入。

    参考代码(C++11)

    #include <bits/stdc++.h>
    using namespace std;
    struct cpp_int {
        string s;
        bool neg;
        cpp_int(long long v = 0) : s(to_string(v)), neg(v < 0) {
            if (neg)
                s.erase(s.begin());
        }
        void norm() {
            auto p = s.find_first_not_of('0');
            s = p == string::npos ? "0" : s.substr(p);
            if (s == "0")
                neg = false;
        }
        static int cmpabs(const cpp_int &a, const cpp_int &b) {
            if (a.s.size() != b.s.size())
                return a.s.size() < b.s.size() ? -1 : 1;
            return a.s == b.s ? 0 : a.s < b.s ? -1 : 1;
        }
        friend bool operator<(const cpp_int &a, const cpp_int &b) {
            if (a.neg != b.neg)
                return a.neg;
            int c = cmpabs(a, b);
            return a.neg ? c > 0 : c < 0;
        }
        friend bool operator==(const cpp_int &a, const cpp_int &b) {
            return a.neg == b.neg && a.s == b.s;
        }
        friend bool operator!=(const cpp_int &a, const cpp_int &b) {
            return !(a == b);
        }
        friend bool operator>(const cpp_int &a, const cpp_int &b) {
            return b < a;
        }
        friend bool operator<=(const cpp_int &a, const cpp_int &b) {
            return !(b < a);
        }
        friend bool operator>=(const cpp_int &a, const cpp_int &b) {
            return !(a < b);
        }
        cpp_int operator-() const {
            cpp_int r = *this;
            if (r.s != "0")
                r.neg = !r.neg;
            return r;
        }
        friend cpp_int operator+(const cpp_int &a, const cpp_int &b) {
            cpp_int r;
            r.s.clear();
            int carry = 0;
            if (a.neg == b.neg) {
                int i = (int)a.s.size() - 1, j = (int)b.s.size() - 1;
                while (i >= 0 || j >= 0 || carry) {
                    int v = carry + (i >= 0 ? a.s[i--] - '0' : 0) + (j >= 0 ? b.s[j--] - '0' : 0);
                    r.s += char('0' + v % 10);
                    carry = v / 10;
                }
                r.neg = a.neg;
            } else {
                const cpp_int *x = &a, *y = &b;
                if (cmpabs(a, b) < 0)
                    swap(x, y);
                int i = (int)x->s.size() - 1, j = (int)y->s.size() - 1;
                while (i >= 0) {
                    int v = x->s[i--] - '0' - carry - (j >= 0 ? y->s[j--] - '0' : 0);
                    carry = v < 0;
                    if (carry)
                        v += 10;
                    r.s += char('0' + v);
                }
                r.neg = x->neg;
            }
            reverse(r.s.begin(), r.s.end());
            r.norm();
            return r;
        }
        friend cpp_int operator-(const cpp_int &a, const cpp_int &b) {
            return a + (-b);
        }
        friend cpp_int operator*(const cpp_int &a, int b) {
            cpp_int r;
            r.s.clear();
            long long carry = 0;
            long long z = b;
            if (z < 0)
                z = -z;
            for (int i = (int)a.s.size() - 1; i >= 0; i--) {
                carry += (a.s[i] - '0') * z;
                r.s += char('0' + carry % 10);
                carry /= 10;
            }
            while (carry) {
                r.s += char('0' + carry % 10);
                carry /= 10;
            }
            reverse(r.s.begin(), r.s.end());
            r.neg = a.neg ^ (b < 0);
            r.norm();
            return r;
        }
        friend cpp_int operator/(const cpp_int &a, int b) {
            cpp_int r;
            r.s.clear();
            long long rem = 0, z = b;
            if (z < 0)
                z = -z;
            for (char c : a.s) {
                rem = rem * 10 + c - '0';
                r.s += char('0' + rem / z);
                rem %= z;
            }
            r.neg = a.neg ^ (b < 0);
            r.norm();
            return r;
        }
        cpp_int &operator+=(const cpp_int &b) {
            *this = *this + b;
            return *this;
        }
        cpp_int &operator-=(const cpp_int &b) {
            *this = *this - b;
            return *this;
        }
        cpp_int &operator*=(int b) {
            *this = *this * b;
            return *this;
        }
        friend istream &operator>>(istream &in, cpp_int &a) {
            string v;
            if (in >> v) {
                a.neg = v[0] == '-';
                if (v[0] == '-' || v[0] == '+')
                    v.erase(v.begin());
                a.s = v;
                a.norm();
            }
            return in;
        }
        friend ostream &operator<<(ostream &out, const cpp_int &a) {
            if (a.neg)
                out << '-';
            return out << a.s;
        }
    };
    int main() {
        ios::sync_with_stdio(false);
        cin.tie(nullptr);
        cpp_int x, sum, mn, mx;
        int n = 0;
        while (cin >> x) {
            if (!n)
                mn = mx = x;
            mn = min(mn, x);
            mx = max(mx, x);
            sum += x;
            n++;
        }
        bool negative = sum < 0;
        if (negative)
            sum = -sum;
        cpp_int scaled = sum * 1000, q = scaled / n;
        int rem = 0;
        for (char c : scaled.s)
            rem = (rem * 10 + c - '0') % n;
        if (rem * 2 > n || (rem * 2 == n && (q.s.back() - '0') % 2))
            q += 1;
        string s = q.s;
        while (s.size() < 4)
            s = '0' + s;
        cout << mn << " " << mx << " " << (negative ? "-" : "") << s.substr(0, s.size() - 3) << "."
             << s.substr(s.size() - 3) << "\n";
    }
    
    • 1

    信息

    ID
    2003
    时间
    1000ms
    内存
    64MiB
    难度
    进阶
    标签
    递交数
    0
    已通过
    0
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