不是很难的一个题目。正确思路是统计每一条边被经过的次数，但我最初由于习惯直接先上了一个前缀和再推的式子，导致极其麻烦难以写对而且会爆\(longlong\)。

#include 
    
     using namespace std;const int N = 100000 + 5;#define ls (p << 1)#define rs (p << 1 | 1)#define mid ((l + r) >> 1)#define int long longint n, m, sum1, sum2, sum3;struct Segment_Tree {    int tag[N << 2], s1[N << 2], s2[N << 2], s3[N << 2];        Segment_Tree () {        memset (s1, 0, sizeof (s1));        memset (s2, 0, sizeof (s2));        memset (s3, 0, sizeof (s3));        memset (tag, 0, sizeof (tag));    }        void push_up (int p) {        s1[p] = s1[ls] + s1[rs];        s2[p] = s2[ls] + s2[rs];        s3[p] = s3[ls] + s3[rs];    }        int F1 (int x, int y) {return y - x + 1;}  // \sum_{i = x} ^ {y} i ^ 0    int F2 (int x, int y) {return (x + y) * (y - x + 1) / 2;} // \sum_{i = x} ^ {y} i ^ 1     int F3 (int x, int y) {         x = x - 1;        int w1 = x * (x + 1) * (2 * x + 1) / 6;        int w2 = y * (y + 1) * (2 * y + 1) / 6;        return w2 - w1;    } // \sum_{i = x} ^ {y} i ^ 2        void work (int p, int l, int r, int val) {        s1[p] += F1 (l, r) * val;        s2[p] += F2 (l, r) * val;        s3[p] += F3 (l, r) * val;        tag[p] += val;    }        void push_down (int p, int l, int r) {        work (ls, l, mid + 0, tag[p]);        work (rs, mid + 1, r, tag[p]);        tag[p] = 0;    }        void modify (int nl, int nr, int w, int l = 1, int r = n, int p = 1) {        if (nl <= l && r <= nr) {            work (p, l, r, w);            return;        }        push_down (p, l, r);        if (nl <= mid) modify (nl, nr, w, l, mid, ls);        if (mid < nr) modify (nl, nr, w, mid + 1, r, rs);        push_up (p); return;    }        void query (int nl, int nr, int l = 1, int r = n, int p = 1) {        if (nl <= l && r <= nr) {            sum1 += s1[p];            sum2 += s2[p];            sum3 += s3[p];            return;        }        push_down (p, l, r);        if (nl <= mid) query (nl, nr, l, mid, ls);        if (mid < nr) query (nl, nr, mid + 1, r, rs);        push_up (p); return;    }}tr; // 维护 \sum_{x = L}^{R} sumd(x) int gcd (int x, int y) {    return y ? gcd (y, x % y) : x;}signed main () {    freopen ("data.in", "r", stdin);    cin >> n >> m;    for (int i = 1; i <= m; ++i) {        char opt; int l, r, v;        cin >> opt;        if (opt == 'C') {            cin >> l >> r >> v; r--;            tr.modify (l, r, v);        } else {            cin >> l >> r; r--;            sum1 = sum2 = sum3 = 0;            tr.query (l, r);            int w1 = (r - l + 1 - r * l);            int w2 = l + r;            int w3 = -1;            int upp = w1 * sum1 + w2 * sum2 + w3 * sum3;            int dwn = (r - l + 2) * (r - l + 1) / 2;            int d = gcd (upp, dwn); upp /= d, dwn /= d;            cout << upp << "/" << dwn << endl;        }    }}