考虑每一段对答案的贡献
用每一段的左端点来表示当前这一段，那么区间就变成了[1,n-1]
如果询问区间[l,r]，其中一个点的位置为x，则它对答案的贡献为(x-l)*(r-x)*s[x]（s[x]为这一段的权值）
化简后得x*s[x]*(l+r-1)-s[x]*(l*r-r)-x*x*s[x]
那么我们就需要维护x*s[x]，s[x]，x*x*s[x]
其中还需要预处理出来x和x*x
然后就ok了
#include <cstdio> #include <cstring> #include <iostream> #define N 500001 #define LL long long #define root 1, 1, n - 1 #define ls now << 1, l, mid #define rs now << 1 | 1, mid + 1, r using namespace std; int n, m; LL s, xs, xxs, ans1, ans2; LL x1[N], x2[N], sum1[N], sum2[N], sum3[N], add[N]; //x1表示 x //x2表示 x^2 //sum1表示 s[x] //sum2表示 x * s[x] //sum3表示 x^2 * s[x] inline int read() { int x = 0, f = 1; char ch = getchar(); for(; !isdigit(ch); ch = getchar()) if(ch == ‘-‘) f = -1; for(; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + ch - ‘0‘; return x * f; } inline void push_down(int now, int l, int r) { if(add[now]) { int mid = (l + r) >> 1; sum1[now << 1] += 1ll * add[now] * (mid - l + 1); sum1[now << 1 | 1] += 1ll * add[now] * (r - mid); sum2[now << 1] += add[now] * x1[now << 1]; sum2[now << 1 | 1] += add[now] * x1[now << 1 | 1]; sum3[now << 1] += add[now] * x2[now << 1]; sum3[now << 1 | 1] += add[now] * x2[now << 1 | 1]; add[now << 1] += add[now]; add[now << 1 | 1] += add[now]; add[now] = 0; } } inline void push_up(int now) { sum1[now] = sum1[now << 1] + sum1[now << 1 | 1]; sum2[now] = sum2[now << 1] + sum2[now << 1 | 1]; sum3[now] = sum3[now << 1] + sum3[now << 1 | 1]; } inline void update(int now, int l, int r, int x, int y, LL z) { if(x <= l && r <= y) { add[now] += z; sum1[now] += 1ll * z * (r - l + 1); sum2[now] += 1ll * z * x1[now]; sum3[now] += 1ll * z * x2[now]; return; } push_down(now, l, r); int mid = (l + r) >> 1; if(x <= mid) update(ls, x, y, z); if(mid < y) update(rs, x, y, z); push_up(now); } inline void build(int now, int l, int r) { if(l == r) { x1[now] += l; x2[now] += 1ll * l * l; return; } int mid = (l + r) >> 1; build(ls); build(rs); x1[now] = x1[now << 1] + x1[now << 1 | 1]; x2[now] = x2[now << 1] + x2[now << 1 | 1]; } inline void query(int now, int l, int r, int x, int y) { if(x <= l && r <= y) { s += sum1[now]; xs += sum2[now]; xxs += sum3[now]; return; } push_down(now, l, r); int mid = (l + r) >> 1; if(x <= mid) query(ls, x, y); if(mid < y) query(rs, x, y); } inline LL gcd(LL x, LL y) { return !y ? x : gcd(y, x % y); } int main() { LL g, z; int i, x, y; char c[10]; n = read(); m = read(); build(root); while(m--) { scanf("%s", c); if(c[0] == ‘C‘) { x = read(); y = read(); z = read(); update(root, x, y - 1, z); } else { x = read(); y = read(); s = xs = xxs = 0; query(root, x, y - 1); ans2 = 1ll * (1 + y - x) * (y - x) / 2; ans1 = 1ll * xs * (x + y - 1) - s * (1ll * x * y - y) - xxs; g = gcd(ans1, ans2); printf("%lld/%lld\n", ans1 / g, ans2 / g); } } return 0; }
一个longlong调了我45min，WNM