题目描述
(Frank) 对天文学非常感兴趣，他经常用望远镜看星星，同时记录下它们的信息，比如亮度、颜色等等，进而估算出星星的距离，半径等等。
(Frank) 不仅喜欢观测，还喜欢分析观测到的数据。他经常分析两个参数之间（比如亮度和半径）是否存在某种关系。
现在 (Frank) 要分析参数 (X) 与 (Y) 之间的关系。他有 (n) 组观测数据，第 (i) 组观测数据记录了 (x_i) 和 (y_i)。他需要一下几种操作
(1 L,R：)
用直线拟合第 (L) 组到第 (R) 组观测数据。用 (overline{x}) 表示这些观测数据中 (x) 的平均数，用 (overline{y}) 表示这些观测数据中 (y) 的平均数，即
(overline{x}={1 over R-L+1} sum _{i=L} ^R x_i)
(overline{y}={1 over R-L+1} sum _{i=L} ^R y_i)
如果直线方程是 (y=ax+b)，那么 (a,b) 应当这样计算：
(a={sum_{i=L} ^R (x_i-overline{x})(y_i-overline{y}) over sum _{i=L} ^R (x_i -overline{x})^2})
你需要帮助 (Frank) 计算 (a)。
(2 L,R,S,T：)
(Frank) 发现测量数据第 (L) 组到第 (R) 组数据有误差，对每个 (i) 满足 (L leq i leq R)，(x_i) 需要加上 (S)，(y_i) 需要加上(T)。
(3 L,R,S,T：)
(Frank)发现第 (L) 组到第 (R) 组数据需要修改，对于每个 (i) 满足 (L leq i leq R)，(x_i)需要修改为 ((S+i))，(y_i) 需要修改为 ((T+i))。
输入格式
第一行两个数 (n,m)，表示观测数据组数和操作次数。
接下来一行 (n) 个数，第 (i) 个数是 (x_i)。
接下来一行 (n) 个数，第 (i) 个数是 (y_i)。
接下来 (m) 行，表示操作，格式见题目描述。
输出格式
对于每个 (1) 操作，输出一行，表示直线斜率 (a)。选手输出与标准输出的绝对误差或相对误差不超过 (10^{-5}) 即为正确。
输入输出样例
输入 #1
3 5
1 2 3
1 2 3
1 1 3
2 2 3 -3 2
1 1 2
3 1 2 2 1
1 1 3
输出 #1
1.0000000000
-1.5000000000
-0.6153846154
说明/提示
对于 (20\%) 的数据 (1 leq n,m leq 1000)
另有 (20\%) 的数据，没有 (3) 操作，且 (2) 操作中 (S=0)
另有 (30\%) 的数据，没有 (3) 操作。
对于 (100\%) 的数据，(1 leq n,m leq 10^5,0 leq |S|,|T| leq 10^5,0 leq |x_i|,|y_i| leq 10^5)
保证 (1) 操作不会出现分母为 (0) 的情况。
时间限制：(1s)
空间限制：(128MB)
分析
把式子化简，就会得到
(egin{aligned} & sum (x_i - ar x)(y_i - ar y) \ = & sum (x_i y_i - x_i ar y - y_i ar x_i + ar x ar y) \ = & sum x_i y_i - ar ysum x_i - ar xsum y_i + nar x ar y \ = & sum x_i y_i - nar x ar y \ \ & sum (x_i - ar x)^2 \ = & sum (x_i^2 + {ar x}^2 - 2x_iar x) \ = & sum x_i^2 +n{ar x}^2 - 2ar xsum x_i \ = & sum x_i^2 -n {ar x}^2\ end{aligned})
那么我们要维护的东西就是 (x_i)、(y_i) 和 (x_iy_i)
对于操作 (2)
(egin{aligned} &sum x_i osum(x_i+S)=sum x_i+nS \ &sum y_i osum(y_i+T)=sum y_i+nT \ &sum x_i^2 osum(x_i+S)^2=sum x_i^2+nS^2+2Ssum x_i\ &sum x_i y_i osum(x_i+S)(y_i+T)=sum x_i y_i+Tsum x_i+Ssum y_i+nST \ end{aligned})
对于操作 (3)
(egin{aligned} &sum x_i osum(i+S)=s_1+nS \ &sum y_i osum(i+T)=s_1+nT \ &sum x_i^2 osum(i+S)^2=s_2+nS^2+2Ss_1\ &sum x_i y_i osum(i+S)(i+T)=s_2+(T+S)s_1+nST\ end{aligned})
其中 (s_1) 是等差数列的求和公式 (frac{n(n+1)}{2})
(s_2) 是 (i^2) 的前缀和 (frac{n(n+1)(2n+1)}{6})
注意下放标记的时候只要有一个不为零就要下放
要先下放覆盖的标记，再下放加的标记
代码
#include <cstdio>
#include <algorithm>
#include <cmath>
#define rg register
const int maxn = 1e5 + 5;
typedef double db;
int n, m;
db jlx[maxn], jly[maxn];
struct trr {
int l, r, siz;
db sumx, sumy, sumxx, sumxy, lazx, lazy, tagx, tagy;
trr() {
tagx = tagy = 1e18;
sumx = sumy = sumxx = sumxy = lazx = lazy = 0;
l = r = siz = 0;
}
} tr[maxn << 2];
db getsum1(int l, int r) { return (db)(r - l + 1.0) * (l + r) / 2.0; }
db getsum2(int r) { return (db)r * (r + 1.0) * (2.0 * r + 1.0) / 6.0; }
void push_up(int da) {
tr[da].sumx = tr[da << 1].sumx + tr[da << 1 | 1].sumx;
tr[da].sumy = tr[da << 1].sumy + tr[da << 1 | 1].sumy;
tr[da].sumxx = tr[da << 1].sumxx + tr[da << 1 | 1].sumxx;
tr[da].sumxy = tr[da << 1].sumxy + tr[da << 1 | 1].sumxy;
}
void push_down(int da) {
if (tr[da].tagx != 1e18 || tr[da].tagy != 1e18) {
tr[da << 1].tagx = tr[da].tagx;
tr[da << 1 | 1].tagx = tr[da].tagx;
tr[da << 1].tagy = tr[da].tagy;
tr[da << 1 | 1].tagy = tr[da].tagy;
tr[da << 1].sumx = tr[da].tagx * tr[da << 1].siz + getsum1(tr[da << 1].l, tr[da << 1].r);
tr[da << 1 | 1].sumx =
tr[da].tagx * tr[da << 1 | 1].siz + getsum1(tr[da << 1 | 1].l, tr[da << 1 | 1].r);
tr[da << 1].sumy = tr[da].tagy * tr[da << 1].siz + getsum1(tr[da << 1].l, tr[da << 1].r);
tr[da << 1 | 1].sumy =
tr[da].tagy * tr[da << 1 | 1].siz + getsum1(tr[da << 1 | 1].l, tr[da << 1 | 1].r);
tr[da << 1].sumxx = tr[da << 1].siz * tr[da].tagx * tr[da].tagx +
2.0 * tr[da].tagx * getsum1(tr[da << 1].l, tr[da << 1].r) +
getsum2(tr[da << 1].r) - getsum2(tr[da << 1].l - 1);
tr[da << 1 | 1].sumxx = tr[da << 1 | 1].siz * tr[da].tagx * tr[da].tagx +
2.0 * tr[da].tagx * getsum1(tr[da << 1 | 1].l, tr[da << 1 | 1].r) +
getsum2(tr[da << 1 | 1].r) - getsum2(tr[da << 1 | 1].l - 1);
tr[da << 1].sumxy = tr[da << 1].siz * tr[da].tagx * tr[da].tagy +
(tr[da].tagx + tr[da].tagy) * getsum1(tr[da << 1].l, tr[da << 1].r) +
getsum2(tr[da << 1].r) - getsum2(tr[da << 1].l - 1);
tr[da << 1 | 1].sumxy = tr[da << 1 | 1].siz * tr[da].tagx * tr[da].tagy +
(tr[da].tagx + tr[da].tagy) * getsum1(tr[da << 1 | 1].l, tr[da << 1 | 1].r) +
getsum2(tr[da << 1 | 1].r) - getsum2(tr[da << 1 | 1].l - 1);
tr[da].tagx = tr[da].tagy = 1e18;
tr[da << 1].lazx = tr[da << 1 | 1].lazx = tr[da << 1].lazy = tr[da << 1 | 1].lazy = 0;
}
if (tr[da].lazx != 0 || tr[da].lazy != 0) {
tr[da << 1].lazx += tr[da].lazx;
tr[da << 1 | 1].lazx += tr[da].lazx;
tr[da << 1].lazy += tr[da].lazy;
tr[da << 1 | 1].lazy += tr[da].lazy;
tr[da << 1].sumxx +=
2.0 * tr[da].lazx * tr[da << 1].sumx + tr[da << 1].siz * tr[da].lazx * tr[da].lazx;
tr[da << 1 | 1].sumxx +=
2.0 * tr[da].lazx * tr[da << 1 | 1].sumx + tr[da << 1 | 1].siz * tr[da].lazx * tr[da].lazx;
tr[da << 1].sumxy += tr[da << 1].sumx * tr[da].lazy + tr[da << 1].sumy * tr[da].lazx +
tr[da << 1].siz * tr[da].lazx * tr[da].lazy;
tr[da << 1 | 1].sumxy += tr[da << 1 | 1].sumx * tr[da].lazy + tr[da << 1 | 1].sumy * tr[da].lazx +
tr[da << 1 | 1].siz * tr[da].lazx * tr[da].lazy;
tr[da << 1].sumx += tr[da << 1].siz * tr[da].lazx;
tr[da << 1 | 1].sumx += tr[da << 1 | 1].siz * tr[da].lazx;
tr[da << 1].sumy += tr[da << 1].siz * tr[da].lazy;
tr[da << 1 | 1].sumy += tr[da << 1 | 1].siz * tr[da].lazy;
tr[da].lazx = tr[da].lazy = 0;
}
}
void build(int da, int l, int r) {
tr[da].l = l, tr[da].r = r, tr[da].siz = r - l + 1;
if (tr[da].l == tr[da].r) {
tr[da].sumx = jlx[l];
tr[da].sumy = jly[l];
tr[da].sumxx = jlx[l] * jlx[l];
tr[da].sumxy = jlx[l] * jly[l];
return;
}
rg int mids = (tr[da].l + tr[da].r) >> 1;
build(da << 1, l, mids);
build(da << 1 | 1, mids + 1, r);
push_up(da);
}
void ad(int da, int l, int r, db valx, db valy) {
if (tr[da].l >= l && tr[da].r <= r) {
tr[da].lazx += valx;
tr[da].lazy += valy;
tr[da].sumxx += 2.0 * valx * tr[da].sumx + tr[da].siz * valx * valx;
tr[da].sumxy += tr[da].sumx * valy + tr[da].sumy * valx + tr[da].siz * valx * valy;
tr[da].sumx += tr[da].siz * valx;
tr[da].sumy += tr[da].siz * valy;
return;
}
push_down(da);
rg int mids = (tr[da].l + tr[da].r) >> 1;
if (l <= mids)
ad(da << 1, l, r, valx, valy);
if (r > mids)
ad(da << 1 | 1, l, r, valx, valy);
push_up(da);
}
void xg(int da, int l, int r, db valx, db valy) {
if (tr[da].l >= l && tr[da].r <= r) {
tr[da].lazx = 0, tr[da].lazy = 0;
tr[da].tagx = valx;
tr[da].tagy = valy;
tr[da].sumx = valx * tr[da].siz + getsum1(tr[da].l, tr[da].r);
tr[da].sumy = valy * tr[da].siz + getsum1(tr[da].l, tr[da].r);
tr[da].sumxx = tr[da].siz * valx * valx + 2.0 * valx * getsum1(tr[da].l, tr[da].r) +
getsum2(tr[da].r) - getsum2(tr[da].l - 1);
tr[da].sumxy = tr[da].siz * valx * valy + (valx + valy) * getsum1(tr[da].l, tr[da].r) +
getsum2(tr[da].r) - getsum2(tr[da].l - 1);
return;
}
push_down(da);
rg int mids = (tr[da].l + tr[da].r) >> 1;
if (l <= mids)
xg(da << 1, l, r, valx, valy);
if (r > mids)
xg(da << 1 | 1, l, r, valx, valy);
push_up(da);
}
db cxx(int da, int l, int r) {
if (tr[da].l >= l && tr[da].r <= r) {
return tr[da].sumx;
}
push_down(da);
rg int mids = (tr[da].l + tr[da].r) >> 1;
rg db nans = 0;
if (l <= mids)
nans += cxx(da << 1, l, r);
if (r > mids)
nans += cxx(da << 1 | 1, l, r);
return nans;
}
db cxy(int da, int l, int r) {
if (tr[da].l >= l && tr[da].r <= r) {
return tr[da].sumy;
}
push_down(da);
rg int mids = (tr[da].l + tr[da].r) >> 1;
rg db nans = 0;
if (l <= mids)
nans += cxy(da << 1, l, r);
if (r > mids)
nans += cxy(da << 1 | 1, l, r);
return nans;
}
db cxxx(int da, int l, int r) {
if (tr[da].l >= l && tr[da].r <= r) {
return tr[da].sumxx;
}
push_down(da);
rg int mids = (tr[da].l + tr[da].r) >> 1;
rg db nans = 0;
if (l <= mids)
nans += cxxx(da << 1, l, r);
if (r > mids)
nans += cxxx(da << 1 | 1, l, r);
return nans;
}
db cxxy(int da, int l, int r) {
if (tr[da].l >= l && tr[da].r <= r) {
return tr[da].sumxy;
}
push_down(da);
rg int mids = (tr[da].l + tr[da].r) >> 1;
rg db nans = 0;
if (l <= mids)
nans += cxxy(da << 1, l, r);
if (r > mids)
nans += cxxy(da << 1 | 1, l, r);
return nans;
}
db getx(int l, int r) { return (db)cxx(1, l, r) / (r - l + 1); }
db gety(int l, int r) { return (db)cxy(1, l, r) / (r - l + 1); }
void solve(int l, int r) {
db ans1 = cxxy(1, l, r) - (db)(r - l + 1) * getx(l, r) * gety(l, r);
db ans2 = cxxx(1, l, r) - (db)(r - l + 1) * getx(l, r) * getx(l, r);
printf("%.10f
", ans1 / ans2);
}
int main() {
scanf("%d%d", &n, &m);
for (rg int i = 1; i <= n; i++) {
scanf("%lf", &jlx[i]);
}
for (rg int i = 1; i <= n; i++) {
scanf("%lf", &jly[i]);
}
build(1, 1, n);
rg int aa, bb, cc;
db dd, ee;
for (rg int i = 1; i <= m; i++) {
scanf("%d%d%d", &aa, &bb, &cc);
if (aa == 1) {
solve(bb, cc);
} else if (aa == 2) {
scanf("%lf%lf", &dd, &ee);
ad(1, bb, cc, dd, ee);
} else {
scanf("%lf%lf", &dd, &ee);
xg(1, bb, cc, dd, ee);
}
}
return 0;
}