codeforces round 512 F. Putting Boxes Together 树状数组维护区间加权平均数

Posted 2021-01-10 bigtom

F. Putting Boxes Together

time limit per test

2.5 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

There is an infinite line consisting of cells. There are $\mathrm{nn boxes\; in\; some\; cells\; of\; this\; line.\; The ii-th\; box\; stands\; in\; the\; cell aiai and\; has\; weight wiwi.\; All aiai are\; distinct,\; moreover, ai?1<aiai?1<ai holds\; for\; all\; valid ii.}$

You would like to put together some boxes. Putting together boxes with indices in the segment $[l,r][l,r] means\; that\; you\; will\; move\; some\; of\; them\; in\; such\; a\; way\; that\; their positions will\; form\; some\; segment [x,x+(r?l)][x,x+(r?l)].$

In one step you can move any box to a neighboring cell if it isn‘t occupied by another box (i.e. you can choose $\mathrm{ii and\; change aiai by 11,\; all\; positions\; should\; remain\; distinct).\; You\; spend wiwi units\; of\; energy\; moving\; the\; box ii by\; one\; cell.\; You\; can\; move\; any\; box\; any\; number\; of\; times,\; in\; arbitrary\; order.}$

Sometimes weights of some boxes change, so you have queries of two types:

1. $\mathrm{idid nwnw —\; weight widwid of\; the\; box idid becomes nwnw.}$
2. $\mathrm{ll rr —\; you\; should\; compute\; the\; minimum\; total\; energy\; needed\; to\; put\; together\; boxes\; with\; indices\; in [l,r][l,r].\; Since\; the\; answer\; can\; be\; rather\; big,\; print\; the\; remainder\; it\; gives\; when\; divided\; by 1000000007=109+71000000007=109+7.\; Note\; that\; the\; boxes\; are\; not\; moved\; during\; the\; query,\; you\; only\; should\; compute\; the\; answer.}$

Note that you should minimize the answer, not its remainder modulo ${\mathrm{109+7109+7.\; So\; if\; you\; have\; two\; possible\; answers 2?109+132?109+13and 2?109+142?109+14,\; you\; should\; choose\; the\; first\; one\; and\; print 109+6109+6,\; even\; though\; the\; remainder\; of\; the\; second\; answer\; is 00.}}^{}$

Input

The first line contains two integers $\mathrm{nn and qq (1\le n,q\le 2?1051\le n,q\le 2?105)\; —\; the\; number\; of\; boxes\; and\; the\; number\; of\; queries.}$

The second line contains $\mathrm{nn integers a1,a2,\dots ana1,a2,\dots an (1\le ai\le 1091\le ai\le 109)\; —\; the\; positions\; of\; the\; boxes.\; All aiai are\; distinct, ai?1<aiai?1<ai holds\; for\; all\; valid ii.}$

The third line contains $\mathrm{nn integers w1,w2,\dots wnw1,w2,\dots wn (1\le wi\le 1091\le wi\le 109)\; —\; the\; initial\; weights\; of\; the\; boxes.}$

Next $\mathrm{qq lines\; describe\; queries,\; one\; query\; per\; line.}$

Each query is described in a single line, containing two integers $\mathrm{xx and yy.\; If x<0x<0,\; then\; this\; query\; is\; of\; the\; first\; type,\; where id=?xid=?x, nw=ynw=y (1\le id\le n1\le id\le n, 1\le nw\le 1091\le nw\le 109).\; If x>0x>0,\; then\; the\; query\; is\; of\; the\; second\; type,\; where l=xl=x and r=yr=y (1\le lj\le rj\le n1\le lj\le rj\le n). xx can\; not\; be\; equal\; to 00.}$

Output

For each query of the second type print the answer on a separate line. Since answer can be large, print the remainder it gives when divided by $\mathrm{1000000007=109+71000000007=109+7.}$

Example

input

Copy

5 8
1 2 6 7 10
1 1 1 1 2
1 1
1 5
1 3
3 5
-3 5
-1 10
1 4
2 5

output

Copy

0
10
3
4
18
7

Note

Let‘s go through queries of the example:

1. $\mathrm{1 11 1 —\; there\; is\; only\; one\; box\; so\; we\; don‘t\; need\; to\; move\; anything.}$
2. $\mathrm{1 51 5 —\; we\; can\; move\; boxes\; to\; segment [4,8][4,8]: 1?|1?4|+1?|2?5|+1?|6?6|+1?|7?7|+2?|10?8|=101?|1?4|+1?|2?5|+1?|6?6|+1?|7?7|+2?|10?8|=10.}$
3. $\mathrm{1 31 3 —\; we\; can\; move\; boxes\; to\; segment [1,3][1,3].}$
4. $\mathrm{3 53 5 —\; we\; can\; move\; boxes\; to\; segment [7,9][7,9].}$
5. $?3 5?3 5 — w3w3 is\; changed\; from 11 to 55.$
6. $?1 10?1 10 — w1w1 is\; changed\; from 11 to 1010.\; The\; weights\; are\; now\; equal\; to w=[10,1,5,1,2]w=[10,1,5,1,2].$
7. $\mathrm{1 41 4 —\; we\; can\; move\; boxes\; to\; segment [1,4][1,4].}$
8. $\mathrm{2 52 5 —\; we\; can\; move\; boxes\; to\; segment [5,8][5,8].}$

$\mathrm{思路：}$

树状数组加二分

  1 #include <iostream>
  2 #include <fstream>
  3 #include <sstream>
  4 #include <cstdlib>
  5 #include <cstdio>
  6 #include <cmath>
  7 #include <string>
  8 #include <cstring>
  9 #include <algorithm>
 10 #include <queue>
 11 #include <stack>
 12 #include <vector>
 13 #include <set>
 14 #include <map>
 15 #include <list>
 16 #include <iomanip>
 17 #include <cctype>
 18 #include <cassert>
 19 #include <bitset>
 20 #include <ctime>
 21 
 22 using namespace std;
 23 
 24 #define pau system("pause")
 25 #define ll long long
 26 #define pii pair<int, int>
 27 #define pb push_back
 28 #define pli pair<ll, int>
 29 #define pil pair<int, ll>
 30 #define clr(a, x) memset(a, x, sizeof(a))
 31 
 32 const double pi = acos(-1.0);
 33 const int INF = 0x3f3f3f3f;
 34 const int MOD = 1e9 + 7;
 35 const double EPS = 1e-9;
 36 
 37 /*
 38 #include <ext/pb_ds/assoc_container.hpp>
 39 #include <ext/pb_ds/tree_policy.hpp>
 40 using namespace __gnu_pbds;
 41 #define TREE tree<pli, null_type, greater<pli>, rb_tree_tag, tree_order_statistics_node_update>
 42 TREE T;
 43 */
 44 
 45 int n, q;
 46 ll a[200015], w[200015];
 47 ll c1[200015], c2[200015];
 48 inline int lb(int x) {return x & -x;}
 49 void add1(int p, ll x) {
 50     for (; p <= n; p += lb(p)) c1[p] += x;
 51 }
 52 ll query1(int p) {
 53     ll res = 0;
 54     for (; p; p -= lb(p)) res += c1[p];
 55     return res;
 56 }
 57 void add2(int p, ll x) {
 58     for (; p <= n; p += lb(p)) {
 59         c2[p] += x;
 60         if (c2[p] >= MOD) c2[p] -= MOD;
 61     }
 62 }
 63 ll query2(int p) {
 64     ll res = 0;
 65     for (; p; p -= lb(p)) res += c2[p];
 66     return res % MOD;
 67 }
 68 int get_p(int l, int r) {
 69     ll r1 = query1(l - 1);
 70     ll r2 = query1(r) - r1;
 71     int mi, res = l;
 72     while (l <= r) {
 73         mi = l + r >> 1;
 74         if ((query1(mi) - r1) * 2 >= r2) {
 75             r = (res = mi) - 1;
 76         } else {
 77             l = mi + 1;
 78         }
 79     }
 80     return res;
 81 }
 82 ll cal(int l, int r, int p) {
 83     ll res1 = query2(r) - query2(p - 1) - (query1(r) - query1(p - 1)) % MOD * a[p];
 84     ll res2 = (query1(p) - query1(l - 1)) % MOD * a[p] - (query2(p) - query2(l - 1));
 85     return ((res1 + res2) % MOD + MOD) % MOD;
 86 }
 87 int main() {
 88     scanf("%d%d", &n, &q);
 89     for (int i = 1; i <= n; ++i) {
 90         scanf("%lld", &a[i]);
 91         a[i] -= i;
 92     }
 93     for (int i = 1; i <= n; ++i) {
 94         scanf("%lld", &w[i]);
 95     }
 96     for (int i = 1; i <= n; ++i) {
 97         add1(i, w[i]);
 98         add2(i, w[i] * a[i] % MOD);
 99     }
100     while (q--) {
101         int l, r;
102         scanf("%d%d", &l, &r);
103         if (l < 0) {
104             add1(-l, r - w[-l]);
105             add2(-l, (r - w[-l]) * a[-l] % MOD);
106             w[-l] = r;
107         } else {
108             int p = get_p(l, r);
109             printf("%lld
", cal(l, r, p));
110         }
111     }
112     return 0;
113 }

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