标签：sts update order iostream define gnu first fstream std
There is an infinite line consisting of cells. There are 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 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:
Note that you should minimize the answer, not its remainder modulo 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.
The first line contains two integers nn and qq (1≤n,q≤2?1051≤n,q≤2?105) — the number of boxes and the number of queries.
The second line contains nn integers a1,a2,…ana1,a2,…an (1≤ai≤1091≤ai≤109) — the positions of the boxes. All aiai are distinct, ai?1<aiai?1<ai holds for all valid ii.
The third line contains nn integers w1,w2,…wnw1,w2,…wn (1≤wi≤1091≤wi≤109) — the initial weights of the boxes.
Next qq lines describe queries, one query per line.
Each query is described in a single line, containing two integers xx and yy. If x<0x<0, then this query is of the first type, where id=?xid=?x, nw=ynw=y (1≤id≤n1≤id≤n, 1≤nw≤1091≤nw≤109). If x>0x>0, then the query is of the second type, where l=xl=x and r=yr=y (1≤lj≤rj≤n1≤lj≤rj≤n). xx can not be equal to 00.
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 1000000007=109+71000000007=109+7.
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
0
10
3
4
18
7
Let‘s go through queries of the example:
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\n", cal(l, r, p)); 110 } 111 } 112 return 0; 113 }
codeforces round 512 F. Putting Boxes Together 树状数组维护区间加权平均数
标签：sts update order iostream define gnu first fstream std
原文地址：https://www.cnblogs.com/BIGTOM/p/9738058.html
