DreamGrid has integers . DreamGrid also has queries, and each time he would like to know the value of
for a given number , where , .
Input
There are multiple test cases. The first line of input is an integer indicating the number of test cases. For each test case:
The first line contains two integers and () -- the number of integers and the number of queries.
The second line contains integers ().
The third line contains integers ().
It is guaranteed that neither the sum of all nor the sum of all exceeds .
Output
For each test case, output an integer , where is the answer for the -th query.
Sample Input
2 3 2 100 1000 10000 100 10 4 5 2323 223 12312 3 1232 324 2 3 5
Sample Output
11366 45619
思路:排序后预处理a[i] / k的前缀和,查询时二分求出相同的一段区间。复杂度O(n*k+ nlogn),k小于32
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
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 mp make_pair 29 #define clr(a, x) memset(a, x, sizeof(a)) 30 31 const double pi = acos(-1.0); 32 const int INF = 0x3f3f3f3f; 33 const int MOD = 1e9; 34 const double EPS = 1e-9; 35 36 /* 37 #include <ext/pb_ds/assoc_container.hpp> 38 #include <ext/pb_ds/tree_policy.hpp> 39 40 using namespace __gnu_pbds; 41 tree<pli, null_type, greater<pli>, rb_tree_tag, tree_order_statistics_node_update> T; 42 */ 43 44 int t, n, m; 45 ll a[100015], sum[100015][30]; 46 int get_pos(int sta, ll tt) { 47 int s = sta, e = n, mi, res = sta - 1; 48 while (s <= e) { 49 mi = s + e >> 1; 50 if (a[mi] <= tt) { 51 s = (res = mi) + 1; 52 } else { 53 e = mi - 1; 54 } 55 } 56 //printf("sta = %d, tt = %lld, res = %d ", sta, tt, res); 57 return res; 58 } 59 int main() { 60 scanf("%d", &t); 61 while (t--) { 62 scanf("%d%d", &n, &m); 63 for (int i = 1; i <= n; ++i) { 64 scanf("%lld", &a[i]); 65 } 66 sort(a + 1, a + n + 1); 67 for (int k = 1; k <= 30; ++k) { 68 for (int i = 1; i <= n; ++i) { 69 sum[i][k] = sum[i - 1][k] + a[i] / k; 70 } 71 } 72 ll res = 0; 73 for (int rep = 1; rep <= m; ++rep) { 74 ll ans = 0, p; 75 scanf("%lld", &p); 76 ll tt = p; 77 for (int i = 1, j = 1; i <= n; ) { 78 int pos = get_pos(i, tt); 79 ans += sum[pos][j] - sum[i - 1][j]; 80 ++j; 81 tt *= p; 82 i = pos + 1; 83 //printf("pos = %d, tt = %lld, i = %d ", pos, tt, i); 84 } 85 res = (res + ans % MOD * rep) % MOD; 86 } 87 printf("%lld ", res); 88 } 89 return 0; 90 }