HDU 5358 (2015 Multi-school Joint Training tournament 1006) first one (interval merge + constant optimization)

HDU 5358

Test instructions

Beg ∑ i = 1 n ? ∑? J=i ? N ??(?Lo g? 2 ??S(I,J)?+1)?(I+J).

Ideas:

S (i,j) < 10^10 < 2^34, so the value of log2 (S) +1 is only possible between 1~35. Due to the slow function of log, we can compute the interval of the same log value by S (i,n), and use the pos[i][j] preprocessing to record the right boundary of each interval where I is the starting point and log (s) value is j to optimize the complexity of the NLOGN.

Mainly write it is more difficult to some, some details more tangled, a certain way of thinking to clarify and then write.

PS. This card constant is not human, you must remember to preprocess the interval mapping, otherwise n (logn) ^2 also have to kneel.

Code

/** @author novicer* language:c++/c*/#include <iostream> #include <sstream> #include <fstream># include<vector> #include <list> #include <deque> #include <queue> #include <stack># include<map> #include <set> #include <bitset> #include <algorithm> #include <cstdio># include<cstdlib> #include <cstring> #include <cctype> #include <cmath> #include <ctime># Include<iomanip> #define INF 2147483647#define CLS (x) memset (x,0,sizeof (x)) #define RISE (I,A,B) for (int i = A; I < = b; i++) using namespace Std;const double eps (1e-8); typedef long LONG lint;const int MAXN = 100000 + 5;lint A[maxn];lint pos[ma Xn][34];lint two[34] = { 2,4,8,16,32,64,128,256,512,1024,2048,4096,8192,16384,32768,65536,131072,262144,524288,1048576,2097152,4194304,8388608,167 77216,33554432,67108864,134217728,268435456,536870912,1073741824,2147483648,4294967296,8589934592,17179869184}; int main () {int t; cin >> T; while (t--) {int n; cin >> n;for (int i = 1; I <= n; i++) {scanf ("%i64d", &a[i]);} for (int i = 0; i <, i++) {Lint sum = a[1];int right = 1;for (int j = 1; J <= N; j + +) {if (J! = 1) Sum-= A[j-1];w Hile (Sum < two[i] && right <= n) sum + = A[++right];p os[j][i] = right;//start with J, Log2 (sum+1 Lint ans = 0;for (int i = 1; I <= n; i++) {Lint L = i;lint r;for (int j = 0; J < G; J + +) {r = Pos[i][j];ans + = (1+j ) * (((i) * (r-l) + (R+L-1) * (R-L)/2);//Add the same log value interval (i+j) up L = r;}} cout << ans << endl;} return 0;}

Copyright notice: Bo Master said authorized all reproduced:)

HDU 5358 (2015 Multi-school Joint Training tournament 1006) first one (interval merge + constant optimization)