hdu 5358 First One（二进制+twopoint）

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题目链接：hdu 5358 First One

枚举二进制位k，考虑每个S(i,j) ≥ 2^k，用twopoint维护。复杂度o(nlogn)

#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>using namespace std;typedef long long ll;const int bitn = 36;const int maxn = 1e5 + 5;int N, A[maxn];ll pre[maxn], bit[maxn];ll solve () {ll ans = 1LL * N * (N+1) / 2 * (N+1), sum;for (int i = 1; i <= bitn; i++) {int r = sum = 0;for (int j = 1; j <= N; j++) {while (r < N && sum < bit[i])sum += A[++r];if (sum < bit[i])break;ll n = N - r + 1;ans += n * j + n * (r + N) / 2;sum -= A[j];}}return ans;}inline int get(ll k) {if (k == 0)return 0;return log2(k);}ll handle () {pre[0] = 0;for (int i = 1; i <= N; i++)pre[i] = pre[i-1] + A[i];ll ans = 0;for (int i = 1; i <= N; i++) {for (int j = i; j <= N; j++) {ans += (ll)(get(pre[j] - pre[i-1]) + 1) * (i + j);}}return ans;}int main () {bit[0] = 1;for (int i = 1; i <= bitn; i++)bit[i] = bit[i-1] * 2;int cas;scanf("%d", &cas);while (cas--) {scanf("%d", &N);for (int i = 1; i <= N; i++)scanf("%d", &A[i]);printf("%lld\n", solve());//printf("%lld\n", handle());}return 0;}

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