Bestcoder round #16

Bestcoder round #16

Question Link

This frustrated, 3 hangs 2, all are very Sb errors 23333 Qaq

A: For each number, the number on the left is multiplied by the number on the right, that is, the number of intervals that can be composed. Then, take note of the modulo operation during the multiplication process. Otherwise, it will pop up.

B: DP, DP [I] [2]: record the first-and second---long Lis. Ah, when the transfer was over, it was a tragedy caused by a point of frustration.

C: first of all, if you know about the NIM game, it is easy to convert the problem into: whether some numbers can be selected to be different or 0, that is, split each number into 40 digits, then, every potential is an exclusive or 0, so that 40 equations can be constructed, and then Gaussian deyuan can be used to solve the problem. If there is a free variable, there is a solution, because there must be a solution that is completely zero, however, this solution is wrong because it cannot be fully obtained, so it becomes a matter of judging the number of free yuan changes. It is still a matter of failure to write down the tragedy.

Code:

A:

#include <cstdio>#include <cstring>#include <vector>#include <set>#include <algorithm>using namespace std;typedef long long ll;const int N = 500005;const ll MOD = 1000000007;int t, n;ll a;int main() {scanf("%d", &t);while (t--) {scanf("%d", &n);ll ans = 0;for (ll i = 0; i < n; i++) {scanf("%I64d", &a);ans = (ans + a * (n - i) % MOD * (i + 1) % MOD) % MOD;}printf("%I64d\n", ans);}return 0;}

B:

#include <cstdio>#include <cstring>#include <vector>#include <algorithm>using namespace std;const int N = 1005;int t, n, a[N], dp[N][2];int main() {scanf("%d", &t);while (t--) {scanf("%d", &n);int ans0 = 0, ans1 = 0;for (int i = 1; i <= n; i++) {scanf("%d", &a[i]);dp[i][0] = 1; dp[i][1] = 0;for (int j = 1; j < i; j++) {if (a[i] <= a[j]) continue;if (dp[i][0] <= dp[j][0] + 1) {dp[i][1] = dp[i][0];dp[i][0] = dp[j][0] + 1;} else if (dp[i][1] < dp[j][0] + 1) {dp[i][1] = dp[j][0] + 1;}dp[i][1] = max(dp[i][1], dp[j][1] + 1);}if (ans0 <= dp[i][0]) {ans1 = ans0;ans0 = dp[i][0];} else if (ans1 < dp[i][0]) ans1 = dp[i][0];ans1 = max(ans1, dp[i][1]);}printf("%d\n", ans1);}return 0;}

C:

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef long long ll;const int N = 1005;int t, n, A[45][N];ll a;int rank(int m, int n) {int i = 0, j = 0, k, r, u;while (i < m && j < n) {r = i;for (k = i; k < m; k++)if (A[k][j]) {r = k; break;}if (A[r][j]) {if (r != i) for(k = 0; k <= n; k++) swap(A[r][k], A[i][k]);for (u = i + 1; u < m; u++) if (A[u][j])for (k = i; k <= n; k++) A[u][k] ^= A[i][k];i++;}j++;}return n - i;}int main() {scanf("%d", &t);while (t--) {scanf("%d", &n);memset(A, 0, sizeof(A));for (int i = 0; i < n; i++) {scanf("%lld", &a);for (int j = 0; j < 40; j++)A[j][i] = (a>>j)&1;}printf("%s\n", rank(40, n) > 0 ? "Yes" : "No");}return 0;}

Bestcoder round #16