Bestcoder round #8 a, B, c

Bestcoder round #8

Question Link

A: I don't have to mention the question of sign-in.

B: The rapid power of the matrix. The formula of the odd number is F (n) = 4 * F (n-1) + 1. the even number is twice that of the odd number, then, construct a matrix of 4 1 0 1 and perform a quick power.

C: DP + tree array acceleration, DP [I] [J] indicates the number of types whose end length is J with I, and then the number is discretization, each status transition requires the previous interval and transfer, so the tree array can be used for maintenance.

Code:

A:

#include <cstdio>#include <cstring>#include <algorithm>#include <set>using namespace std;typedef long long ll;ll num[105];set<ll> save;int n;int main() {while (~scanf("%d", &n)) {save.clear();for (int i = 0; i < n; i++) {scanf("%I64d", &num[i]);for (int j = 0; j < i; j++) {save.insert(num[i] + num[j]);}}ll ans = 0;for (set<ll>::iterator it = save.begin(); it != save.end(); it++) {ans += *it;}printf("%I64d\n", ans);}return 0;}

B:

#include <cstdio>#include <cstring>#include <cmath>#include <iostream>using namespace std;typedef long long ll;ll n, m;struct mat {ll v[2][2];mat() {memset(v, 0, sizeof(v));}mat operator * (mat c) {mat ans;for (int i = 0; i < 2; i++) {for (int j = 0; j < 2; j++) {for (int k = 0; k < 2; k++) {ans.v[i][j] = (ans.v[i][j] + v[i][k] * c.v[k][j]) % m;}}}return ans;}};mat pow_mod(mat A, ll k) {mat ans;for (int i = 0; i < 2; i++) ans.v[i][i] = 1;while (k) {if (k&1) ans = ans * A;A = A * A;k >>= 1;}return ans;}int main() {while (cin >> n >> m) {mat A;A.v[0][0] = 4; A.v[0][1] = 1;A.v[1][0] = 0; A.v[1][1] = 1;A = pow_mod(A, (n + 1) / 2);ll ans = A.v[0][1];ans %= m;if (n % 2 == 0) ans = ans * 2 % m;cout << ans << endl;}return 0;}

C:

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;#define lowbit(x) (x&(-x))const int N = 10005;const int M = 105;const int MOD = 123456789;int n, m, dp[N][M], bit[105][N];void add(int *bit, int x, int v) {while (x < N) {bit[x] = (bit[x] + v) % MOD;x += lowbit(x);}}int get(int *bit, int x) {int ans = 0;while (x) {ans = (ans + bit[x]) % MOD;x -= lowbit(x);}return ans;}struct Num {int val, rank, id;} num[N];bool cmpid(Num a, Num b) {return a.id < b.id;}bool cmpval(Num a, Num b) {return a.val < b.val;}int main() {while (~scanf("%d%d", &n, &m)) {memset(bit, 0, sizeof(bit));for (int i = 1; i <= n; i++) {scanf("%d", &num[i].val);num[i].id = i;}sort(num + 1, num + n + 1, cmpval);num[1].rank = 2;for (int i = 2; i <= n; i++) {num[i].rank = num[i - 1].rank;if (num[i].val > num[i - 1].val)num[i].rank++;}sort(num + 1, num + n + 1, cmpid);add(bit[0], 1, 1);memset(dp, 0, sizeof(dp));for (int i = 1; i <= n; i++) {for (int j = min(m, i); j >= 1; j--) {int tmp = get(bit[j - 1], num[i].rank - 1);dp[i][j] = (dp[i][j] + tmp) % MOD;add(bit[j], num[i].rank, tmp);}}int ans = 0;for (int i = 1; i <= n; i++)ans = (ans + dp[i][m]) % MOD;printf("%d\n", ans);}return 0;}

Bestcoder round #8 a, B, c