A
#include <bits/stdc++.h> #define PI acos(-1.0) #define mem(a,b) memset((a),b,sizeof(a)) #define TS printf("!!! ") #define pb push_back #define inf 1e9 //std::ios::sync_with_stdio(false); using namespace std; //priority_queue<int,vector<int>,greater<int>> que; const double eps = 1.0e-6; typedef pair<int, int> pairint; typedef long long ll; typedef unsigned long long ull; const int maxn = 1e6 + 10; const int maxm = 300; const int turn[4][2] = {{1, 0}, { -1, 0}, {0, 1}, {0, -1}}; //next_permutation ll mod = 1e9 + 7; int n, aim; int v[105]; int main() { //freopen("in.txt", "r", stdin); //freopen("out1.txt", "w", stdout); cin >> n >> aim; int pa = 0, pb = 0; v[0] = 1; for (int i = 1; i <= n; i++) { cin >> pa >> pb; if (!v[pa]) { cout << "NO" << endl; exit(0); } for (int j = pa; j <= pb; j++) { //cout<<j<<endl; v[j] = 1; } if (v[aim]) { cout << "YES" << endl; exit(0); } } cout << "NO" << endl; }
B
#include <bits/stdc++.h> #define PI acos(-1.0) #define mem(a,b) memset((a),b,sizeof(a)) #define TS printf("!!! ") #define pb push_back #define inf 0x3f3f3f3f //std::ios::sync_with_stdio(false); using namespace std; //priority_queue<int,vector<int>,greater<int>> que; get min const double eps = 1.0e-10; const double EPS = 1.0e-4; typedef pair<int, int> pairint; typedef long long ll; typedef unsigned long long ull; //const int maxn = 3e5 + 10; const int turn[4][2] = {{0, 1}, { 1, 0}, { 0, -1}, { -1, 0}}; //priority_queue<int, vector<int>, less<int>> que; //next_permutation int n; vector<int> tree[10005]; int anser = 0; int color[10005]; int aim[10005]; void dfs(int x, int now) { if (now != aim[x]) { anser++; now = aim[x]; } int len = tree[x].size(); for (int i = 0; i < len; i++) { int to = tree[x][i]; dfs(to, now); } } int main() { //freopen("out.txt","w",stdout); cin >> n; for (int i = 2; i <= n; i++) { int now; cin >> now; tree[now].pb(i); } for (int i = 1; i <= n; i++) { cin >> aim[i]; } dfs(1, 0); cout << anser << endl; return 0; }
C
#include <bits/stdc++.h> #define PI acos(-1.0) #define mem(a,b) memset((a),b,sizeof(a)) #define TS printf("!!! ") #define pb push_back #define inf 1e9 //std::ios::sync_with_stdio(false); using namespace std; //priority_queue<int,vector<int>,greater<int>> que; const double eps = 1.0e-6; typedef pair<int, int> pairint; typedef long long ll; typedef unsigned long long ull; const int maxn = 1e6 + 10; const int maxm = 300; const int turn[4][2] = {{1, 0}, { -1, 0}, {0, 1}, {0, -1}}; //next_permutation ll mod = 1e9 + 7; int n; int h[100085]; int main() { //freopen("in.txt", "r", stdin); //freopen("out1.txt", "w", stdout); cin >> n; int flag = 0; int now = 0; for (int i = 1; i <= n + 1; i++) { scanf("%d", h + i); if (h[i - 1] > 1 && h[i] > 1) { flag = 1; } } if (!flag) { cout << "perfect" << endl; } else { cout << "ambiguous" << endl; for (int i = 1; i <= n + 1; i++) { for (int j = 1; j <= h[i]; j++) { cout << now << " "; } now += h[i]; } cout << endl; int flag2 = 1; now = 0; for (int i = 1; i <= n + 1; i++) { if (h[i - 1] > 1 && h[i] > 1 && flag2) { flag2 = 0; cout << now - 1 <<" "; for (int j = 1; j <= h[i] - 1; j++) { cout << now << " "; } } else { for (int j = 1; j <= h[i]; j++) { cout << now << " "; } } now += h[i]; } cout << endl; } }
D
要求你给出两个多项式 使得A,B辗转相除N次得到答案(B为0)
仿照斐波那契数列使得Fn=x*Fn-1±Fn-2 后面的±的作用是使得给出的多项式满足系数绝对值不超过1的条件
数组可以滚动
#include <bits/stdc++.h> #define PI acos(-1.0) #define mem(a,b) memset((a),b,sizeof(a)) #define TS printf("!!! ") #define pb push_back #define inf 0x3f3f3f3f //std::ios::sync_with_stdio(false); using namespace std; //priority_queue<int,vector<int>,greater<int>> que; get min const double eps = 1.0e-10; const double EPS = 1.0e-4; typedef pair<int, int> pairint; typedef long long ll; typedef unsigned long long ull; //const int maxn = 3e5 + 10; const int turn[4][2] = {{0, 1}, { 1, 0}, { 0, -1}, { -1, 0}}; //priority_queue<int, vector<int>, less<int>> que; //next_permutation int anser[155][155]; int main() { anser[0][0] = 1; anser[1][0] = 0; anser[1][1] = 1; int n; int aim = 1; cin >> n; for (int i = 2; i <= n; i++) { aim ^= 1; for (int j = 1; j <= i; j++) { anser[aim][j] = (anser[aim][j] + anser[aim ^ 1][j - 1]) % 2; } } cout << n << endl; for (int i = 0; i <= n; i++) { cout << anser[aim][i] << " "; } cout << endl << n - 1 << endl; for (int i = 0; i < n; i++) { cout << anser[aim ^ 1][i] << " "; } cout << endl; }