2020hdu多校第八场比赛及补题
1003 Clockwise or Counterclockwise
给A B C三个点，这三个点都在一个以原点为圆心的圆上，问ABC方向是顺时针方向还是逆时针方向
挺烦的模拟
#include <bits/stdc++.h>
using namespace std;
#define int long long
// const int MAXN = ;
// const int MOD = ;
// const int INF = ;
// const double eps = ;
const double PI = acos(-1);
// const int DIRX[] = {};
// const int DIRY[] = {};
struct Node
{
int x, y;
double s;
double c;
double the;
}node[5];
int T;
double r;
int32_t main(void)
{
ios::sync_with_stdio(false);
cin.tie(0); cout.tie(0);
cin >> T;
while (T--)
{
for (int i = 1; i <= 3; ++i)
{
cin >> node[i].x >> node[i].y;
r = sqrt((long long)node[1].x * node[1].x + (long long) node[1].y * node[1].y);
node[i].s = (double) node[i].y / r;
node[i].c = (double) node[i].x / r;
node[i].the = asin(node[i].s);
// cout << node[i].s << " " << node[i].c << " " << node[i].the << endl;
}
for (int i = 1; i <= 3; ++i)
{
if (node[i].s > 0 && node[i].c < 0)
node[i].the = PI - node[i].the;
if (node[i].s <= 0 && node[i].c < 0)
node[i].the = PI - node[i].the;
if (node[i].s < 0 && node[i].c > 0)
node[i].the += 2 * PI;
// cout << (node[i].the / PI) << " ";
}
if (node[1].the > node[2].the)
{
if (node[3].the > node[1].the || node[3].the < node[2].the)
{
cout << "Clockwise" << endl;
}
else
{
cout << "Counterclockwise" << endl;
}
}
else
{
if (node[3].the > node[2].the || node[3].the < node[1].the)
{
cout << "Counterclockwise" << endl;
}
else
{
cout << "Clockwise" << endl;
}
}
}
return 0;
}
1006 Fluctuation Limit
问是否存在这样一个数列，li ≤ ai ≤ ri (1 ≤ i ≤ n) and | ai − ai+1 | ≤ k (1 ≤ i≤ n−1)
若存在，输出这样的一个数列
简单的线段合并模拟
a1的范围是[l1,r1],那么a2的范围是被包含在[l1-k,ri+k]内的，那么a2的范围就是[l1-k,ri+k]和[l2,r2]的交集,这样再推出a3,a4...an的范围
然后在an选一个数，再确定an-1，再确定an-2...从后往前确定数
#include<iostream>
#include<algorithm>
#include<cstdio>
using namespace std;
const int MAXN = 1e5+7;
struct LINE{
long long l,r;
}line[MAXN];
long long l, r;
long long ans[MAXN];
bool merge(LINE a,LINE b){
if(a.r<b.l||a.l>b.r) return false;
l = max(a.l,b.l);
r = min(a.r,b.r);
return true;
}
int main()
{
int T;
int n;
long long k;
cin>>T;
while(T--){
cin>>n>>k;
for(int i = 1;i <= n;i++){
scanf("%lld%lld",&line[i].l,&line[i].r);
}
bool flag = true;
LINE pre, aft;
for(int i = 2;i <= n;i++){
pre.l = line[i-1].l-k;
pre.r = line[i-1].r+k;
if(merge(pre,line[i])){
line[i].l = l;
line[i].r = r;
}
else {
flag = false;
break;
}
}
if(!flag) cout<<"NO"<<endl;
else{
cout<<"YES"<<endl;
ans[n] = line[n].l;
for(int i = n - 1; i;i--){
aft.l = ans[i + 1]-k;
aft.r = ans[i + 1]+k;
merge(line[i],aft);
ans[i] = l;
}
for(int i = 1;i < n;i++) printf("%lld ",ans[i]);
printf("%lld\n",ans[n]);
}
}
return 0;
}
1008 Hexagon
阴间构造蜂巢题，感觉比第十场的扫雷恶心多了
#include<iostream>
#include<algorithm>
#include<cstdio>
using namespace std;
void gon(int n){
if(n == 1) return;
if(n == 2) {
printf("543215");
return;
}
for(int i = 1;i <= n - 2;i++) printf("42");
printf("3");
for(int i = 1;i <= n - 2;i++) printf("53");
printf("4");
for(int i = 1;i <= n - 2;i++) printf("64");
printf("5");
for(int i = 1;i <= n - 2;i++) printf("15");
printf("6");
for(int i = 1;i <= n - 2;i++) printf("26");
printf("1");
for(int i = 1;i <= n - 3;i++) printf("31");
printf("2");
printf("43");
gon(n-2);
}
int main()
{
int T, n;
scanf("%d",&T);
while(T--){
scanf("%d",&n);
gon(n);
printf("\n");
}
return 0;
}
1009 Isomorphic Strings
哈希
加一个小剪枝:总串一共有x种字母，第一段子串有y种字母,x!=y就直接return false;
第一次写双哈希，三哈希
#include<iostream>
#include<algorithm>
#include<cstring>
#include<vector>
using namespace std;
const int MAXN = 5e6+7;
const int INF = 1e7;
const int BASE = 29;
const long long MOD1 = 99824435;
const long long MOD2 = 1e9+7;
//const long long MOD3 = 1e9+9;
char s[MAXN];
int su[26];
long long ten[3][MAXN];
int n;
int kind;
void pre(int n){
ten[0][0] = ten[1][0] = ten[2][0] = 1;
for(int i = 1;i <= n;i++){
ten[0][i] = ten[0][i-1] * 10 % MOD1;
ten[1][i] = ten[1][i-1] * 10 % MOD2;
//ten[2][i] = ten[2][i-1] * 10 % MOD3;
}
}
int gcd(int a,int b){
if(a<b) swap(a,b);
while(b){
a = a % b;
swap(a,b);
}
return a;
}
bool check(int len){
vector<long long>hash_1,hash_2
//,hash_3
;
bool vis[26]={false};//这里一定要初始化，我这里wa了找了好久qwq
long long res1 = 0, res2 = 0
//,res3 = 0
;
int kd = 0;
for(int i = 1;i <= len;i++){
res1 = res1 * 10 + s[i] - 'a';
res2 = res2 * 10 + s[i] - 'a';
//res3 = res3 * 10 + s[i] - 'a';
res1 %= MOD1;
res2 %= MOD2;
//res3 %= MOD3;
if(!vis[s[i]-'a']){
kd++;
vis[s[i]-'a'] = true;
}
}
if(kd!=kind) return false;
hash_1.push_back(res1);
hash_2.push_back(res2);
//hash_3.push_back(res3);
for(int i = 1;i < len;i++){
res1 -= (long long)1 * (s[i] - 'a') * ten[0][len-1];
res1 = (res1 % MOD1 + MOD1) * 10 % MOD1;
res1 += s[i] - 'a';
res1 %= MOD1;
res2 -= (long long)1 * (s[i] - 'a') * ten[1][len-1];
res2 = (res2 % MOD2 + MOD2) * 10 % MOD2;
res2 += s[i] - 'a';
res2 %= MOD2;
//res3 -= (long long)1 * (s[i] - 'a') * ten[2][len-1];
//res3 = (res3 % MOD3 + MOD3) * 10 % MOD3;
//res3 += s[i] - 'a';
//res3 %= MOD3;
hash_1.push_back(res1);
hash_2.push_back(res2);
//hash_3.push_back(res3);
}
for(int pos = 1;pos < n / len;pos++){
bool flag = false;
res1 = 0;res2 = 0;
//res3 = 0;
int st = pos * len;
for(int i = 1;i<=len;i++){
res1 = res1 * 10 + s[st + i] - 'a';
res1 %= MOD1;
res2 = res2 * 10 + s[st + i] - 'a';
res2 %= MOD2;
//res3 = res3 * 10 + s[st + i] - 'a';
//res3 %= MOD3;
}
for(int i = 0;i < hash_1.size();i++){
if(hash_1[i]==res1&&hash_2[i]==res2
//&&hash_3[i]==res3
)
{
flag = true;
break;
}
}
if(!flag) return false;
}
return true;
}
int main()
{
pre(5e6);
int T;
scanf("%d",&T);
while(T--){
scanf("%d",&n);
getchar();
kind = 0;
for(int i = 0;i < 26;i++) su[i] = 0;
for(int i = 1;i <= n;i++){
scanf("%c",&s[i]);
su[s[i]-'a']++;
}
int gg;
int first;
int mi = INF;
for(int i = 0;i < 26;i++)
if(su[i]) {
kind++;
mi = min(mi,su[i]);
}
for(int i = 0;i < 26;i++){
if(su[i]) {
gg = su[i];
first = i;
break;
}
}
for(int i = first + 1;i < 26;i++){
if(su[i]) gg = gcd(gg,su[i]);
}
bool flag = false;
for(int k = 2;k <= gg;k ++){
if(n%k) continue;
int len = n / k;
if(check(len)) {
flag = true;
break;
}
}
if(flag) cout<<"Yes"<<endl;
else cout<<"No"<<endl;
}
return 0;
}
