The Preliminary Contest for ICPC Asia Xuzhou 2019 部分题解

B. so easy
题目链接

#include 
using namespace std;
typedef long long LL;
unordered_map mp;
int x,n,q,op;
int Find(int x){
    return mp.count(x) ? mp[x] = Find(mp[x]) : x;
}
int main(){
    scanf("%d %d",&n,&q);
    while(q--){
        scanf("%d %d",&op,&x);
        if(op == 1) mp[x] = x + 1;
        else printf("%d\n", Find(x));
    }
    return 0;
}

C. Buy Watermelon
题目链接

#include
#include
#include
#include
#define LL long long
#define uLL unsigned long long 
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fix fixed << setprecision(2)
int max(int x, int y) {if(x > y) return x; return y;}
int min(int x, int y) {if(x > y) return y; return x;}
using namespace std;
int mod = 1e9 + 7;
const int INF = 1e9 + 7;
const double PI = 3.1415927;
int main() {
  int n;
  cin >> n;
  if(n % 2 == 0 && n != 2) cout << "YES" << endl;
  else cout << "NO" << endl;
  return 0;
}

D. Carneginon
题目链接

#include
#include
#include
#include
#define LL long long
#define uLL unsigned long long 
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fix fixed << setprecision(2)
int max(int x, int y) {if(x > y) return x; return y;}
int min(int x, int y) {if(x > y) return y; return x;}
using namespace std;
int mod = 1e9 + 7;
const int INF = 1e9 + 7;
const double PI = 3.1415927;
char st[100005];
char dt[100005];
int Next[100005];
void getnext(char b[], int len_b) {//求的是模式串的next 
  int i, j = 0;
  for(i = 1; i < len_b; ++i) {
    while(j && b[i] != b[j]) j = Next[j];
    if(b[i] == b[j]) j += 1;
    Next[i + 1] = j;
  }
}
int kmp(char b[], int len_b, char a[], int len_a) {
  int i, j = 0;
  for(i = 0; i < len_a; ++i) {
    while(j && a[i] != b[j]) j = Next[j];
    if(a[i] == b[j]) j += 1;
    if(j == len_b) {
      return 1;
//      j = next[j];
    }
  }
  return 0;
}
int main() {
  scanf("%s", st);
  int q, len = strlen(st);
  scanf("%d", &q);
  while(q--) {
    scanf("%s", dt);
    int l = strlen(dt);
    memset(Next, 0, sizeof(Next));
    if(l == len) {
      if(strcmp(dt, st) == 0) printf("jntm!\n");
      else printf("friend!\n");
    } else if(len > l) {
      getnext(dt, l);
      if(kmp(dt, l, st, len)) printf("my child!\n");
      else printf("oh, child!\n");
    } else {
      getnext(st, len);
      if(kmp(st, len, dt, l)) printf("my teacher!\n");
      else printf("senior!\n");
    }
  }
  return 0;
}

E. XKC’s basketball team
题目链接

#include
#include
#include
#include
#define LL long long
#define uLL unsigned long long 
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fix fixed << setprecision(2)
int max(int x, int y) {if(x > y) return x; return y;}
int min(int x, int y) {if(x > y) return y; return x;}
using namespace std;
int mod = 1e9 + 7;
const int INF = 1e9 + 7;
const double PI = 3.1415927;
struct node {
  int l, r, m;
};
node f[500005 * 4];
int a[500005];
int b[500005];
int n, m;
void Build(int i, int l, int r) {
  f[i].l = l, f[i].r = r;
  if(l == r) {
    f[i].m = a[l];
    return ;
  }
  int mid = l + r >> 1;
  Build(i * 2, l, mid);
  Build(i * 2 + 1, mid + 1, r);
  f[i].m = max(f[i * 2].m, f[i * 2 + 1].m);
}
int Query(int i, int l, int r, int k) {
  if(f[i].l == f[i].r) return f[i].l;
  int mid = f[i].l + f[i].r >> 1;
  if(f[i * 2 + 1].m >= k) {
    return Query(i * 2 + 1, l, r, k);
  } else if(l <= mid && f[i * 2].m >= k) {
    return Query(i * 2, l, r, k);
  } else 
    return -1;
}
int main() {
  scanf("%d %d", &n, &m);
  for(int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
  }
  Build(1, 1, n);
  int cnt;
  for(int i = 1; i <= n; ++i) {
    if(i == n) cnt = -1;
    else cnt = Query(1, i + 1, n, a[i] + m);
    if(cnt != -1)
      cnt -= i + 1;
    printf("%d%c", cnt, i == n ? '\n' : ' ');
  }
  return 0;
}