# 2018 National multi-school algorithm winter training Camp Practice Competition (fifth)

Source: Internet
Author: User

A-Reverse order number

Classic problem, there are many methods, such as tree array, line segment tree, merge sort, etc. The code is not posted.

B-big Water problem

Single-point modification for interval and tree-like array or line tree can be.

`#include <bits/stdc++.h>using namespace std; const int MAXN = 1e5 + 10;long long C[MAXN]; int lowbit (int x) {  return x & (-X);} long Long (int p) {  long long res = 0;  while (p) {    res + = c[p];    P-= Lowbit (P);  }  return res;} void update (int x, long long y) {while  (x < MAXN) {    c[x] + = y;    x + = Lowbit (x);  }} int main () {     int n, m;  scanf ("%d%d", &n, &m);  for (int i = 1; I <= n; i + +) {    long long x;    scanf ("%lld", &x);    Update (i, x);  }  while (M-) {    int op;    scanf ("%d", &op);    if (op = = 1) {      int x;      Long long y;      scanf ("%d%lld", &x, &y);      Update (x, y);    } else {      int L, R;      scanf ("%d%d", &l, &r);      printf ("%lld\n", SUM (R)-sum (L-1));    }  }  return 0;}`

C-Problems with strings

The violence in the game from the big to the small enumeration prefix length, KMP verify the AC? Also think about the complexity of the right.

`#include <iostream> #include <cstring>using namespace std; const int N = 1000010;int Nx[n];char s[n], t[n];int Slen, Tlen;  void GetNext () {int J, K; j = 0; K =-1;  Nx[0] =-1; while (J < Tlen) if (k = =-1 | |    T[J] = = T[k]) nx[++j] = ++k;   else k = Nx[k];  }/* Returns the number of times the pattern string appears in the main string s */int kmp_count () {int ans = 0;     int I, j = 0;    if (Slen = = 1 && tlen = = 1) {if (s[0] = = T[0]) return 1;  else return 0;  } getNext ();    for (i = 0; i < Slen; i++) {while (J > 0 && s[i]! = t[j]) j = nx[j];    if (s[i] = = T[j]) j + +;      if (j = = Tlen) {ans++;    j = Nx[j]; }} return ans;  /*int Main () {int TT;  int i, CC;  cin>>tt;    while (tt--) {cin>>s>>t;    Slen = strlen (S);    Tlen = strlen (T);    The first occurrence of the cout<< "pattern string T in the main string s is:" <<kmp_index () <<endl;  cout<< "mode string T appears in the main string s in the number of times:" <<kmp_count () <<endl; } return 0;}  */int check (int x) {int p1 = 0; int P2 = SLEn-x;    while (P1 <= x-1) {if (S[P1]! = s[p2]) return 0;    P1 + +;  P2 + +;    } for (int i = 0; i < x; i + +) {T[i] = s[i];  T[i + 1] = 0;  } Tlen = x;  if (Kmp_count () >= 3) return 1; return 0;}  int main () {scanf ("%s", s);  Slen = strlen (S);  int ans = 0;  int L = 1, R = Slen;      for (int i = slen-1; I >= 1; I-) {if (check (i)) {ans = i;    Break  }} if (ans = = 0) printf ("Just a legend\n");    else {for (int i = 0; i < ans; i + +) {printf ("%c", S[i]);  } printf ("\ n"); }}`

D-Collection Issues

Write a bunch of violent simulation rules to get past, see and check the collection of AC, to learn.

`#include <bits/stdc++.h>using namespace std; const int MAXN = 110000;int n;int A, B;int a[maxn];map<int, int> belong;map<int, bool> flag;  int main () {int fail = 0;  scanf ("%d%d%d", &n, &a, &b);    for (int i = 1; I <= n; i + +) {scanf ("%d", &a[i]);  Flag[a[i]] = 1;  } sort (A + 1, a + 1 + N);        if (A = = B) {for (int i = 1; I <= n; i + +) {if (flag[b-a[i]] = = 0) {fail = 1;      Break    } Belong[a[i]] = 2;      }} else {while (1) {if (fail);      int ss = 0;        for (int i = 1; I <= n; i + +) {if (Belong[a[i]]) continue;          if (flag[a-a[i] [= 0 && flag[b-a[i]] = = 0) {fail = 1;        Break             } else if (flag[a-a[i]] = = 1 && (flag[b-a[i]] = = 0 | | belong[b-a[i]]) {if (belong[a-a[i]] = = 2) {            Fail = 1;          Break          } Belong[a[i]] = 1;          Belong[a-a[i]] = 1;        SS = 1; } else if ((flag[a-a[i]] = = 0 | |            Belong[a-a[i]]) && flag[b-a[i]] = = 1) {if (belong[b-a[i]] = = 1) {fail = 1;          Break          } Belong[a[i]] = 2;          Belong[b-a[i]] = 2;        SS = 1; } else {if (a-a[i]! = A[i] && b-a[i]! = A[i]) {if (belong[a-a[i]] = = 0 && belong            [B-a[i]] = = 0) {continue;              } else if (Belong[a-a[i]]) {Belong[a[i]] = belong[a-a[i]];            SS = 1;              } else {Belong[a[i]] = belong[b-a[i]];            SS = 1;              }} else if (a-a[i] = = A[i]) {if (belong[b-a[i] = = 1) {fail = 1;            Break            } Belong[a[i]] = 2;            Belong[b-a[i]] = 2;          SS = 1;              } else {if (belong[a-a[i]] = = 2) {fail = 1;            Break            } Belong[a[i]] = 1;     Belong[a-a[i]] = 1;       SS = 1;      }}} if (fail) break;    if (ss = = 0) break; }} if (fail = = 0) {for (int i = 1; I <= n; i + +) {if (belong[a[i] = = 1 && belong[a-a[i]]! = 1)      Fail = 1;    if (belong[a[i] [= 2 && belong[b-a[i]]! = 2) fail = 1;  }} if (fail) {printf ("no\n");    } else {printf ("yes\n");      for (int i = 1; I <= n; i + +) {printf ("%d", Belong[a[i]]-1);      if (i < n) printf ("");    else printf ("\ n"); }} return 0;}`

E-The light bulb of Valentine's Day

Single-point update, sub-matrix and two-dimensional tree array can be.

`#include <bits/stdc++.h>using namespace std; const int MAXN = 1010;int n, m; int C[maxn][maxn];int A[MAXN][MAXN]; int lowbit (int x) {return x & (-X);} void Add (int x, int y, int c) {for (int i = x; i <= n; i + = Lowbit (i)) {F    or (int j = y; J <= N; j + = Lowbit (j)) {C[i][j] + = C;  }}} int sum (int x,int y) {int ret = 0;    for (int i = x; i > 0; I-= Lowbit (i)) {for (int j = y;j > 0;j = Lowbit (j)) {ret + = C[i][j]; }} return ret;}  int main () {scanf ("%d%d", &n, &m);      for (int i = 1; I <= n; i + +) {for (int j = 1; J <= N; j + +) {scanf ("%d", &a[i][j]);    Add (I, J, A[i][j]);    }} while (M-) {int op;    scanf ("%d", &op);      if (op = = 1) {int x, y;      scanf ("%d%d", &x, &y);      if (a[x][y] = = 1) {Add (x, Y,-1);      } else {Add (x, y, 1);    } A[x][y] = (A[x][y] ^ 1);      } else {int x1, y1, x2, y2;      scanf ("%d%d%d%d", &x1, &y1, &x2, &y2); Printf ("%d\n", sum (x2, y2)-sum (x1-1, y2)-sum (x2, y1-1) + sum (x1-1, y1-1)); }} return 0;}`

F-the Biggest water problem

Just a mock-up.

`#include <bits/stdc++.h>using namespace Std;long long N;long long work (long long x) {  long long res = 0;  while (x) {    res + = x%;    x = X/10;  }  return res;} int main () {  scanf ("%d", &n);  while (1) {    if (n <= 9) {      printf ("%lld\n", n);      break;    }    n = Work (n);  }  return 0;}`

G-Send out-qaq

25,161 samples of the problem and HDU

H-tree Recovery

Interval Plus, interval summation, line tree can be.

`#include <bits/stdc++.h>using namespace std; const int MAXN = 1e5 + 10; int n, Q;long long sum[4 * Maxn];long long f[4 * MAXN];  void pushdown (int l, int r, int rt) {if (f[rt] = = 0) return;  int mid = (L + r)/2;  int len_left = (mid-l + 1);  int len_right = (r-mid);  F[2 * RT] + = F[rt];  f[2 * RT + 1] + = F[rt];  Sum[2 * RT] + = f[rt] * LEN_LEFT;  sum[2 * RT + 1] + = f[rt] * len_right; F[RT] = 0;} void pushup (int rt) {SUM[RT] =sum[2 * RT] + sum[2 * RT + 1];} void build (int l, int r, int rt) {if (L = = r) {scanf (    "%lld", &sum[rt]);  Return  } int mid = (L + r)/2;  Build (L, Mid, 2 * RT);  Build (mid + 1, r, 2 * RT + 1); Pushup (RT);}  void Update (int l, int R, Long long val, int l, int r, int rt) {if (L <= l && R <= R) {Sum[rt] + = val *    (r-l + 1);    F[RT] + = val;  Return  } pushdown (L, R, RT);  int mid = (L + r)/2;  if (l <= mid) Update (L, R, Val, L, Mid, 2 * RT);  if (R > Mid) Update (L, R, Val, mid + 1, r, 2 * RT + 1); Pushup (RT);} Long long query (int l, int r, int l, int r, int rt) {if (L <= l && R <= R) {return SUM[RT];  } pushdown (L, R, RT);  int mid = (L + r)/2;  Long long left = 0;  Long long right = 0;  if (l <= mid) left = query (L, R, L, Mid, 2 * RT);  if (R > Mid) Right = query (L, R, Mid + 1, r, 2 * RT + 1);  Pushup (RT); return left + right;}  int main () {scanf ("%d%d", &n, &q);  Build (1, N, 1);    while (Q-) {char op[10];    scanf ("%s", op);      if (op[0] = = ' Q ') {int L, R;      scanf ("%d%d", &l, &r);    printf ("%lld\n", Query (L, R, 1, N, 1));      } else {int L; int R;      A long long Val;      scanf ("%d%d%lld", &l, &r, &val);    Update (L, R, Val, 1, N, 1); }} return 0;}`

2018 National multi-school algorithm winter training Camp Practice Competition (fifth)

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