Codeforces Round #366 (Div. 2)

CF Avengers Federation Field ...

Water Problem A-hulk (Hulk)

Output Love hate ...

#include <bits/stdc++.h>typedef long long ll;const int n = 1e5 + 5;int main () {    int n;    scanf ("%d", &n);    printf ("I hate");    for (int i=1, i<n; ++i) {        if (I & 1) printf ("That's Love");        else printf ("that I hate");    }    Puts ("it");    return 0;}

Game + Play table to find the Law B-spider Man (Spiderman)

Notice that each time more than one lap, the previous or initial state, the groups are independent, so to play a table or analysis can find the law.

#include <bits/stdc++.h>typedef long long ll;const int N = 1e5 + 5;int a[n];int sg[105];int sg (int x) {    if (x &L T 2) return sg[x] = 0;    BOOL vis[105];    Memset (Vis, false, sizeof (VIS));    for (int i=1; i<x; ++i) {        vis[sg (i) ^ (SG (x-i))] = true;    }    int &ret = Sg[x] = 0;    while (Vis[ret]) ret++;    return ret;} void Test () {    memset (SG,-1, sizeof (SG));    for (int i=1; i<=20; ++i) {        printf ("sg[%d]=%d\n", I, SG (i));}    } int main () {    //test ();    int n;    scanf ("%d", &n);    for (int i=1; i<=n; ++i) scanf ("%d", a+i);    int ans = 0;    for (int i=1; i<=n; ++i) {        int res = a[i] & 1? 0:1;        Ans ^= Res;        printf ("%d\n", ans? 1:2);    }    return 0;}

Construction C-thor (Thor)

Test instructions: A mobile phone has n applications, there are three kinds of operations:

1. The first X application has an unread message;

2. Read all unread information for the current x application;

3. Read the first T message (information may be reread);

The number of bars for which the current unread information is output after each operation.

Idea: The 3rd operation "First T" is critical, so as long as the Max (T) operation. If the information to be read is read at a later point in time (the 2nd operation) then it is not updated, so maintain the latest "empty" app number and time. Before there was a place to write the continue, the answer did not output, WA for a long time.

#include <bits/stdc++.h>typedef long long ll;const int N = 3e5 + 5;int cnt[n];int clear_time[n];std::p air<int, in    t> Que[n];int Main () {int N, q;    scanf ("%d%d", &n, &q);    int m = 0, ans = 0;    int TP, X, T, maxt = 0;        for (int i=1; i<=q; ++i) {scanf ("%d", &AMP;TP);            if (tp = = 1 | | tp = = 2) {scanf ("%d", &x);                if (tp = = 1) {cnt[x]++;                ans++;            Que[++m] = {x, i};                } else {ans-= cnt[x];                Cnt[x] = 0;            CLEAR_TIME[X] = i;            }} else {scanf ("%d", &t); if (T > Maxt) {for (int j=maxt+1; j<=t; ++j) {int pos = Que[j].first, Tim = que[                    J].second;                    if (Clear_time[pos] >= Tim) continue;                    cnt[pos]--;                ans--;            } maxt = t; }} printf ("%d\n", ans); } return 0;}

Codeforces Round #366 (Div. 2)