CodeForces Round #565 Div.3

A. Divide it!

#include <bits/stdc++.h>
using namespace std;
 
int N;
map<long long, int> mp;
 
void init() {
    mp.clear();
    long long u[62];
    u[1] = 2;
    mp[2] = 1;
    
    for(int i = 2; i <= 62; i ++) {
        u[i] = 2 * u[i - 1];
        mp[u[i]] = i;
    }
    
}
 
int main() {
    scanf("%d", &N);
    init();
    while(N --) {
        long long x;
        scanf("%lld", &x);
        long long ans = 0;
        if(x == 1) ans = 0;
        else {
                while(x % 3 == 0) {
                    x /= 3;
                    x *= 2;
                    ans ++;
                }
                        
                while(x % 5 == 0) {
                    x /= 5;
                    x *= 4;
                    ans ++;
                }
            
            if(!mp[x]) ans = -1;                
            else ans += mp[x];
                        
        }
         
        printf("%lld\n", ans);
        
    }
    return 0;
}

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B. Merge it!

#include <bits/stdc++.h>
using namespace std;
 
const int maxn = 110;
int T;
int a[maxn];
 
int main() {
    scanf("%d", &T);
    while(T --) {
        int N;
        scanf("%d", &N);
        int one = 0, two = 0, ans = 0;
        for(int i = 1; i <= N; i ++) {
            scanf("%d", &a[i]);
            if(a[i] % 3 == 0) ans ++;
            if(a[i] % 3 == 1) one ++;
            if(a[i] % 3 == 2) two ++;
        }
                
        if(one == two) ans += one;
        else if(one > two) {
            one -= two;
            ans += two;
            ans += (one / 3);
        } else {
            two -= one;
            ans += one;
            ans += (two / 3);
        }
        
        printf("%d\n", ans);
        
    }
    return 0;
}

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C. Lose it!

#include <bits/stdc++.h>
using namespace std;
 
const int maxn = 5e5 + 10;
int N;
int a[maxn], pos[maxn];
int num[7] = {0, 4, 8, 15, 16, 23, 42};
 
int main() {
    
    memset(pos, 0, sizeof(pos));
    
    scanf("%d", &N);
    for(int i = 1; i <= N; i ++) {
        scanf("%d", &a[i]);
        
        int t = lower_bound(num + 1, num + 1 + 6, a[i]) - num;
        
        if(t == 1) ++ pos[1];
        else {
            if(pos[t - 1] > 0) {
                -- pos[t - 1];
                ++ pos[t];
            }
        }
        
    }
    
    printf("%d\n", N - 6 * pos[6]);
    
    return 0;
}

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E. Cover it!

#include <bits/stdc++.h>
using namespace std;
 
const int maxn = 2e5 + 10;
int T;
int N, M;
int vis[maxn];
vector<int> ans[2];
vector<int> v[maxn];
 
void dfs(int st, int cnt, int lev) {
    vis[st] = 1;
    ans[lev].push_back(st);
    for(int i = 0; i < v[st].size(); i ++) {
        if(vis[v[st][i]]) continue;
        vis[v[st][i]] = 1;
        dfs(v[st][i], cnt + 1, !lev);
    }
}
 
int main() {
    scanf("%d", &T);
    while(T --) {
        scanf("%d%d", &N, &M);
 
        for(int i = 1; i <= N; i ++)
            v[i].clear(), vis[i] = 0;
 
        ans[0].clear(), ans[1].clear();
 
        for(int i = 0; i < M; i ++) {
            int st, en;
            scanf("%d%d", &st, &en);
            v[st].push_back(en);
            v[en].push_back(st);
        }
 
        dfs(1, 1, 0);
 
        int temp;
        printf("%d\n", min(ans[0].size(), ans[1].size()));
        ans[0].size() < ans[1].size() ? temp = 0 : temp = 1;
 
        for(int i = 0; i < ans[temp].size(); i ++)
            printf("%d%s", ans[temp][i], i != ans[temp].size() - 1 ? " " : "\n");
    }
    return 0;
}

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E 每次初始化的时候 vis 不能 memset 会 TLE 在 15 组样例（两次！！）！！！

好久好久才更了一组 最近可能是有点忙？还是不能丢下当初辛辛苦苦学了那么久的东西 熬过去每一段觉得筋疲力尽的日子就会好起来了 

也许我知道现在还不是最糟糕 那就等 down 到谷底再触底反弹的瞬间 8！