Round G 2019 - Kick Start 2019

Posted 2021-03-25 163467wyj

https://codingcompetitions.withgoogle.com/kickstart/round/0000000000050e02/000000000018fd0d

Book Reading (10pts, 15pts)

1 ≤ T ≤ 100.
1 ≤ P₁ < P₂ < ... < P_M ≤ N.
1 ≤ R_i ≤ N, for all i.

Test set 1 (Visible)

1 ≤ M ≤ N ≤ 1000.
1 ≤ Q ≤ 1000.

Test set 2 (Hidden)

1 ≤ M ≤ N ≤ 10⁵.
1 ≤ Q ≤ 10⁵.

题意：一本书有 1……n页，撕去 p 页，有 q 个读者， 问读者一共能阅读的页数

如果直接枚举暴力的话， 10ps 到手

#include<bits/stdc++.h>

using namespace std;
#define _for(i,a,b) for(int i = (a); i < (b); i++)
#define _rep(i,a,b) for(int i = (a); i <= (b); i++)
const int N = 1e5+100;
int a[N],b[N],vis[N];

int main(){
    ios::sync_with_stdio(0); cin.tie(0);
    int t,n,m,q,kase = 1; cin >>t ;
    while(t--) {
        cin >> n >> m >> q; int x,cnt = 0;
        memset(vis, 0, sizeof(vis));
        _for(i,0,m) cin >> x, vis[x] = 3;
        _for(i,0,q) {
            cin >> x;
            for(int j = x; j <= n; j += x)
                if(vis[j] != 3) cnt++;
        }
        
        cout << "Case #"<< kase++ << ": ";
        cout << cnt << endl;
    }
    return 0;
}

View Code

但是很明显， 会 TLE， 所以需要优化下

#include<bits/stdc++.h>

using namespace std;
#define _for(i,a,b) for(int i = (a); i < (b); i++)
#define _rep(i,a,b) for(int i = (a); i <= (b); i++)
#define ll long long
const int N = 1e5+100;

int main(){
    ios::sync_with_stdio(0); cin.tie(0);
    int t,n,m,q,kase = 1; cin >>t ;
    while(t--) {
        vector<int> a(N, 0);
        vector<bool> vis(N, false);
        cin >> n >> m >> q; int x;
        ll cnt = 0;
        _for(i,0,m) cin >> x, vis[x] = true;
        _for(i,0,q) {
            cin >> x;
            if(a[x]){
                cnt += a[x];
                continue;
            }
            
            for(int j = x; j <= n; j += x)
                if(!vis[j]) a[x]++;
            cnt += a[x];
        }
        
        cout << "Case #"<< kase++ << ": ";
        cout << cnt << endl;
    }
    return 0;
}

 

https://codingcompetitions.withgoogle.com/kickstart/round/0000000000050e02/000000000018fe36
The Equation (12pts, 20pts)

直接暴力 12pts可到手

#include<bits/stdc++.h>
  
using namespace std;
  
const double eps = 1e-10;
const double pi = 3.1415926535897932384626433832795;
const double eln = 2.718281828459045235360287471352;

#define scld(x) scanf("%lld", &x)
#define prcas printf("Case #%d: ", cas)
#define pncas printf("Case #%d:
", cas)
#define lowbit(x) ((x) & (-(x)))
#define fi first
#define se second
typedef long long LL;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<LL, LL> pll;
typedef vector<int> vi;

const int maxn = 25;
int dp[1048550];
LL a[maxn], b[maxn];
int n, T, h;

int main()
{   
    scanf("%d", &T);
    for(int cas = 1; cas <= T; ++ cas)
    {
        scanf("%d %d", &n, &h);
        for(int i = 0; i < n; ++ i) scld(a[i]);
        //for(int i = 0; i < n; ++ i) scld(b[i]);
        prcas; int flag = 0;
        for(int i = 128; i >= 0; i--){
            int sum = 0;
            for(int j = 0; j < n; j++)
                sum += a[j]^i;
            if(sum <= h) {
                flag = 1; printf("%d
", i);break;
            }
        }
        if(!flag) printf("-1
");
    } 
    return 0;
}

通过对 k 的二进制位数进行枚举， cnt[i][0]   cnt[i][1]分别记录 数组a的 第i+1位的 0 与 1的个数

再通过 ret[i][0]  ret[i][1] 分别表示对前 i 位的a数组的 0 与 1 表示的和

然后 通过 suf[i] 记录前 i 项 min(ret[i][0], ret[i][1])的和， （？？？）
然后枚举 位数 i 51~0 如果符合 ret[i][1]+suf[i-1]  k |= 1<<i , 即 k 的这个二进制位为 1， 否则为 0  m-= ( 为1 ? ret[i][1] : ret[i][0])
若枚举结束  m < 0， 则不存在 k

#include<bits/stdc++.h>
  
using namespace std;
  
const double eps = 1e-10;
const double pi = 3.1415926535897932384626433832795;

#ifdef __LOCAL_DEBUG__
# define _debug(fmt, ...) fprintf(stderr, "[%s] " fmt "
",     __func__, ##__VA_ARGS__)
#else
# define _debug(...) ((void) 0)
#endif

#define IN freopen("B.in", "r", stdin)
#define OUT freopen("B.out", "w", stdout)
#define scd(x) scanf("%d", &x)
#define scld(x) scanf("%lld", &x)
#define scs(x) scanf("%s", x)
#define mp make_pair
#define pb push_back
#define sqr(x) (x) * (x)
#define prcas printf("Case #%d: ", cas)
#define pncas printf("Case #%d:
", cas)
#define lowbit(x) ((x) & (-(x)))
#define fi first
#define se second
typedef long long LL;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<LL, LL> pll;
typedef vector<int> vi;

const int maxk = 51;
int cnt[65][2];
int n, T;
LL m, ans, ret[65][2], suf[65];

int main()
{    
    scd(T);
    for(int cas = 1; cas <= T; ++ cas)
    {
        for(int i = 0; i <= maxk; ++ i) cnt[i][0] = cnt[i][1] = 0;
        scanf("%d%lld", &n, &m);
        while(n --)
        {
            LL x; scanf("%lld", &x); ans = 0;
            for(int i = 0; i <= maxk; ++ i, x >>= 1) cnt[i][x & 1] ++;
        }
        for(int i = 0; i <= maxk; ++ i) 
            ret[i][0] = 1ll * cnt[i][1] * (1ll << i), ret[i][1] = 1ll * cnt[i][0] * (1ll << i);
        suf[0] = min(ret[0][0], ret[0][1]); 
        for(int i = 1; i <= maxk; ++ i) 
            suf[i] = suf[i - 1] + min(ret[i][0], ret[i][1]);
        for(int i = maxk; i >= 0 && m >= 0; -- i)
        {
            LL t = 0; if(i > 0) t = suf[i - 1];
            if(ret[i][1] + t <= m)
            {
                ans |= 1ll << i;
                m -= ret[i][1];
            }else m -= ret[i][0];
        }
        if(m < 0) ans = -1;
        prcas; printf("%lld
", ans);
    }
    return 0;
}
/*
Input 
4
3 27
8 2 4
4 45
30 0 4 11
1 0
100
6 2
5 5 1 5 1 0

  
Case #1: 12
Case #2: 14
Case #3: 100
Case #4: -1
*/

View Code

 

 

https://codingcompetitions.withgoogle.com/kickstart/round/0000000000050e02/000000000018fd5e
Shifts (20pts, 23pts)

题意：有n个班次， h 快乐值要求， 问有多少种情况 a 与 b 都快乐（快乐值大于等于h）

如果直接暴力的话， 0~ 3^n-1,　暴力枚举 每个班次的三种情况（单选a， 单选b， 同时选ab），n<=12，5e5的样子，能过 Test set 1， 但Test set 2 3e8（n<=20）就会TLE

 TLE

  #include<bits/stdc++.h>
  using namespace std;
  #define rep(x,y) for(long long i =x;i<=y;i++)
  #define pb push_back
  #define mp make_pair
  #define ALL(x) (x).begin(),(x).end()
  #define FastIO  ios_base::sync_with_stdio(false); cin.tie(NULL);
  typedef long long ll;
  typedef pair< ll, ll> pll;
  typedef vector< ll > vll;
  typedef map < ll, ll > mll;

int main()
{
  ll i,j,t,test,ans,n,num,s1,s2,a[30],b[30],h,ways;
  test=1;
  cin >> t;
  while(t--)
  {
    cout << "Case #" << test << ": " ;
    ans=0;
    test++;
    cin >> n >> h;
    rep(1,n)
      cin >> a[i];
    rep(1,n)
      cin >> b[i];
    ways= pow(3,n)-1;
    rep(0,ways)
    {
      num=i;
      j=1;
      s1=0;
      s2=0;
      while(j<=n)
      {
        if(num%3==0)
        {
          s1+=a[j];
        }
        else if(num%3==1)
          s2+=b[j];
        else
        {
          s1+=a[j];
          s2+=b[j];
        }
        num/=3;
        j++;
      }
    //  cout << s1 << " " << s2 << endl;
      if(s1>=h && s2>=h)  ans++;
    }
    cout << ans << endl;
  }
  return 0;
}

View Code

dp[mask]  mask 0~2^n-1, 以0 1 分 n个a是否被取
then  mask 0~2^n-1  i 0~n  dp[mask^(1<<i)] += dp[mask]

then  x 0~2^n-1  dp[x]  2^n-x-1(二进制01 代表n个b是否被取)    神奇的穷举操作
最后会发现 所有的可能 3^n 都被枚举了

比如说 n = 2, 
        　　　　a           b

dp[0] = 00 01 10 11    11

dp[1] = 01 11 　　　  10

dp[2] = 10 11 　　　  01

dp[3] = 11 　　　　   00

#include<bits/stdc++.h>
  
using namespace std;
  
const double eps = 1e-10;
const double pi = 3.1415926535897932384626433832795;
const double eln = 2.718281828459045235360287471352;

#define scld(x) scanf("%lld", &x)
#define prcas printf("Case #%d: ", cas)
#define pncas printf("Case #%d:
", cas)
#define lowbit(x) ((x) & (-(x)))
#define fi first
#define se second
typedef long long LL;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<LL, LL> pll;
typedef vector<int> vi;

const int maxn = 25;
int dp[1048550];
LL a[maxn], b[maxn];
int n, T, h;

int main()
{   
    scanf("%d", &T);
    for(int cas = 1; cas <= T; ++ cas)
    {
        scanf("%d %d", &n, &h);
        for(int i = 0; i < n; ++ i) scld(a[i]);
        for(int i = 0; i < n; ++ i) scld(b[i]);
        for(int mask = 0; mask < (1 << n); ++ mask)
        {
            LL sum = 0;
            for(int i = 0; i < n; ++ i) 
                if((mask >> i) & 1) sum += a[i];
            if(sum >= h) dp[mask] = 1; else dp[mask] = 0;
        }
//for(int mask = 0; mask < (1 << n); ++ mask) printf("dp[%d]= %d  ", mask, dp[mask]);printf("
");
        for(int i = 0; i < n; ++ i)
            for(int mask = 0; mask < (1 << n); ++ mask)
                if((mask >> i) & 1) dp[mask ^ (1 << i)] += dp[mask];
        LL ans = 0;
//for(int mask = 0; mask < (1 << n); ++ mask) printf("dp[%d]= %d  ", mask, dp[mask]);printf(" 	...
");
        for(int mask = 0; mask < (1 << n); ++ mask)
        {
            LL sum = 0, task = 0;
            for(int i = 0; i < n; ++ i)
                if((mask >> i) & 1) sum += b[i];
                else task |= (1 << i);
            if(sum < h) continue;
            ans += dp[task];
        }
        prcas; printf("%lld
", ans);
    } 
    return 0;
}