Codeforces Round #568 (Div. 2) AB C1 C2 题解

A. Ropewalkers

题意：给三个数，每次可以对一个数+1或-1，问最少多少次可以使得三个数两两之间距离>=d。

思路：水题，存进来的排个序，abc依次表示从小到大的。只要考虑b往两边距离满足题意即可。

view code

#include<iostream>

#include<string>

#include<algorithm>

#include<cstdio>

#include<cstring>

#include<cmath>

#include<map>

#include <queue>

#include<sstream>

#include <stack>

#include <set>

#include <bitset>

#include<vector>

#define FAST ios::sync_with_stdio(false)

#define abs(a) ((a)>=0?(a):-(a))

#define sz(x) ((int)(x).size())

#define all(x) (x).begin(),(x).end()

#define mem(a,b) memset(a,b,sizeof(a))

#define max(a,b) ((a)>(b)?(a):(b))

#define min(a,b) ((a)<(b)?(a):(b))

#define rep(i,a,n) for(int i=a;i<=n;++i)

#define per(i,n,a) for(int i=n;i>=a;--i)

#define pb push_back

#define mp make_pair

#define fi first

#define se second

using namespace std;

typedef long long ll;

typedef pair<ll,ll> PII;

const int maxn = 1e5+200;

const int inf=0x3f3f3f3f;

const double eps = 1e-7;

const double pi=acos(-1.0);

const int mod = 1e9+7;

inline int lowbit(int x){return x&(-x);}

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}

void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){d=a,x=1,y=0;}else{ex_gcd(b,a%b,d,y,x);y-=x*(a/b);}}//x=(x%(b/d)+(b/d))%(b/d);

inline ll qpow(ll a,ll b,ll MOD=mod){ll res=1;a%=MOD;while(b>0){if(b&1)res=res*a%MOD;a=a*a%MOD;b>>=1;}return res;}

inline ll inv(ll x,ll p){return qpow(x,p-2,p);}

inline ll Jos(ll n,ll k,ll s=1){ll res=0;rep(i,1,n+1) res=(res+k)%i;return (res+s)%n;}

inline ll read(){ ll f = 1; ll x = 0;char ch = getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1; ch = getchar();}while(ch>='0'&&ch<='9') x = (x<<3) + (x<<1) + ch - '0',  ch = getchar();return x*f; }

int dir[4][2] = { {1,0}, {-1,0},{0,1},{0,-1} };

ll arr[5];

int main()

{

    ll a, b , c ,d;

    cin>>arr[1]>>arr[2]>>arr[3]>>d;

    sort(arr+1,arr+1+3);

    a = arr[1] , b = arr[2], c = arr[3];

    ll d1 = 0;

    if(b - a < d) d1 += d - (b - a);

    if(c - b < d) d1 += d - (c - b);

    cout<<d1<<endl;

    return 0;

}

B. Email from Polycarp

题意：s串为原串，判断t串是否是在s的基础上在使某若干位置的字符重复出现得到的。

思路：双指针简单模拟一下。p指向原串，q指向目标串。如果s[p] == t[q]，就继续往下匹配。若不一样，只有两种可能：1.t串在重复上一个字符。 2. 完全失配，直接退出。如果p走完了s串且t剩下来的全是重复的，就输出"YES"。否则都为"NO" 。

view code

#include<iostream>

#include<string>

#include<algorithm>

#include<cstdio>

#include<cstring>

#include<cmath>

#include<map>

#include <queue>

#include<sstream>

#include <stack>

#include <set>

#include <bitset>

#include<vector>

#define FAST ios::sync_with_stdio(false)

#define abs(a) ((a)>=0?(a):-(a))

#define sz(x) ((int)(x).size())

#define all(x) (x).begin(),(x).end()

#define mem(a,b) memset(a,b,sizeof(a))

#define max(a,b) ((a)>(b)?(a):(b))

#define min(a,b) ((a)<(b)?(a):(b))

#define rep(i,a,n) for(int i=a;i<=n;++i)

#define per(i,n,a) for(int i=n;i>=a;--i)

#define pb push_back

#define mp make_pair

#define fi first

#define se second

using namespace std;

typedef long long ll;

typedef pair<ll,ll> PII;

const int maxn = 1e6+200;

const int inf=0x3f3f3f3f;

const double eps = 1e-7;

const double pi=acos(-1.0);

const int mod = 1e9+7;

inline int lowbit(int x){return x&(-x);}

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}

void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){d=a,x=1,y=0;}else{ex_gcd(b,a%b,d,y,x);y-=x*(a/b);}}//x=(x%(b/d)+(b/d))%(b/d);

inline ll qpow(ll a,ll b,ll MOD=mod){ll res=1;a%=MOD;while(b>0){if(b&1)res=res*a%MOD;a=a*a%MOD;b>>=1;}return res;}

inline ll inv(ll x,ll p){return qpow(x,p-2,p);}

inline ll Jos(ll n,ll k,ll s=1){ll res=0;rep(i,1,n+1) res=(res+k)%i;return (res+s)%n;}

inline ll read(){ ll f = 1; ll x = 0;char ch = getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1; ch = getchar();}while(ch>='0'&&ch<='9') x = (x<<3) + (x<<1) + ch - '0',  ch = getchar();return x*f; }

int dir[4][2] = { {1,0}, {-1,0},{0,1},{0,-1} };

ll Map[500];

int main()

{

    int kase;

    cin>>kase;

    while(kase--)

    {

        mem(Map,0);

        string s;  cin>>s;

        string t;  cin>>t;

        for(int i=0; i<s.size();i++) Map[s[i]] ++;

        int p = 0, q = 0;

        while(p<s.size()&&q<t.size())

        {

            if(Map[t[q]]==0) break;

            if(t[q] == s[p]) p++, q++;

            else

            {

                if(q>0&&t[q] == t[q-1])

                {

                    while(q<t.size()&&t[q]!=s[p]&&t[q]==t[q-1]) q++;

                }

                else break;

            }

        }

        if(p<s.size())

        {

                cout<<"NO"<<'\n';

        }

        else

        {

            int flag = 1;

            if(q<t.size())

            for(int i=q; i<t.size();i++) if(t[i]!=t[i-1]) flag = 0;

            if(!flag)

            cout<<"NO"<<'\n';

            else

            cout<<"YES"<<'\n';

        }

    }

    return 0;

}

C1. Exam in BerSU (easy version)

题意：给出n个数，每个数会消耗t[i]的时间，问在m时限的情况下i前面要最少去掉多少人才能保证t[i]放得下

思路：贪心一下，要走的人最少的话肯定是让前面的用时大的人走最好。所以读入一个数就将前i-1个数排序一下。从大往小选即可。

view code

#include<iostream>

#include<string>

#include<algorithm>

#include<cstdio>

#include<cstring>

#include<cmath>

#include<map>

#include <queue>

#include<sstream>

#include <stack>

#include <set>

#include <bitset>

#include<vector>

#define FAST ios::sync_with_stdio(false)

#define abs(a) ((a)>=0?(a):-(a))

#define sz(x) ((int)(x).size())

#define all(x) (x).begin(),(x).end()

#define mem(a,b) memset(a,b,sizeof(a))

#define max(a,b) ((a)>(b)?(a):(b))

#define min(a,b) ((a)<(b)?(a):(b))

#define rep(i,a,n) for(int i=a;i<=n;++i)

#define per(i,n,a) for(int i=n;i>=a;--i)

#define pb push_back

#define mp make_pair

#define fi first

#define se second

using namespace std;

typedef long long ll;

typedef pair<ll,ll> PII;

const int maxn = 1e5+200;

const int inf=0x3f3f3f3f;

const double eps = 1e-7;

const double pi=acos(-1.0);

const int mod = 1e9+7;

inline int lowbit(int x){return x&(-x);}

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}

void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){d=a,x=1,y=0;}else{ex_gcd(b,a%b,d,y,x);y-=x*(a/b);}}//x=(x%(b/d)+(b/d))%(b/d);

inline ll qpow(ll a,ll b,ll MOD=mod){ll res=1;a%=MOD;while(b>0){if(b&1)res=res*a%MOD;a=a*a%MOD;b>>=1;}return res;}

inline ll inv(ll x,ll p){return qpow(x,p-2,p);}

inline ll Jos(ll n,ll k,ll s=1){ll res=0;rep(i,1,n+1) res=(res+k)%i;return (res+s)%n;}

inline ll read(){ ll f = 1; ll x = 0;char ch = getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1; ch = getchar();}while(ch>='0'&&ch<='9') x = (x<<3) + (x<<1) + ch - '0',  ch = getchar();return x*f; }

int dir[4][2] = { {1,0}, {-1,0},{0,1},{0,-1} };

ll t[maxn];

int main()

{

    ll n = read(), m = read();

    ll cur = 0;

    rep(i,1,n)

    {

        ll x = read();

        sort(t+1,t+i,greater<ll>());

        ll p = 1; ll tmp = m - cur;

        while(tmp < x) tmp += t[p], p++;

        cout<<p-1<<' ';

        t[i] = x;

        cur += x;

    }

    cout<<'\n';

    return 0;

}

C2. Exam in BerSU (hard version)

题意：如上题。

思路：这题数据量加大很多，不能想C1那样莽了。仔细看数据发现t[i]<=100，这是个破题口。也就是说我们完全可以把上一题的思路顺延下来，排序的时候只需要在1~100出现过的数字进行排序就行了。所以我们用一个Map记录一下当前数有没有出现过，没出现过就加入all数组，Map同时计数这个数出现了多少次。然后拿到一个数就对前面的all数组排序（长度不会超过100），从大到小挑就行了。而且all[j]有多少个我们也是知道的，可以在O（1）时间内知道all[j]要挑多少。详见代码。

view code

#include<iostream>

#include<string>

#include<algorithm>

#include<cstdio>

#include<cstring>

#include<cmath>

#include<map>

#include <queue>

#include<sstream>

#include <stack>

#include <set>

#include <bitset>

#include<vector>

#define FAST ios::sync_with_stdio(false)

#define abs(a) ((a)>=0?(a):-(a))

#define sz(x) ((int)(x).size())

#define all(x) (x).begin(),(x).end()

#define mem(a,b) memset(a,b,sizeof(a))

#define max(a,b) ((a)>(b)?(a):(b))

#define min(a,b) ((a)<(b)?(a):(b))

#define rep(i,a,n) for(int i=a;i<=n;++i)

#define per(i,n,a) for(int i=n;i>=a;--i)

#define pb push_back

#define mp make_pair

#define fi first

#define se second

using namespace std;

typedef long long ll;

typedef pair<ll,ll> PII;

const int maxn = 2e5+200;

const int inf=0x3f3f3f3f;

const double eps = 1e-7;

const double pi=acos(-1.0);

const int mod = 1e9+7;

inline int lowbit(int x){return x&(-x);}

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}

void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){d=a,x=1,y=0;}else{ex_gcd(b,a%b,d,y,x);y-=x*(a/b);}}//x=(x%(b/d)+(b/d))%(b/d);

inline ll qpow(ll a,ll b,ll MOD=mod){ll res=1;a%=MOD;while(b>0){if(b&1)res=res*a%MOD;a=a*a%MOD;b>>=1;}return res;}

inline ll inv(ll x,ll p){return qpow(x,p-2,p);}

inline ll Jos(ll n,ll k,ll s=1){ll res=0;rep(i,1,n+1) res=(res+k)%i;return (res+s)%n;}

inline ll read(){ ll f = 1; ll x = 0;char ch = getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1; ch = getchar();}while(ch>='0'&&ch<='9') x = (x<<3) + (x<<1) + ch - '0',  ch = getchar();return x*f; }

int dir[4][2] = { {1,0}, {-1,0},{0,1},{0,-1} };

ll t[maxn];

vector<ll> all;

map<ll,ll> Map;

int main()

{

    ll n = read(), m = read();

    ll cur = 0;

    rep(i,1,n)

    {

        ll x = read();

        sort(all.begin(), all.end(), greater<ll>());

        ll tmp = m - cur;

        ll need = 0;

        for(int j=0; j< all.size(); j++)

        {

            if(tmp >= x) break;

            ll num = min(Map[all[j]] , ( x - tmp+all[j]-1)/all[j] );

            need += num;

            tmp += num * all[j];

        }

        cout<<need<<' ';

        if(!Map[x]) all.pb(x);

        Map[x]++;

        cur += x;

    }

    cout<<'\n';

    return 0;

}