Codeforces Global Round 1

1110D. Jongmah

大意: 给定$n$个数, (x,x,x)或者(x,x+1,x+2)可以组成三元组, 每个数最多用一次, 求最多组成多少三元组.

核心观察是以$x$开头的$(x,x+1,x+2)$最多出现两次, 然后暴力dp即可. 

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <cstring>
#include <bitset>
#include <functional>
#include <random>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<',';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head



#ifdef ONLINE_JUDGE
const int N = 1e6+50;
#else
const int N = 1e2+10;
#endif


int n,m,a[N],dp[N][3][3];
void chkmax(int &a, int b) {a<b?a=b:0;}
int main() {
    scanf("%d%d", &n, &m);
    REP(i,1,n) {
        int t;
        scanf("%d", &t);
        ++a[t];
    }
    memset(dp,0xef,sizeof dp);
    dp[0][0][0] = 0;
    //f[i][x][y] = (i,i+1,i+2)使用y次, (i-1,i,i+1)使用x次的答案
    REP(i,1,m) REP(x,0,2) REP(y,0,2) REP(z,0,2) {
        int &r = dp[i-1][x][y];
        if (r<0||x+y+z>a[i]) continue;
        //x个(i-2,i-1,i),y个(i-1,i,i+1),z个(i,i+1,i+2)
        chkmax(dp[i][y][z],r+(a[i]-x-y-z)/3+z);
    }
    printf("%d\n",dp[m][0][0]);
}

 

 

1110E. Magic Stones

大意: 给定序列$c,t$, 每次操作选择一个$i$, 将$c_i$变为$c_{i+1}+c_{i-1}-c_{i}$, 求判断$c$是否能变为$t$.

这种题一看就没什么思路, 然后开始打表找规律, 跑了一下发现长$n$的序列能得到$n!$种不同的序列, 似乎没什么用处. 搞了半个小时, 觉得似乎可以只判断第二个数的情况, 这样肯定很难卡掉. 交了一发就A了..... 具体证明的话, 考虑每次操作对差分序列的影响, 可以发现相当于是交换了差分序列的两个数, 而第一个数是不会变的, 所以只要差分序列相同即可, 也就等价于第二个数所有可能值相同.

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <cstring>
#include <bitset>
#include <functional>
#include <random>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<',';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head



const int N = 200;
typedef vector<int> node[N];
int n;
node f;

vector<int> add(vector<int> a, vector<int> b) {
    REP(i,1,n) a[i]+=b[i];
    return a;
}
vector<int> sub(vector<int> a, vector<int> b) {
    REP(i,1,n) a[i]-=b[i];
    return a;
}

typedef vector<vector<int> > vv;
set<vv> s;

void dfs(node x) {
    vv t(n+1);
    REP(i,1,n) t[i] = x[i];
    if (s.count(t)) return;
    s.insert(t);
    REP(i,2,n-1) {
        auto t = x[i];
        x[i] = sub(add(x[i-1],x[i+1]),x[i]);
        dfs(x);
        x[i] = t;
    }
}

int main() {
    scanf("%d",&n);
    REP(i,1,n) f[i].resize(n+1);
    REP(i,1,n) f[i][i] = 1;
    dfs(f);
    printf("tot=%d\n",(int)s.size());
    int p=0;
    set<vector<int> > qq;
    for(auto &e:s) {
        qq.insert(e[2]);
//        REP(i,1,n) {
//            REP(j,1,n) cout<<e[i][j]<<' ';hr;
//        }hr;
    }
    for(auto &e:qq) {
        REP(i,1,n) cout<<e[i]<<' ';hr;
        hr;
    }
}

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <cstring>
#include <bitset>
#include <functional>
#include <random>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<',';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head


const int N = 1e6+50;
int n;
ll c[N],t[N];

int main() {
    scanf("%d", &n);
    REP(i,1,n) scanf("%lld",c+i);
    REP(i,1,n) scanf("%lld",t+i);
    if (c[1]!=t[1]||c[n]!=t[n]) return puts("No"),0;
    //求出c[2]的所有可能值
    multiset<ll> L;
    L.insert(c[2]);
    REP(i,2,n-1) L.insert(c[1]-c[i]+c[i+1]);
    multiset<ll> R;
    R.insert(t[2]);
    REP(i,2,n-1) R.insert(t[1]-t[i]+t[i+1]);
    puts(L==R?"Yes":"No");
}

 

1110F. Nearest Leaf

大意: 给定一棵树, 节点编号与$dfs$序一致, $q$个询问, 每次询问给出$(v,l,r)$, 求区间$[l,r]$内的所有叶子到$v$的最短距离.

询问离线, 然后$dfs$一遍, 线段树节点$x$维护编号为$x$的叶子到当前遍历到的点的距离, 每次移动相当于是子树加减.