nowcoder16732 序列

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题解

这题就是要在 $n-1$个位置放一些 $1,0.5,-2$，使得乘积等于 $1$，那么显然 $0.5$和 $-2$的数量要相同，且 $-2$要有偶数个，枚举 $-2$的数量 $i$，那么这种情形对答案的贡献就是 $C_{n-1}^iC_{n-1-i}^i$

代码

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define iinf 0x3f3f3f3f
#define linf (1ll<<60)
#define eps 1e-8
#define maxn 1000010
#define maxe 1000010
#define cl(x) memset(x,0,sizeof(x))
#define rep(_,__) for(_=1;_<=(__);_++)
#define em(x) emplace(x)
#define emb(x) emplace_back(x)
#define emf(x) emplace_front(x)
#define fi first
#define se second
#define de(x) cerr<<#x<<" = "<<x<<endl
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
ll read(ll x=0)
{
    ll c, f(1);
    for(c=getchar();!isdigit(c);c=getchar())if(c=='-')f=-f;
    for(;isdigit(c);c=getchar())x=x*10+c-0x30;
    return f*x;
}
vector<ll> v[maxn];
ll n, m, sta[maxn], tb[maxn];
pll check(ll Min)
{
    ll i, j;
    rep(i,n)
    {
        sta[i]=0;
        for(j=0;j<m;j++)sta[i] |=  ( ll(v[i][j]>=Min) << j );
    }
    ll mask=(1<<m)-1;
    for(i=0;i<=mask;i++)tb[i]=0;
    rep(i,n)
    {
        tb[sta[i]]=i;
        for(j=0;j<=mask;j++)if( (j|sta[i]) == mask )
        {
            if(tb[j])return pll(tb[j],i);
        }
    }
    return pll(-1,-1);
}
int main()
{
    ll i, j, k, l, r, mid;
    pll ans;
    n=read(), m=read();
    rep(i,n)rep(j,m)v[i].emb(read());
    l=0, r=1e9;
    while(l<r)
    {
        mid=(l+r+1>>1);
        auto pr=check(mid);
        if(pr.first==-1)r=mid-1;
        else l=mid, ans=pr;
    }
    ans = check(l);
    cout<<ans.first<<' '<<ans.second;
    return 0;
}