C++14(g++5.4) 解法, 执行用时: 4ms, 内存消耗: 508K, 提交时间: 2020-02-10 15:36:49

#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 110
#define maxe 1000010
#define cl(x) memset(x,0,sizeof(x))
#define rep(i,a,b) for(i=a;i<=b;i++)
#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;
}
ll dig[maxn], c[maxn][maxn], ans[maxn], tot;
ll C(ll n, ll m)
{
    if(n<0 or m<0 or m>n)return 0;
    return c[n][m];
}
ll count()
{
    ll ans=0, i, c[]={0,0};
    rep(i,1,tot)c[dig[i]]++;
    c[1]--;
    for(i=tot-1;i;i--)
    {
        if(dig[i])ans+=C(c[0]-1+c[1],c[1]);
        c[dig[i]]--;
    }
    return ans+1;
}
ll construct(ll res)
{
    ll i, c[]={0,0};
    rep(i,1,tot)c[dig[i]]++;
    c[1]--;
    if(C(c[0]+c[1],c[0])<res)return -1;
    ans[tot]=1;
    for(i=tot-1;i;i--)
    {
        ll t=C(c[0]-1+c[1],c[1]);
        if(res>t)
        {
            res-=t;
            ans[i]=1;
            c[1]--;
        }
        else
        {
            ans[i]=0;
            c[0]--;
        }
    }
    return 0;
}
int main()
{
    ll n, k, T, i, j, kase;
    rep(i,0,maxn-1)c[i][0]=1;
    rep(i,1,maxn-1)rep(j,1,i)c[i][j]=c[i-1][j-1]+c[i-1][j];
    T=read();
    rep(kase,1,T)
    {
        n=read(), k=read();
        tot=0;
        while(n)
        {
            dig[++tot]=n&1;
            n>>=1;
        }
        ll t=count();
        ll ret=construct(t+k);
        // de(t);
        printf("Case #%lld: ",kase);
        if(ret==-1)printf("IMPOSSIBLE\n");
        else
        {
            ll t=0;
            rep(i,1,tot)t=t+(ans[i]<<i-1);
            // rep(i,1,tot)printf("%lld",ans[i]); putchar(32);
            printf("%lld\n",t);
        }
    }
    return 0;
}

C++11(clang++ 3.9) 解法, 执行用时: 5ms, 内存消耗: 488K, 提交时间: 2018-03-29 21:26:42

#include<bits/stdc++.h>

using namespace std;
typedef long long ll;
ll query,kase,n,k,c[64][64],p,q,d[64],r,w,s,t,ans;

int main()
{
    c[0][0]=1;
    for (int i=1;i<64;i++)
    {
        c[i][0]=c[i][i]=1;
        for (int j=1;j<i;j++)
            c[i][j]=c[i-1][j-1]+c[i-1][j];
    }
    scanf("%d",&query);
    while (query--)
    {
        scanf("%lld%lld",&n,&k); r=0; p=q=0;
        printf("Case #%d: ",++kase);
        while (n) {
            d[r++]=n&1;
            n>>=1;
        }
        for (int i=0;i<r-1;i++) if (d[i]) p++; else q++;
        s=p; t=q; w=1;
        for (int i=r-2;i>=1;i--)
            if (d[i]) w+=c[p+q-1][p],p--; else q--;
        w+=k;
        if (w>c[s+t][s]) printf("IMPOSSIBLE\n");
        else {
            ans=1ll<<(r-1);
            for (int i=r-2;i>=0;i--)
                if (t==0||c[s+t-1][s]<w) w-=c[s+t-1][s],s--,ans+=(1ll<<i); else t--;
            cout << ans << endl;
        }
    }
    return 0;
}