BZOJ4725: [POI2017]Reprezentacje ró?nicowe

这题远没有它看起来那么可怕..
容易发现当n>70时a[n]就已远大于1e9，此时和a[n]有关的差只有a[n]-a[n-1]在n为偶数的情况下在1e9之内
于是打个70的表，如果当前的x在打表时没出现，就二分找已经出现的差有多少个

#include<set>#include<map>#include<deque>#include<queue>#include<stack>#include<cmath>#include<ctime>#include<bitset>#include<string>#include<vector>#include<cstdio>#include<cstdlib>#include<cstring>#include<climits>#include<complex>#include<iostream>#include<algorithm>#define ll long long#define inf 1e9using namespace std;const int maxn = 110000;int n;ll a[maxn];struct node{    int p,q;    ll x;};inline bool operator <(const node x,const node y){return x.x<y.x;}set<node>S;set<node>::iterator it,it1;ll t[maxn]; int tp;int main(){    a[1]=1; S.insert((node){1,1,0});    a[2]=2; S.insert((node){2,1,1}); it1=S.find((node){2,1,1}); t[++tp]=1;    ll now=1; int u=70;    for(int i=3;i<=u;i++)    {        if(i&1)        {            a[i]=a[i-1]*2ll;            for(int j=i-1;j>=1;j--)            {                node tmp=(node){i,j,a[i]-a[j]};                it=S.find(tmp);                if(it==S.end()) S.insert(tmp),t[++tp]=tmp.x;            }        }        else        {            it=it1;            while(it!=S.end())                it=S.find((node){0,0,++now});            a[i]=a[i-1]+now;            for(int j=i-1;j>=1;j--)            {                node tmp=(node){i,j,a[i]-a[j]};                it=S.find(tmp);                if(it==S.end()) S.insert(tmp),t[++tp]=tmp.x;            }        }    }    /*for(int i=1;i<=u;i++) printf("%lld\n",a[i]);    puts(""); puts("");puts("");puts("");puts("");    for(int i=1;i<=tp;i++) printf("%lld\n",t[i]);*/    sort(t+1,t+tp+1);    scanf("%d",&n);    while(n--)    {        ll k; scanf("%lld",&k);        it=S.find((node){0,0,k});        if(it==S.end())        {            int l=1,r=tp;            while(l<=r)            {                int mid=l+r>>1;                if(t[mid]<k) l=mid+1;                else r=mid-1;            }l--;            ll p=u+(k-l)*2ll;            printf("%lld %lld\n",p,p-1);        }        else        {            printf("%d %d\n",(*it).p,(*it).q);        }    }    return 0;}