Bzoj 3131 Gold Rush

Considering that the things that come out are actually not much ... Make a note of the direct map.

And then, for example, the X-bit count, multiply the case of ceil (X/2) and Trunc (X/2).

Then there is a ai,j=tab[i]*tab[j], asking for the K-big question before the tables.

This can be sorted, then the monotone queue, the state slowly backward push on the line. (Note that ai,j do not have to take a mold to do so, otherwise ...) I don't know how to do it.

#include <iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<map>#include<queue>#include<cmath>#defineMAXN 1000050#defineMoD 1000000007using namespacestd;Long Longn,k,bit[ -],ret=0, a[maxn],tot=0;Long Longans=0; Map<Long Long,Long Long> tab[7],mp;map<pair<Long Long,Long Long,Long Long>Vis;map<Long Long,Long Long>:: iterator it1,it2;structstatus{Long LongP1,P2; Long LongVal; Status (Long LongP1,Long LongP2,Long Longval):p 1 (p1), p2 (p2), Val (val) {} status () {} friendBOOL operator< (ConstStatus &x,ConstStatus &y) {returnx.val<Y.val; }};p Riority_queue<status>Q;BOOLcmpLong LongXLong LongY) {returnX>y;} voidPre_dfs (Long LongNowLong Longret) {    if(now-1) tab[now-1][ret]++; if(now==7)return;  for(Long LongI=1; i<=9; i++) Pre_dfs (now+1, ret*i);}voidGet_bit (Long Longx) {ret=0;  while(x) {bit[++ret]=x%Ten; x/=Ten;}}voidUpdateLong LongRetLong Longx) {    if(!x) {mp[ret]++;return;} if(! (X-1)) { for(Long LongI=1; i<=9; i++) mp[ret*i]++;return;} Long LongL=ceil ((Double) x/2), R=trunc ((Double) x/2);  for(It1=tab[l].begin (); It1!=tab[l].end (); it1++)         for(It2=tab[r].begin (); It2!=tab[r].end (); it2++) Mp[ret* (It1->first) * (It2->first)]+=it1->second*it2->second;}voidDIGIT_DP () { for(Long LongI=1; i<=ret-1; i++) Update (1, i); Long Longrets=1;  for(Long Longi=ret;i>=1; i--)    {         for(Long Longj=1; j<bit[i];j++) Update (RETs*j,i-1); RETs*=Bit[i]; } Mp[rets]++;}intMain () {scanf ("%lld%lld",&n,&k); Pre_dfs (1,1);    Get_bit (n);    DIGIT_DP ();  for(It1=mp.begin (); It1!=mp.end (); it1++)    {        if(!it1->first)Continue; a[++tot]=it1->second; } sort (A+1, a+tot+1, CMP); Q.push (Status (1,1,(Long Long) a[1]*a[1])); Long LongTr=0;  for(Long LongI=1; i<=k;i++) {TR++; Status Now=q.top (); Q.pop (); if(Vis[make_pair (NOW.P1,NOW.P2)]) {i--;Continue;} Ans= (ans+now.val%mod)%mod;vis[make_pair (NOW.P1,NOW.P2)]=1; Q.push (Status (NOW.P1,NOW.P2+1,(Long Long) a[now.p1]*a[now.p2+1])); Q.push (Status (Now.p1+1, NOW.P2, (Long Long) a[now.p1+1]*A[NOW.P2])); } printf ("%lld\n", ans%MoD); return 0;}

Bzoj 3131 Gold Rush