birthday present for girlfriend

if the number exceeds 1000, the output is "alternative!". (because there may be many options for buying a book, so there is no need to export a specific plan)Sample inputSample Input2 32 1 24 8 45 5Sample outputSample Output361#include 2#include 3 using namespacestd;4 inta[Ten][ A],f[ A],sum[ A],lim=1001;//limit the number of boundaries the topic requires no more than5 intn,m;6 voidDfsints) {7 for(intI=0; i){8 for(intk=1; k)9Sum[k]+=a[s][k]*i;//The number of books for each color;T

1293: [SCOI2009] Birthday present time limit:10 Sec Memory limit:162 MB submit:1096 solved:584 [Submit] [Status] Description Little West has a very long ribbon, ribbon hanging on a variety of color beads. Known color beads have n, divided into K species. Simply put, the ribbon can be considered as the x-axis, each color bead has a corresponding coordinate (i.e. position). There can be no color bea

, thus the maximum value of all colors computed by POS, and then the shortest in 1~n. 1#include 2#include 3#include 4#include 5#include 6#include 7#include 8#include 9 using namespacestd;TentypedefLong LongLL; One Const intmaxn=1000010; A intn,k,maxpos,cnt; - intkin[maxn],a[maxn],next[maxn],head[ -],tmp[ -]; - intans=1e9; theInlinevoidCalcintPOS) { - intans=0; - for(intI=1; i){ - while(kin[head[i]]>POS) { + if(next[head[i]]==0)return ; -head[i]=Next[head[i]]; +

Title: http://www.lydsy.com/JudgeOnline/problem.php?id=1293Record the color of each point and the coordinates of the previous color point, and then, after discretization, the endpoint of the enumeration is swept forward again.#include #include#include#include#include#include#defineRep (i,l,r) for (int i=l;i#defineDown (i,l,r) for (int i=l;i>=r;i--)#defineCLR (x, y) memset (x,y,sizeof (×))#defineLow (x) (x (×))#defineMAXN 1005000#defineINF Int (1E9)#defineMM 1000000007#definell Long Longusing nam

A year of Sichuan province election.Enumerate each bead and look for all the kinds of beads in front of it and its maximum distance (this distance must include all the beads) and then update the answer.For convenience, we sort the beads and form the list of links (also in reverse order).However, this water problem ziliuziliu still not very good to think out. I don't know what's going on. This problem is actually quite simple ah.#include #include #include #include #define INF 0X7FFFFFFF#define MA

[i]=sum,a[sum]=y; A } the } +Sort (A +1, a+sum+1); - for(intI=sum;i>0; i--) $ { $ if(a[i]!=a[i+1]) - if(!js (A[i])) Break; - } theprintf"%d", ans); - return 0;Wuyi}---restore content ends---Topic linksMonotonic Queue OptimizationA lifting point to find the maximum value of the distance from the first position of each color after this position.Then find out the minimum value of each start result.1#include 2#include 3#include 4#include 5 using namespace

=GetChar (); - while(rx>='0'rx'9') ra*=Ten, ra+=rx- -, Rx=getchar ();returnTalfh; - } - BOOL operatorreturnABS (V[a.id]) >ABS (V[b.id]);} - intMain () { -N=read (), M=read ();if(!m) {Puts ("0");return 0;} in for(i=1; i0; - for(i=1; i]) to if(LL) a[i]* (LL) v[cnt]0||! CNT) cnt++; + for(i=1; iif(v[i]>0) zsnum++,ans+=V[i]; - if(zsnum>m) { the for(i=1; i1, next[i]=i+1;//, printf ("%d", V[i]);p UTS (""); *pre[1]=next[cnt]=0; $ for(i=zsnum-m;i;i--){Pan