TC SRM 673 Div1

TC srm.673 300

Time Limit:20 Sec

Memory limit:256 MB

Topic Connection

Description

Give you n (n<=50) horse and n person, a horse and a person can form a cavalry, this cavalry's fighting power equals the combat effectiveness of the horse, ask you how many combinations of ways to meet the other cavalry's combat effectiveness is no more than the No. 0 cavalry.

Input

Output

Sample Input

Sample Output

HINT

Test instructions

Exercises

Probably on the violence enumeration which horse and the No. 0 match, the cavalry behind us according to the combat effectiveness from the big to the small after the order, for each cavalry two points (or violence) to find the maximum number of horses can be selected, and then can be a little bit to do it!

Code

#include <stdio.h>#include<string.h>#include<iostream>#include<algorithm>using namespacestd;Const intMoD = 1e9+7;BOOLcmpintAintb) {    returnA>b;}classbearcavalry{ Public:    intCountassignments (Vector <int> Warriors, Vector <int>horses) {Sort (Warriors.begin ()+1, Warriors.end (), CMP);        Sort (Horses.begin (), Horses.end ()); Long LongAns =0;  for(intI=0; I) {sort (Horses.begin (), Horses.end ()); intp = warriors[0]*Horses[i]; intK =Horses[i]; Horses.erase (Horses.begin ()+i); Long LongAns2 =1; intFlag =0;  for(intj=1; J<warriors.size (); j + +)            {                Long LongTMP =-1;  for(intt=0; T)                {                    if(warriors[j]*horses[t]>=p) Break; TMP=T; }                //cout<<i<< "" <<j<< "" <<tmp<<endl;TMP = tmp +2-J; //cout<<i<< "" <<j<< "" <<tmp<<endl;                if(tmp<=0) {flag=1;  Break; } ans2= (ANS2 * tmp)%MoD; }            if(flag==0) ans= (ans + ans2)%MoD;        Horses.push_back (k); }        returnans; }};intMain () {bearcavalry C; Vector<int>A; Vector<int>b; A.push_back (5); A.push_back (8); A.push_back (4); A.push_back (8); B.push_back ( +); B.push_back ( +); B.push_back ( -); B.push_back ( -); printf ("%d\n", C.countassignments (A, b));}

TC SRM 673 Div1