Edited by moderator 23.08.2020 01:25

var
a:array[1..15000] of integer;
i:integer;
m,n,k:integer;
b:array[1..1000000] of longint;
j:integer;
Begin
k:=0;
readln(n);
for i:=1 to n do read(a[i]);
readln(m);
for j:=1 to m do readln(b[j]);
for i:=1 to n do
for j:=1 to m do
if b[j]=a[i] then k:=k+1;
writeln(k);
end.

using int gives AC

why authors didn't rejudge???
some authors got accepted more than 0.5 sec....
i don't understand why?

1+sqrt(2)

Edited by author 26.06.2015 05:22

I've just looked through the submission list. Almost everyone had WA #11. Maybe someone with AC can give me some hint about it?

Can somebody give some hints about that test?

Dear All,

I am a student and I got an assignment from my teacher to run "1409 Two Gangsters". I write Python program v.2.7 as per following codes. I run the program many times, but it still showed "Runtime Error". I could not fix it.

It would be appreciated if someone can help me. Thanks.


# 1409 Two Gangsters (Created: 24th June 2015 / Python 2.7)


# -------------------- Start of Program -------------------- #

# Set initial values
total = 0    # total number of cans shot by Harry and Larry
Harry = 0    # the number of cans shot by Harry
Larry = 0    # the number of cans shot by Larry
h = 0            # the number of cans that were not shot by Harry
l = 0           # the number of cans that were not shot by Larry


Harry = int(input("Please enter the number of cans shot by Harry: ")
Larry = int(input("Please enter the number of cans shot by Larry: ")


# Calculation of "The number of cans that were not shot by Harry and Larry"
total = Harry + Larry -1
h = total - Harry
l = total - Larry


# Output
print "-"*100
print "Total number of cans are : ", total
print "The number of cans that were not shot by Harry and Larry are ", h, "and", l, "respectively."
print "-"*100




# -------------------- End of Program -------------------- #

My code got WA7.
const Nmax=100500;
var
 Dp:array[1..Nmax] of longint;
 A,V:array[1..Nmax] of longint;
 N,i:longint;

BEGIN
 readln(N);
 for i:=1 to N do
 begin
  read(A[i]);
  Dp[a[i]]:=Nmax+5;
 end;

 Dp[a[n]]:=0;
 V[a[n]]:=0;
 for i:=N-1 downto 1 do
 begin
  if(Dp[a[i]]>Dp[a[i+1]]+1) then begin
   Dp[a[i]]:=Dp[a[i+1]]+1;
   V[a[i]]:=a[i+1];
  end;
 end;

 write(a[1],' ');
 i:=V[a[1]];
 while(i<>0) do begin
  write(i,' ');
  i:=V[i];
 end;
END.

#include<iostream>

int main()
{
 using namespace std;
 unsigned long long n,id[10000],x;
 int m[100],y,i,j;
 cin>>n;
 for(i=0;i<n;i++)
  cin>>id[i]>>m[i];
 for(i=0;i<n;i++)
 {
  for(j=i;j<n;j++)
  {
   if(m[i]<m[i+1])
   {
    x=id[i];
    id[i]=id[i+1];
    id[i+1]=x;
    y=m[i];
    m[i]=m[i+1];
    m[i+1]=y;
   }
  }
 }

 for(i=0;i<n;i++)
  cout<<id[i]<<"  "<<m[i]<<"\n";
 return 0;
}

i try use hungary to get some line use more student，than try the no use student，but wa on #12，pls help me！

[delete]

Edited by author 23.06.2015 16:37

Edited by author 23.06.2015 16:37

var
   n,m,i,j:integer;
   a:array[1..1000]of integer;
   b:array[1..1000]of real;
begin
read(n,m);
for i:=1 to m do
readln(a[i]);

for j:=1 to n do begin
   for i:=1 to m do begin
   if j=a[i] then b[j]:=b[j]+1;
   end;
end;

for i:=1 to n do begin
b[i]:=(b[i]/m)*100;
writeln(b[i]:0:2,'%');
end;
end.

Just use it to solve this problem.

import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;


public class T1880 {

    public static void main(String[] args) {
        // TODO Auto-generated method stub
        Scanner sc=new Scanner(System.in);
        List<Integer> an=new ArrayList<Integer>();
        List<Integer> bn=new ArrayList<Integer>();
        List<Integer> cn=new ArrayList<Integer>();
        int a=sc.nextInt();
        int count=0;
        int a1=a;
        while(a1-->0)
        {
            int n=sc.nextInt();
            an.add(n);
        }

        int b=sc.nextInt();
        int b1=b;
        while(b1-->0)
        {
            int n=sc.nextInt();
            bn.add(n);
        }

        int c=sc.nextInt();
        int c1=c;
        while(c1-->0)
        {
            int n=sc.nextInt();
            cn.add(n);
        }

        if(a>=b && a>=c)
        {
            for (Integer integer : an) {
                if(bn.contains(integer)&&cn.contains(integer))
                    count++;
            }
        }

        else if(b>=a && b>=c)
        {
            for (Integer integer : bn) {
                if(an.contains(integer)&&cn.contains(integer))
                    count++;
            }
        }
        else if(c>=b && c>=a)
        {
            for (Integer integer : cn) {
                if(bn.contains(integer)&&bn.contains(integer))
                    count++;
            }
        }

        System.out.println(count);

    }

}

Test 1:
150 50 1000 100
200 900
250 800
300 700
350 600
400 500
450 400
Answer:450
Test 2:
1 2 12 3
3 9
5 6
7 3
Answer:6
Test 3:
4 3 6 1
Answer:6
Test 4:
11 1 10 2
Answer:11
Test 5:
1000 2 2000 3
.............
.............
1388 1418
1390 1415
1392 1412
1394 1409
1396 1406
1398 1403
1400 1400
Answer:1400

Можно решить задачу эмпирически. В начале найдем решения по-честному,когда пешка стоит на 1 горизонтали(т.е. игра почти завершена). Все состояния A[x,y,1] нам уже известны,а состояния A[x,y,i] i>1 неизвестны. Из очередного состояния (т.е. увеличив i и рассмотрев любые x,y-коорд. слона) будем выбирать макс. значение среди всех переходов. Если состояние,в которое мы попадаем неизвестно,то снова вызываем рекурсию,пока не попадем в известное(такое всегда будет,т.к. на каждом шаге пешка сдвигается хотя бы на 1,а состояния А[x,y,1] известны). Как только из данного состоя ния мы попадаем во все известные, то данное состояние становится известным.