562 - Dividing Coins

[md_yusuf]

why wa i have cant understand

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//C++
#include<iostream>
#include<algorithm>
using namespace std;
#define max_(a,b)(a>b?a:b)

int a[110];
int mark[10000][10000];
int store[10000][10000];
int n,t;

int dp(int i, int w)
{
   if(i==0 || w==0)
      return 0;
   else if(mark[i][w]==t)
      return store[i][w];
   int res;
   if(a[i]<=w)
   {
      res=max_(dp(i-1,w),dp(i-1,w-a[i])+a[i]);
   }
   else
      res=dp(i-1,w);
   mark[i][w]=t;
   store[i][w]=res;
   return res;
}
int main()
{
   int i,w,cs,tt,sum,x;
   //freopen("1.txt","r",stdin);
   cin>>tt;
   for(cs=1;cs<=tt;cs++)
   {
      t++;
      cin>>n;
      sum=0;
      for(i=1;i<=n;i++)
      {
         scanf("%d",&a[i]);
         sum+=a[i];
      }
      w=sum/2;
      x=dp(n,w);
      printf("%d\n",(sum-2*x));
   }
}


plz help me. im getting frustrated..
plz...

[sazzadcsedu]

you should get memory limit exceed.How much memory you used-

Code: Select all

int mark[10000][10000];
int store[10000][10000];

Its a simple 1-dimentional DP problem.look at the sample input
4
1 2 4 6

Declare an array A[50000] because highest value may be 100*500=50000 and and fill it by 0.

now add all element .Sum=1+2+4+6=13.
as sum is odd you can't divide it evenly,so look for number closet to half-6.
so try to make sum=6 by adding some element.

For 1st element we have sum- 1. set A[1]=1;
For 2nd element we have sum-2,(2+1=3).set A[2]=1,A[3]=1.
For 3rd element we have sum-4,(1+4=5),(2+4=6) stop as you find 6 from element 2,4=6 and other set certainly contain=13-6=7.

SO diff is 1.

Hope you understand .

[md_yusuf]

thx via. i cant understand why im getting wa. I should get re or ce.. bt making the the arry mark[101][25001] and mark[101][25001] get accepted its too weird to me.

[Shafaet_du]

I got a wa for a simple but tricky case.

Code: Select all

1
0

output is 0



another case that may help you:

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1
13
1 4 6 8 5 3 4 7 5 3 5 2 100

ans: 47

[mdpallob]

got two times wrong answer.
Can anyone help me?


#include<iostream>


using namespace std;

int main()
{
int n;
cin>>n;
int value=0;
int diff;

for(int i=1;i<=n;i++)
{
int coin_no;
cin>>coin_no;

for(int x=1;x<=coin_no;x++)
{
int val;
cin>>val;
value=value+val;
}
diff=value%2;
cout<<"\n"<<diff;

}

return 0;
}

[Imti]

Hello
My code passing all the input found here in this board and also my hand-generated one's(compared with toolkit output).I used simple DP here.
I worked like..
1)Sum up all the coin first
2)Then set S=sum/2
3)Now I used 0-1 knapsack here.That is Fill knapsack with size S as maximum as possible
4)then output sum-2*p[n][S]
But I'm getting WA..Here is my code

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//Got Accepted.


Edit:
A silly Bug.Algorithm was okay.allocated memory less than program need..wondered to see no one here in this board to help....its dead one..

[Imti]

mr. mdpallob
What you did???You really understood what problem wants??I don't think so
Your algorithm is totally wrong.Your code always produce minimal difference as 0 or 1.But answer may be greater than 0 or 1 too. Read carefully problem statement again and try learning Dynamic programming technique.

[mostafiz93]

I used 0-1 knapsack to solve this problem.
can anyone find any bug in my code:

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removed after AC


[brianfry713]

see I/O #7 at http://ideone.com/XgQ97, AC output is 1.

[mostafiz93]

thanks brianfry!
got AC.

[shopnobaj_raju]

Runtime error. Why????

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AC

[brianfry713]

For one thing, this declares more memory than you need to solve this problem.
long int coins[102],opt[102][25010];

Your RE is here:
for(i=0;i<103;i++)
for(j=0;j<25011;j++) opt[i][j]=-100;

[shopnobaj_raju]

Thanks

[connor]

I try your input and got right
but still get wa on oj
i don't know why==
plz help me
3ks

here is my codes:

#include<stdlib.h>
#include<stdio.h>
int min(int a,int b) {if(a>b)return b;return a;}
int abs(int a) {if(a>0)return a;return -a;}
int f(int sum,int a) {return abs(sum-a-a);}
int fmin(int a,int b,int sum)
{
int res=min(f(sum,a),f(sum,b));
if(res==f(sum,a))return a;
return b;
}
int main()
{
int k,z;
scanf("%d",&z);
for(k=1;k<=z;k++)
{
int m;
scanf("%d",&m);
int i,j,a[110],sum=0;
int dp[110][110];
for(i=1;i<=m;i++)
{
scanf("%d",&a[i]);
sum+=a[i];
}
for(i=0;i<=m;i++)for(j=0;j<=m;j++)dp[i][j]=0;
for(j=1;j<=m;j++)
{
int mm=a[1];
for(i=2;i<=j;i++)mm=fmin(mm,a[i],sum);
dp[1][j]=mm;
}
for(i=2;i<=m;i++)
{
for(j=1;j<=i;j++)
{
dp[i][j]=fmin(dp[i-1][j-1]+a[i],dp[i-1][j],sum);
}
}
int res=dp[m][1];
for(i=2;i<=m;i++)res=fmin(res,dp[m][i],sum);
printf("%d\n",f(sum,res));
}
// return main();
return 0;}

[brianfry713]

Input:

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1
10
500 499 500 488 467 500 499 488 499 487

AC output 1