Answer:
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int main()
{
ll i,n,j,s;
cout<<"Enter size of array: ";
cin>>n;
ll a[n];
cout<<"Enter elements of array: ";
s=0;
for(i=0;i<n;i++)
{
cin>>a[i];
s+=a[i];
}
if(s%3!=0)
{
cout<<"No";
return 0;
}
ll dp[n][s/3+1];
for(i=0;i<=s/3;i++)
{
if(i==a[0])
dp[0][i]=1;
else
dp[0][i]=0;
}
for(i=0;i<n;i++)
dp[i][0]=1;
for(i=1;i<n;i++)
for(j=0;j<=s/3;j++)
{
if(j<a[i])
dp[i][j]=dp[i-1][j];
else
dp[i][j]=dp[i-1][j]||dp[i-1][j-a[i]];
}
if(dp[n-1][s/3]==0)
{
cout<<"No";
return 0;
}
ll vis[n];
for(i=0;i<n;i++)
vis[i]=0;
ll curr=s/3;
ll indx=n-1;
ll c=0;
while(indx>0)
{
if(dp[indx][curr]==1&&(curr-a[indx]>=0&&dp[indx-1][curr-a[indx]]==1))//include indx
{
vis[indx]=1;
curr=curr-a[indx];
c++;
}
indx--;
}
if(dp[0][curr]==1){
vis[0]=1;
c++;}
ll b[n-c];
ll k=0;
for(i=0;i<n;i++)
{
if(vis[i]==0)
b[k++]=a[i];
}
ll dp1[k][s/3+1];
for(i=0;i<=s/3;i++)
{
if(i==b[0])
dp1[0][i]=1;
else
dp1[0][i]=0;
}
for(i=0;i<k;i++)
dp1[i][0]=1;
for(i=1;i<k;i++)
for(j=0;j<=s/3;j++)
{
if(j<b[i])
dp1[i][j]=dp1[i-1][j];
else
dp1[i][j]=dp1[i-1][j]||dp1[i-1][j-b[i]];
}
if(dp1[k-1][s/3]==0)
{
cout<<"No";
return 0;
}
else
cout<<"Yes";
}
Explanation:
Answer:
Private key
Explanation:
From the question, we understand that the encryption type is a public encryption.
For better analysis, I will create the following scenario.
Encryption keys are stored in an encrypted form in a PGP (i.e. Pretty Good Privacy). This PGP homes the public and the private keys.
However, for Nancy to access the public keyring, she needs the private keys.
If by chance she lost/forgot the private keys, she will not be able to decrypt any information she receives.
Hence, the solution to this question is: <em>private key</em>
This has been "somewhat" answered on Brainly already by me, look around and I'm sure you can find the other question that was asked that I answered. Thanks!
Answer:
the rain hyperlink and water from the ground
Explanation: