Please do share the editorial here itself @regex0754
General term for Die 1: 5i-3
General term for Die 2: 5j-2
So for any throw sum of both dice would be: 5(i+j)-5
Hence any valid X should be a multiple of 5 and also be between 5m & 55m
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How to do the Johto League poblem??? any hint
Brother I too did the same, can you once check my code
https://www.codechef.com/viewsolution/47746937
access denied
Code isn’t visible, share ide1 link or copy paste it here after formatting.
How to solve Pickachu and Stones??
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pow(2, N) - 0 facing upward - 1 facing upward - 0 facing downward - 1 facing downward.
pow(2, n) - 2 * n - 2
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I used binary search but it failed
Total number of ways with no restrictions = 2^N
Number of ways with either all front faced or back faced =2
Number of ways with either exactly 1 front faced or back faced =2N
So answer is just 2^N-2(N+1)
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Are you dividing instead of multiplying? It will give WA if dividing because some cases will be missed
Please tell how to do pikachu and stone ?
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#include<stdio.h>
#define ll long long int
int main()
{
ll t;
scanf("%lld",&t);
while(t--)
{
ll n,m,count=0,a[10009],j;
scanf("%lld %lld",&n,&m);
for(j=0;j<n;j++)
scanf("%lld",&a[j]);
for(j=0;j<n;j++)
{
if(a[j]%5==0&&a[j]>=5*m&&a[j]<=m*50)
count++;
}
printf("%lld\n",count);
}
return 0;
}
what if there 3 front and 3 backward would have been given
Pikachu and Stones -
Dp -
Maintain the number of stones till current index. Maintain the highest stone.
Do it once up to down. Do it once down to up.
if The highest stones mismatch. choose the once till the given range with highest and the one from other.
Create a prefix max and suffix max array and notice that any valid peak would have some prefix from the prefix max array rest of the suffix from the suffix max array. So for any index j to be the peak we’ll have to add \displaystyle \sum_{i=0}^{j-1}(P_i-A_i)+\displaystyle \sum_{i=j+1}^N(S_i-A_i)+\max(P_j,S_j)-A_j and this could be easily calculate by prefix & suffix sums .
CODE
#include "bits/stdc++.h"
#define int long long
using namespace std ;
void solve(){
int ans=1e10,c=0,d=0,n ;
cin >> n ;
vector<int>a(n),p(n),s(n) ;
for(int &x:a)
cin >> x ;
p[0]=a[0] ;
for(int i=1;i<n;i++)
p[i]=max(p[i-1]+1,a[i]) ;
s[n-1]=a[n-1] ;
for(int i=n-2;~i;--i){
s[i]=max(s[i+1]+1,a[i]) ;
c+=s[i]-a[i] ;
}
c-=s[0]-a[0] ;
d+=p[0]-a[0] ;
for(int i=1;i<n-1;i++){
c-=s[i]-a[i] ;
ans=min(ans,c+d+max(p[i],s[i])-a[i]) ;
d+=p[i]-a[i] ;
}
cout << ans << '\n' ;
}
signed main(){
ios_base::sync_with_stdio(false);
cin.tie(NULL);
int TC ;
cin >>TC ;
while(TC--)
solve() ;
}
2 Likes
55 is the upper limit. You are using 50.
hoooo, my bad
Is this approach correct?
Can you suggest test case on which it will not work?
#include <bits/stdc++.h>
using namespace std;
typedef long long int ll;
#define rep(i,n) for (ll i = 0; i < n; i++)
#define ysno(ok) if(ok){cout<<"YES"<<endl;}else{cout<<"NO"<<endl;}
#define testloop ll t;cin>>t;while(t--)
int main(){
ios::sync_with_stdio(false);
cin.tie(0);
testloop{
ll n;
cin>>n;
ll a[n];
ll dp2[n]; dp2[n] = -100;ll dp1[n+1]; dp1[0] = -100;
rep(i,n){cin>>a[i];dp1[i+1] = a[i]; dp2[i+1] = a[i];}
rep(i,n){
dp1[i+1] = max(dp1[i]+1,dp1[i+1]);
}
for(int i = n-1;i >=0;i--){
dp2[i] = max(dp2[i+1]+1,dp2[i]);
}
rep(i,n+1){
dp1[i] = max(dp1[i],dp2[i]);
}
ll mini = 1e18;
ll r = 0;
for(int i = 1 ; i < n-1;i++){
if(dp1[i+1] < mini){mini = dp1[i+1];r = i;}
}
ll ans = mini -a[r];
a[r] = mini;
ll x = mini;
for(int i =1;i < r ;i++){
ll y = max(a[i-1]+1,a[i]);
ans += abs(a[i] - y);
a[i] = y;
}
for(int i =n-2;i >r ;i--){
ll y = max(a[i+1]+1,a[i]);
ans += abs(a[i] - y);
a[i] = y;
}
rep(i,n){cout<<a[i]<<" ";}
cout<<endl;
cout<<ans<<endl;
}
}
Paste full code or share a link It won’t compile in my compiler.