PROBLEM LINK:
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Setter: Abhinav Gupta
Tester: Takuki Kurokawa
Editorialist: Yash Kulkarni
DIFFICULTY:
1193
PREREQUISITES:
PROBLEM:
Chef has an array A (1 \leq A_i \leq 10^5) of length N.
Let
pref_i = A_1 + A_2 + \dots A_i
suff_i = A_i + A_{i+1} + \dots A_N
Now Chef created another array B of length N such that B_i = pref_i + suff_i.
Chef lost the original array A and wants to recover it with the help of B. Help Chef to recover the array.
It is guaranteed in the input that there exists a valid array A for the given array B. In case of multiple valid arrays A, output any valid array.
EXPLANATION:
Hint
There exists a unique solution.
Solution
Let us have a look at the elements of array B.
B_i = pref_i + suff_i = (A_1 + A_2 + ... A_i) + (A_i + A_{i+1} + ... A_N) = (A_1 + A_2 + ... A_N) + A_i. Hence B_i = SUM_A + A_i, where SUM_A is the sum of the original array (A).
Now, let us have a look at the sum of the elements of the array B and use the above observation.
SUM_B = B_1 + B_2 + ... B_N = (SUM_A + A_1) + (SUM_A + A_2) + ... (SUM_A + A_N).
Hence SUM_B = (N + 1) \times SUM_A.
So, we can find out SUM_A using SUM_B i.e. SUM_A = SUM_B / (N+1). After knowing SUM_A and using the first observation (A_i = B_i - SUM_A), we can find all the elements of the original array (A), which is the required answer.
TIME COMPLEXITY:
O(N) for each test case.
SOLUTION:
Setter's solution
#include <bits/stdc++.h>
#define ll long long int
#define pb push_back
#define mp make_pair
#define mod 1000000007
#define vl vector <ll>
#define all(c) (c).begin(),(c).end()
using namespace std;
ll power(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll modInverse(ll a){return power(a,mod-2);}
const int N=500023;
bool vis[N];
vector <int> adj[N];
long long readInt(long long l,long long r,char endd){
long long x=0;
int cnt=0;
int fi=-1;
bool is_neg=false;
while(true){
char g=getchar();
if(g=='-'){
assert(fi==-1);
is_neg=true;
continue;
}
if('0'<=g && g<='9'){
x*=10;
x+=g-'0';
if(cnt==0){
fi=g-'0';
}
cnt++;
assert(fi!=0 || cnt==1);
assert(fi!=0 || is_neg==false);
assert(!(cnt>19 || ( cnt==19 && fi>1) ));
} else if(g==endd){
if(is_neg){
x= -x;
}
if(!(l <= x && x <= r))
{
cerr << l << ' ' << r << ' ' << x << '\n';
assert(1 == 0);
}
return x;
} else {
assert(false);
}
}
}
string readString(int l,int r,char endd){
string ret="";
int cnt=0;
while(true){
char g=getchar();
assert(g!=-1);
if(g==endd){
break;
}
cnt++;
ret+=g;
}
assert(l<=cnt && cnt<=r);
return ret;
}
long long readIntSp(long long l,long long r){
return readInt(l,r,' ');
}
long long readIntLn(long long l,long long r){
return readInt(l,r,'\n');
}
string readStringLn(int l,int r){
return readString(l,r,'\n');
}
string readStringSp(int l,int r){
return readString(l,r,' ');
}
int sumN=0;
void solve()
{
int n=readInt(1,100000,'\n');
sumN+=n;
assert(sumN<=100000);
ll B[n+1]={0};
for(int i=1;i<=n;i++)
{
if(i==n)
B[i]=readInt(1,(ll)200000 * 1000000,'\n');
else
B[i]=readInt(1,(ll)200000 * 1000000,' ');
}
ll sum=0;
for(int i=1;i<=n;i++)
sum+=B[i];
assert((sum%(n+1))==0);
sum/=(n+1);
ll A[n+1]={0};
for(int i=1;i<=n;i++)
{
A[i]=B[i]-sum;
assert(A[i]>=1 && A[i]<=100000);
cout<<A[i]<<' ';
}
cout<<'\n';
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif
ios_base::sync_with_stdio(false);
cin.tie(NULL),cout.tie(NULL);
int T=readInt(1,1000,'\n');
while(T--)
solve();
assert(getchar()==-1);
cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC << "ms\n";
}
Editorialist's Solution
using namespace std;
int main() {
int T;
cin >> T;
while(T--){
int n;
cin >> n;
vector<long long>b(n);
long long big_sum=0;
for(int i=0;i<n;i++){
cin >> b[i];
big_sum+=b[i];
}
long long small_sum=big_sum/(n+1);
for(int i=0;i<n;i++){
cout << b[i]-small_sum << " ";
}
cout << endl;
}
return 0;
}