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Practice
Setter: Utkarsh Gupta
Tester: Samarth Gupta
Editorialist: Taranpreet Singh
DIFFICULTY
Simple
PREREQUISITES
Basic Math
PROBLEM
Given N and S, consider an array of length N where A_i = i, find position x(1 \leq x \leq N) such that
\displaystyle \sum_{i = 1}^{x-1} A_i + \sum_{i = x+1}^N A_i = S.
If there are multiple such x, print any one of them. If no such x exists, print -1.
QUICK EXPLANATION
- For chosen x, \displaystyle \sum_{i = 1}^{x-1} A_i + \sum_{i = x+1}^N A_i can be written as \displaystyle \sum_{i = 1}^N A_i - x. Since A_i = i, \displaystyle \sum_{i = 1}^N A_i = \frac{N*(N+1)}{2}
- So, we can only choose x = \displaystyle \frac{N*(N+1)}{2} - S.
- If 1 \leq x \leq N is satisfied, chosen x is the answer, otherwise no such x exists.
EXPLANATION
Let’s focus on the expression first.
\displaystyle \sum_{i = 1}^{x-1} A_i + \sum_{i = x+1}^N A_i = S.
If we consider both summations, we can see that only element not added is A_x. All elements before x-th element is included in the first summation, while all elements after x-th element are covered in second summation.
So we can write \displaystyle \sum_{i = 1}^{x-1} A_i + \sum_{i = x+1}^N A_i = \sum_{i = 1}^N A_i - A_x.
We can also replace A_i by i, so we have \displaystyle \sum_{i = 1}^{N}i - x =S.
The first summation is just sum of first N natural numbers, which is \displaystyle \frac{N*(N+1)}{2}. Hence the expression becomes \displaystyle x = \frac{N*(N+1)}{2} - S. This way, we have computed the required x.
We also require 1 \leq x \leq N, since the element with value x has to exist in the array. Hence, if the condition is not satisfied, there’s no such x which can satisfy the requirements. Otherwise found x is the required answer.
TIME COMPLEXITY
The time complexity is O(1) per test case.
SOLUTIONS
Setter's Solution
//Utkarsh.25dec
#include <bits/stdc++.h>
#include <chrono>
#include <random>
#define ll long long int
#define ull unsigned long long int
#define pb push_back
#define mp make_pair
#define mod 1000000007
#define rep(i,n) for(ll i=0;i<n;i++)
#define loop(i,a,b) for(ll i=a;i<=b;i++)
#define vi vector <int>
#define vs vector <string>
#define vc vector <char>
#define vl vector <ll>
#define all(c) (c).begin(),(c).end()
#define max3(a,b,c) max(max(a,b),c)
#define min3(a,b,c) min(min(a,b),c)
#define deb(x) cerr<<#x<<' '<<'='<<' '<<x<<'\n'
using namespace std;
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
#define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>
// ordered_set s ; s.order_of_key(val) no. of elements strictly less than val
// s.find_by_order(i) itertor to ith element (0 indexed)
typedef vector<vector<ll>> matrix;
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];
void solve()
{
ll n,s;
cin>>n>>s;
ll total_sum=(n*(n+1))/2;
ll ele=(total_sum-s);
if(ele>=1 && ele<=n)
cout<<ele<<'\n';
else
cout<<-1<<'\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);
int T=1;
cin>>T;
int t=0;
while(t++<T)
{
//cout<<"Case #"<<t<<":"<<' ';
solve();
//cout<<'\n';
}
cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC << "ms\n";
}
Tester's Solution
#include <bits/stdc++.h>
using namespace std;
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;
}
assert(l<=x&&x<=r);
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,' ');
}
void readEOF(){
assert(getchar()==EOF);
}
int main() {
// your code goes here
ios_base::sync_with_stdio(false);
cin.tie(0);
int t;
t = readIntLn(1, 1000);
int sum = 0;
while(t--){
int n;
n = readIntSp(2, 1e9);
long long s = readIntLn(1, 1e18);
long long sum = n * 1ll * (n + 1)/2;
if(sum - s <= n && sum - s > 0)
cout << sum - s << '\n';
else
cout << -1 << '\n';
}
readEOF();
return 0;
}
Editorialist's Solution
import java.util.*;
import java.io.*;
class BIGARRAY{
//SOLUTION BEGIN
void pre() throws Exception{}
void solve(int TC) throws Exception{
long N = nl(), S = nl();
long sum = (N*N+N)/2;
if(sum-S >= 1 && sum-S <= N)pn(sum-S);
else pn(-1);
}
//SOLUTION END
void hold(boolean b)throws Exception{if(!b)throw new Exception("Hold right there, Sparky!");}
static boolean multipleTC = true;
FastReader in;PrintWriter out;
void run() throws Exception{
in = new FastReader();
out = new PrintWriter(System.out);
//Solution Credits: Taranpreet Singh
int T = (multipleTC)?ni():1;
pre();for(int t = 1; t<= T; t++)solve(t);
out.flush();
out.close();
}
public static void main(String[] args) throws Exception{
new BIGARRAY().run();
}
int bit(long n){return (n==0)?0:(1+bit(n&(n-1)));}
void p(Object o){out.print(o);}
void pn(Object o){out.println(o);}
void pni(Object o){out.println(o);out.flush();}
String n()throws Exception{return in.next();}
String nln()throws Exception{return in.nextLine();}
int ni()throws Exception{return Integer.parseInt(in.next());}
long nl()throws Exception{return Long.parseLong(in.next());}
double nd()throws Exception{return Double.parseDouble(in.next());}
class FastReader{
BufferedReader br;
StringTokenizer st;
public FastReader(){
br = new BufferedReader(new InputStreamReader(System.in));
}
public FastReader(String s) throws Exception{
br = new BufferedReader(new FileReader(s));
}
String next() throws Exception{
while (st == null || !st.hasMoreElements()){
try{
st = new StringTokenizer(br.readLine());
}catch (IOException e){
throw new Exception(e.toString());
}
}
return st.nextToken();
}
String nextLine() throws Exception{
String str = "";
try{
str = br.readLine();
}catch (IOException e){
throw new Exception(e.toString());
}
return str;
}
}
}
Feel free to share your approach. Suggestions are welcomed as always. 