Pigeonhole

Author: ivorynoise

DIVSUBS-d.cpp

@@ -0,0 +1,59 @@
+//codechef : pigeonhole
+#include <bits/stdc++.h>
+using namespace std;
+typedef long long ll;
+typedef vector<ll> vi;
+typedef vector<vi> vvi;
+const int mod = 1000000007;
+void printMat(vvi& a){for(int i=0;i<a.size();++i){for(int j=0;j<a[0].size();++j)cout<<a[i][j]<<" ";}cout<<"\n";}
+const int sze = 100010;
+
+int main(){
+	int t; cin >> t;
+	// Each element of the multiset is a positive integer, not exceeding 109.
+	//Alert sum can exceed int range
+
+	int arr[sze];
+	ll sum[sze] = {};
+
+	while(t--){
+		int n;
+		cin >> n;
+
+		for(int i = 0; i < n; ++i){
+			cin >> arr[i];
+			sum[i+1] = arr[i] + sum[i];	//prev + cur
+		}
+
+		int* remainder = new int[n];
+		fill(remainder, remainder + n, -1);
+
+		int be = -1;
+		int en = -1;
+
+		for(int i = 0; i <= n; ++i){
+			int& curIdx = remainder[sum[i] % n]; 
+
+			if (curIdx != -1) {
+				//this remainder has been previously visited
+				be = curIdx;
+				en = i;
+				break;
+			}
+			else {
+				curIdx = i;
+			}
+		}
+
+		cout << en-be <<"\n";	//no of elements in a range
+		++be;
+		while(be <= en) {
+			cout << be << " ";
+			be++;
+		}
+
+		cout << "\n";
+		delete [] remainder;
+
+	}
+}

SPP-d.cpp

@@ -0,0 +1,84 @@
+#include <bits/stdc++.h>
+using namespace std;
+typedef long long ll;
+typedef vector<ll> vi;
+typedef vector<vi> vvi;
+ll mod;// = 1000000007;
+void printMat(vvi& a) {
+	for (int i = 0; i < a.size(); ++i)
+	{	for (int j = 0; j < a[0].size(); ++j)cout << a[i][j] << " ";
+		cout << "\n";
+	}
+}
+
+vvi mul(const vvi& a, const vvi& b) {
+	vvi c(a.size(), vi(b[0].size(), 0));
+
+	for (int r = 0; r < a.size(); ++r) {
+		for (int j = 0; j < b[0].size(); ++j) {
+			for (int x = 0; x < b.size(); ++x) {
+				c[r][j] += (a[r][x] % mod) * (b[x][j] % mod);
+				c[r][j] %= mod;
+			}
+		}
+	}
+	return c;
+}
+
+vvi pow(const vvi& a, int n) {
+	if (n <= 1) return a;
+	if (n & 1) return mul(a, pow(a, n - 1));
+
+	vvi ans = pow(a, n / 2);
+	ans = mul(ans, ans);
+	return ans;
+}
+
+ll sum(int n, vi& b, vi& c) {
+	if (n < 1) return 0;
+
+	int k = b.size();
+	vvi t(k + 1, vi(k + 1, 0));
+
+	for (int i = 0; i <= k; ++i) {
+		for (int j = 0; j <= k; ++j) {
+			//for last row
+			if (i != k && j == i + 1) t[i][j] = 1;
+			else if (i == k && j != 0) t[i][j] = c[j-1];
+			// else if (i == 0 && j == 1) t[i][j] = 1;
+		}
+	}
+	t[0][0] = t[0][1] = 1;
+
+	vvi tn = pow(t, n);
+	// printMat(tn);
+	ll ans  = 0;
+	//the R0b0 is the answer
+	for (int i = 0; i < k; ++i) {
+		// cout << t[0][1+i] << endl;
+		ans += (tn[0][i + 1] * b[i]) % mod;
+		ans %= mod;
+	}
+	// cout << ans << endl;
+	return ans;
+}
+
+int main() {
+	int t;
+	cin >> t;
+	while (t--) {
+		int k; cin >> k;
+		vi b(k);
+		for (int i = 0; i < k; ++i) cin >> b[i];
+		vi c(k);
+		for (int i = 0; i < k; ++i) cin >> c[k - i - 1];
+		ll m, n;
+		cin >> m >> n >> mod;
+		ll m_ = sum(m - 1, b, c);
+		ll n_ = sum(n, b, c);
+		// int ans = (n==1&& m<=1) ? b[0] : (n_ - m_ + mod);
+		ll ans = (n_- m_ + mod) % mod;
+		//ALWAYS handle negative mods
+		cout << ans << "\n";
+	}
+}

SUMSUMS-d_1.cpp

@@ -0,0 +1,82 @@
+#include <bits/stdc++.h>
+using namespace std;
+
+typedef long long ll;
+const ll N = 50000;
+ll n, t;
+ll cows[N];
+const ll m = 98765431;
+
+//mult(ll[][] ll[][])
+void mul(ll a[][2], ll b[][2]){
+	ll c[2][2]={};
+	for(ll i = 0; i < 2; ++i){
+		for(ll j = 0; j < 2; ++j){
+			for(ll x = 0; x < 2; ++x){
+				c[i][j] += (a[i][x] * b[x][j]) % m;
+				c[i][j] %= m;	
+			}
+		}
+	}
+
+	for(ll i = 0; i < 2; ++i){
+		for(ll j = 0; j < 2; ++j)
+			a[i][j] = c[i][j];
+	}
+}
+
+
+
+//pow(ll[][], t)
+void pow(ll mat[][2], ll n){
+	if (n <= 1) return;
+
+	if (n&1) {
+		ll tmp[2][2] = {
+			{mat[0][0], mat[0][1]},
+			{mat[1][0], mat[1][1]}
+		};
+		pow(mat, n-1);
+		mul(mat, tmp);
+		return;
+	}
+
+	pow(mat, n/2);
+	mul(mat, mat);
+}
+
+void transform(ll tn[][2]){
+	if (t == 0) return;
+	pow(tn, t);
+}
+
+
+int main(){
+	cin >> n >> t;
+	ll s =0;
+
+	for(ll i = 0; i < n; ++i){
+		cin >> cows[i];
+		s += cows[i] % m;
+		s %= m;
+	}	
+		ll tn[2][2] = {
+			{n-1, 0},
+			{1, -1}
+		};
+
+		transform(tn);
+		// cout << tn[0][0] << " " << tn[0][1] << " " << tn[1][0] << " " << tn[1][1] << " " << endl;;
+
+	for(ll i = 0; i < n; ++i){
+		//S1 = (n - 1)S0
+		//c[1] = S0 - c(0)
+
+		ll f[2][1] = {{s}, {cows[i]}};
+		ll ans = (tn[1][0]*f[0][0]%m) + (tn[1][1] * f[1][0])%m;
+
+		//this might be negative
+		cout << ((ans+m) % m) << "\n";
+
+	}
+}

SUMSUMS.cpp

@@ -0,0 +1,21 @@
+5
+2
+1 1
+1 1
+1 1 1000003
+2
+1 1
+1 1
+1 2 1000003
+2
+1 1
+1 1
+1 3 1000003
+2
+1 1
+1 1
+2 3 1000003
+2
+1 1
+1 1
+2 4 1000003