Submission #832624

#	Time	Username	Problem	Language	Result	Execution time	Memory
832624		AkramElOmrani	Horses (IOI15_horses)	C++17	34 / 100	33 ms	24436 KiB

horses

#include <stdio.h>
#include <stdlib.h>
#include <bits/stdc++.h>
using namespace std;

#include <bits/stdc++.h>
using namespace std;
#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("avx,avx2,bmi,bmi2,popcnt,lzcnt")

#define ll long long

template<typename A, typename B> ostream& operator<<(ostream &os, const pair<A, B> &p) { return os << '(' << p.first << ", " << p.second << ')'; }
template<typename T_container, typename T = typename enable_if<!is_same<T_container, string>::value, typename T_container::value_type>::type> ostream& operator<<(ostream &os, const T_container &v) { os << '{'; string sep; for (const T &x : v) os << sep << x, sep = ", "; return os << '}'; }
 
void dbg_out() { cerr << endl; }
template<typename Head, typename... Tail> void dbg_out(Head H, Tail... T) { cerr << ' ' << H; dbg_out(T...); }
#define dbg(...) cerr << "(" << #__VA_ARGS__ << "):", dbg_out(__VA_ARGS__);


const int N = 1e5 + 100, mod = 1e9 + 7;
struct Node {
	int ans = 0, x = 1, y = 1; // ans is the answer, x is the value of x, y is the value of y
	double X = 0, Y = 0, m = 0; // X = sum of logs of X
};

Node tree[N * 4];

int mul(int a, int b) {
	a = (ll)a * b % mod;
	return a;
}

void upd(int node, int l, int r, int p, int x, int y) {
	// dbg(node)	
	if(r < p || p < l) return;
	if(l == r) {
		if(x > 0) {
			tree[node].X = log10(x); tree[node].x = x; tree[node].m = tree[node].X + tree[node].Y;
		}
		if(y > 0) {
			tree[node].Y = log10(y); tree[node].y = y; tree[node].m = tree[node].X + tree[node].Y;
		}
		tree[node].ans = mul(tree[node].x, tree[node].y);
		// dbg(node, tree[node].x)
		return;
	}
	int mid = (l + r) / 2;
	upd(node * 2, l, mid, p, x, y);
	// dbg(node * 2, tree[node * 2].x)
	upd(node * 2 + 1, mid + 1, r, p, x, y);
	tree[node].X = tree[node * 2 + 1].X + tree[node * 2].X;
	tree[node].x = mul(tree[node * 2 + 1].x, tree[node * 2].x);

	// check if the answer is on the left or on the right
	if(tree[node * 2].m < tree[node * 2 + 1].m + tree[node * 2].X) {
		tree[node].m = tree[node * 2 + 1].m + tree[node * 2].X;
		tree[node].ans = mul(tree[node * 2 + 1].ans, tree[node * 2].x);
	} else {
		tree[node].m = tree[node * 2].m;
		tree[node].ans = tree[node * 2].ans;
	}
}

int n;
int init(int n_, int x[], int y[]) {
	n = n_;
	for(int i = 0; i < n; ++i) {
		upd(1, 0, n - 1, i, x[i], y[i]);
	}

	// dbg(tree[1].m, tree[1].x, tree[1].ans, tree[1].X);
	return tree[1].ans;
}
int updateX(int pos, int val) {
	upd(1, 0, n - 1, pos, val, 0);
	return tree[1].ans;
}
int updateY(int pos, int val) {
	upd(1, 0, n - 1, pos, 0, val);
	return tree[1].ans;
}





























// static char buffer[1024];
// static int currentChar = 0;
// static int charsNumber = 0;

// static inline int read() {
//   if (charsNumber < 0) {
// 	exit(1);
//   }
//   if (!charsNumber || currentChar == charsNumber) {
// 	charsNumber = (int)fread(buffer, sizeof(buffer[0]), sizeof(buffer), stdin);
// 	currentChar = 0;
//   }
//   if (charsNumber <= 0) {
// 	return -1;
//   }
//   return buffer[currentChar++];
// }

// static inline int readInt() {
//   int x; cin >> x;
//   return x;
// }

// int main() {
//   int N;
//   cin >> N;

//   int *X = (int *)malloc(sizeof(int) * (unsigned int)N);
//   int *Y = (int *)malloc(sizeof(int) * (unsigned int)N);

//   for (int i = 0; i < N; i++) {
// 	X[i] = readInt();
//   }

//   for (int i = 0; i < N; i++) {
// 	Y[i] = readInt();
//   }

//   printf("%d\n", init(N, X, Y));

//   int M;
//   M = readInt();

//   for (int i = 0; i < M; i++) {
// 	int type;
// 	type = readInt();
// 	int pos;
// 	pos = readInt();
// 	int val;
// 	val = readInt();

// 	if (type == 1) {
// 	  printf("%d\n", updateX(pos, val));
// 	} else if (type == 2) {
// 	  printf("%d\n", updateY(pos, val));
// 	}
//   }

//   return 0;
// }

Compilation message (stderr)

horses.cpp: In function 'int mul(int, int)':
horses.cpp:31:16: warning: conversion from 'long long int' to 'int' may change value [-Wconversion]
   31 |  a = (ll)a * b % mod;
      |      ~~~~~~~~~~^~~~~

• Subtask 1: 17 / 17
• Subtask 2: 17 / 17
• Subtask 3: 0 / 20
• Subtask 4: 0 / 23
• Subtask 5: 0 / 23