Kattis Problem 'Mali'
$begingroup$
I wrote a solution to the Mali problem on Kattis:
Given inputs a1, a2, a3, …, an and b1, b2, b3, …, bn, determine n pairings (ai, bj) such that each number in the A sequence is used in exactly one pairing, each number in the B sequence is used in exactly one pairing, and the maximum of all sums ai + bj is minimal.
Input
The first line of input contains a single integer N (1 ≤ N ≤ 100000), the number of rounds.
The next N lines contain two integers A and B (1 ≤ A, B ≤ 100), the numbers given in that round.
Output
The output consists of N lines, one for each round. Each line should contain the smallest maximal sum for that round.
I'm fairly sure I can get the right answer 100% of the time, but the code exceeds the time limit of one second. I was wondering if you could help me optimize my code to help it run in time, or if you could explain why my code is inefficient.
#include <iostream>
#include <vector>
int main() {
int x, c = 0, big = 0;
std::cin >> x;
std::vector<int> as, bs;
for (int i = 0; i < x; i++) {
bool founda = false, foundb = false;
int a, b;
std::cin >> a >> b;
c++;
if (c == 1) {
as.push_back(a);
bs.push_back(b);
}
else {
for (int i = 0; i < c; i++) {
if (as[i] < a || i == c-1) {
as.insert(as.begin()+i, a);
founda = true;
}
if (bs[i] < b || i == c-1) {
bs.insert(bs.begin()+i, b);
foundb = true;
}
if (founda == true && foundb == true)
break;
}
}
for (int i = 0; i < c; i++) {
if (as[i] + bs[c-1-i] > big)
big = as[i] + bs[c-1-i];
}
std::cout << big << std::endl;
}
}
c++ programming-challenge time-limit-exceeded
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add a comment |
$begingroup$
I wrote a solution to the Mali problem on Kattis:
Given inputs a1, a2, a3, …, an and b1, b2, b3, …, bn, determine n pairings (ai, bj) such that each number in the A sequence is used in exactly one pairing, each number in the B sequence is used in exactly one pairing, and the maximum of all sums ai + bj is minimal.
Input
The first line of input contains a single integer N (1 ≤ N ≤ 100000), the number of rounds.
The next N lines contain two integers A and B (1 ≤ A, B ≤ 100), the numbers given in that round.
Output
The output consists of N lines, one for each round. Each line should contain the smallest maximal sum for that round.
I'm fairly sure I can get the right answer 100% of the time, but the code exceeds the time limit of one second. I was wondering if you could help me optimize my code to help it run in time, or if you could explain why my code is inefficient.
#include <iostream>
#include <vector>
int main() {
int x, c = 0, big = 0;
std::cin >> x;
std::vector<int> as, bs;
for (int i = 0; i < x; i++) {
bool founda = false, foundb = false;
int a, b;
std::cin >> a >> b;
c++;
if (c == 1) {
as.push_back(a);
bs.push_back(b);
}
else {
for (int i = 0; i < c; i++) {
if (as[i] < a || i == c-1) {
as.insert(as.begin()+i, a);
founda = true;
}
if (bs[i] < b || i == c-1) {
bs.insert(bs.begin()+i, b);
foundb = true;
}
if (founda == true && foundb == true)
break;
}
}
for (int i = 0; i < c; i++) {
if (as[i] + bs[c-1-i] > big)
big = as[i] + bs[c-1-i];
}
std::cout << big << std::endl;
}
}
c++ programming-challenge time-limit-exceeded
New contributor
$endgroup$
add a comment |
$begingroup$
I wrote a solution to the Mali problem on Kattis:
Given inputs a1, a2, a3, …, an and b1, b2, b3, …, bn, determine n pairings (ai, bj) such that each number in the A sequence is used in exactly one pairing, each number in the B sequence is used in exactly one pairing, and the maximum of all sums ai + bj is minimal.
Input
The first line of input contains a single integer N (1 ≤ N ≤ 100000), the number of rounds.
The next N lines contain two integers A and B (1 ≤ A, B ≤ 100), the numbers given in that round.
Output
The output consists of N lines, one for each round. Each line should contain the smallest maximal sum for that round.
I'm fairly sure I can get the right answer 100% of the time, but the code exceeds the time limit of one second. I was wondering if you could help me optimize my code to help it run in time, or if you could explain why my code is inefficient.
#include <iostream>
#include <vector>
int main() {
int x, c = 0, big = 0;
std::cin >> x;
std::vector<int> as, bs;
for (int i = 0; i < x; i++) {
bool founda = false, foundb = false;
int a, b;
std::cin >> a >> b;
c++;
if (c == 1) {
as.push_back(a);
bs.push_back(b);
}
else {
for (int i = 0; i < c; i++) {
if (as[i] < a || i == c-1) {
as.insert(as.begin()+i, a);
founda = true;
}
if (bs[i] < b || i == c-1) {
bs.insert(bs.begin()+i, b);
foundb = true;
}
if (founda == true && foundb == true)
break;
}
}
for (int i = 0; i < c; i++) {
if (as[i] + bs[c-1-i] > big)
big = as[i] + bs[c-1-i];
}
std::cout << big << std::endl;
}
}
c++ programming-challenge time-limit-exceeded
New contributor
$endgroup$
I wrote a solution to the Mali problem on Kattis:
Given inputs a1, a2, a3, …, an and b1, b2, b3, …, bn, determine n pairings (ai, bj) such that each number in the A sequence is used in exactly one pairing, each number in the B sequence is used in exactly one pairing, and the maximum of all sums ai + bj is minimal.
Input
The first line of input contains a single integer N (1 ≤ N ≤ 100000), the number of rounds.
The next N lines contain two integers A and B (1 ≤ A, B ≤ 100), the numbers given in that round.
Output
The output consists of N lines, one for each round. Each line should contain the smallest maximal sum for that round.
I'm fairly sure I can get the right answer 100% of the time, but the code exceeds the time limit of one second. I was wondering if you could help me optimize my code to help it run in time, or if you could explain why my code is inefficient.
#include <iostream>
#include <vector>
int main() {
int x, c = 0, big = 0;
std::cin >> x;
std::vector<int> as, bs;
for (int i = 0; i < x; i++) {
bool founda = false, foundb = false;
int a, b;
std::cin >> a >> b;
c++;
if (c == 1) {
as.push_back(a);
bs.push_back(b);
}
else {
for (int i = 0; i < c; i++) {
if (as[i] < a || i == c-1) {
as.insert(as.begin()+i, a);
founda = true;
}
if (bs[i] < b || i == c-1) {
bs.insert(bs.begin()+i, b);
foundb = true;
}
if (founda == true && foundb == true)
break;
}
}
for (int i = 0; i < c; i++) {
if (as[i] + bs[c-1-i] > big)
big = as[i] + bs[c-1-i];
}
std::cout << big << std::endl;
}
}
c++ programming-challenge time-limit-exceeded
c++ programming-challenge time-limit-exceeded
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edited 13 mins ago
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asked 33 mins ago
Griffin WongGriffin Wong
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Griffin Wong is a new contributor. Be nice, and check out our Code of Conduct.
Griffin Wong is a new contributor. Be nice, and check out our Code of Conduct.
Griffin Wong is a new contributor. Be nice, and check out our Code of Conduct.
Griffin Wong is a new contributor. Be nice, and check out our Code of Conduct.
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