few days ago I wrote a blog asking about if vectors can cause you TLE. and now I have faced such problem in solving this problem. and now I show you the codes that their solution is totally the same but the first one gets AC and the other one gets TLE. I only changed the vectors of the second one to arrays and it passed. please tell me if I'm wrong and TLE was for smth else.
AC solution:
//\\//\\ * * * //\\// ||
#include <bits/stdc++.h>
#define debug(x) cerr << #x << ": " << x << endl
using namespace std;
typedef long long ll;
int md;
inline void add(int& a, int b) {
a += b;
if (a >= md) {
a -= md;
}
}
inline int mul(int a, int b) {
return (int) ((ll) a * b % md);
}
const int N = (int) 1e4 + 10;
const int M = 2000;
bool is_prime[N];
int dp[N][M];
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
freopen("exercise.in", "r", stdin);
freopen("exercise.out", "w", stdout);
int n;
cin >> n >> md;
for (int i = 0; i <= n; i++) {
is_prime[i] = true;
}
is_prime[1] = is_prime[0] = false;
for (int i = 2; i <= n; i++) {
if (!is_prime[i]) {
continue;
}
for (int j = i + i; j <= n; j += i) {
is_prime[j] = false;
}
}
vector<int> p;
for (int i = 0; i <= n; i++) {
if (is_prime[i]) {
p.push_back(i);
}
}
int m = (int) p.size();
for (int i = 0; i <= m - 1; i++) {
dp[0][i] = 0;
dp[1][i] = 0;
}
for (int i = 2; i <= n; i++) {
int x = 2;
while (x <= i) {
x *= 2;
}
x /= 2;
if (x == i) {
add(dp[i][0], x);
}
for (int j = 1; j < m; j++) {
add(dp[i][j], dp[i][j - 1]);
int y = p[j];
while (y <= i) {
if (y == i) {
add(dp[i][j], y);
break;
}
add(dp[i][j], mul(dp[i - y][j - 1], y));
y *= p[j];
}
}
}
int ans = 0;
for (int i = 0; i <= n; i++) {
add(ans, dp[i][m - 1]);
}
add(ans, 1);
cout << ans << '\n';
return 0;
}
TLE solution:
/**
@@@@@@@ @@@@ @@@@ @@@@ @@@@ @
@ @ @ @ @ @ @ @ @ @ @ @
@ @ @ @ @ @ @ @ @ @ @ @
@ @ @ @ @ @ @ @ @ @ @ @
@ @ @ @@@@ @@@@ @@@@ @ @ @
**/
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#define db double
#define ld long double
#define str string
#define pi pair
#define tup tuple
#define vec vector
#define temp template
#define tn typename
#define ff first
#define ss second
#define SZ(x) (ll) x.size()
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define rsz resize
#define ins insert
#define fr front
#define bc back
#define pf push_front
#define pb push_back
#define eb emplace_back
#define lb lower_bound
#define ub upper_bound
#define mk make_pair
#define FOR(i, a, b) for (ll i = (a); i <= (b); i++)
#define ROF(i, a, b) for (ll i = (a); i >= (b); i--)
#define trav(a, x) for (auto &a : x)
#define popcount __builtin_popcountll
#define ctz __builtin_ctzll
#define clz __builtin_clzll
#define gcd __gcd
#define max(a, b) max((ll) a, (ll) b)
#define min(a, b) min((ll) a, (ll) b)
// DEBUG
#define ts to_string
temp<tn A, tn B> str ts(pi<A, B> p);
temp<tn A, tn B, tn C> str ts(tup<A, B, C> p);
str ts(const str& s) { return '"' + s + '"'; }
str ts(const char* s) { return ts((str) s); }
str ts(bool b) { return (b ? "true" : "false"); }
str ts(vec<bool> v) { bool f = true; str res = "{";
FOR(i, 0, static_cast<ll>(v.size()) - 1)
{ if (!f) res += ", "; res += ts(v[i]);
f = false; } res += "}"; return res; }
temp<size_t N> str ts(bitset<N> v) { str res = "";
trav(a, v) { res += (char) ('0' + a); } return res; }
temp<tn A> str ts(A v) { bool f = true; str res = "{";
trav(x, v) { if (!f) res += ", "; res += ts(x);
f = false; } res += "}"; return res; }
temp<tn A, tn B> str ts(pi<A, B> p) {
return "(" + ts(p.ff) + ", " + ts(p.ss) + ")"; }
temp<tn A, tn B, tn C> str ts(tup<A, B, C> t) {
return "(" + ts(get<0>(t)) + ", " + ts(get<1>(t)) + ", " + ts(get<2>(t)) + ")"; }
void DBG() { cerr << endl; }
temp<tn H, tn... T> void DBG(H h, T... t) { cerr << " " << ts(h); DBG(t...); }
#ifdef LOCAL
#define dbg(...) cerr << "[" << #__VA_ARGS__ << "]:", DBG(__VA_ARGS__);
#else
#define dbg(...) 42
#endif
// INPUT AND OUTPUT
temp<tn T> void re(T& x) { cin >> x; };
temp<tn H, tn... T> void re(H& h, T&... t) { re(h); re(t...); }
temp<tn T> void pr(T x) { cout << x; };
temp<tn H, tn... T> void pr(H h, T... t) { pr(h); pr(t...); }
temp<tn T> void pri(T x) { cout << x; cout << endl; }; // used for interactives
temp<tn H, tn... T> void pri(H h, T... t) { pri(h); pri(t...); }
/**
* Name: Modular operations
* Description:
* used for doing modular operations
**/
ll md;
ll nrm(ll x) { while (x < (ll) 0) x += md; x %= md; return x; }
ll sum() { return (ll) 0; }
temp <tn H, tn... T> ll sum(H h, T... t) { return nrm(nrm(h) + sum(t...)); }
ll mul() { return (ll) 1; }
temp <tn H, tn... T> ll mul(H h, T... t) { return nrm(nrm(h) * mul(t...)); }
ll power(ll a, ll b) {
ll res = 1;
while (b) { if (b & 1) res = mul(res, a); a = mul(a, a); b >>= 1; }
return res;
}
ll inv(ll x) { return power(x, md - 2); }
// pay attention to md's value first!
// use normalized form of modular integers for output!
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
freopen("exercise.in", "r", stdin);
freopen("exercise.out", "w", stdout);
ll n;
re(n, md);
vec<ll> is_prime(n + 1, true);
is_prime[1] = is_prime[0] = false;
FOR(i, 2, n) {
if (!is_prime[i]) {
continue;
}
for (ll j = i + i; j <= n; j += i) {
is_prime[j] = false;
}
}
vec<ll> p;
FOR(i, 0, n) {
if (is_prime[i]) {
p.pb(i);
}
}
// dbg(p);
ll m = SZ(p);
vec<vec<ll>> dp(n + 1, vec<ll>(m));
FOR(i, 0, m - 1) {
dp[0][i] = 0;
dp[1][i] = 0;
}
FOR(i, 2, n) {
ll x = 2;
while (x <= i) {
x *= 2;
}
x /= 2;
if (x == i) {
dp[i][0] = sum(dp[i][0], x);
}
FOR(j, 1, m - 1) {
dp[i][j] = sum(dp[i][j], dp[i][j - 1]);
ll y = p[j];
while (y <= i) {
if (y == i) {
dp[i][j] = sum(dp[i][j], y);
break;
}
dp[i][j] = sum(dp[i][j], mul(dp[i - y][j - 1], y));
y *= p[j];
}
}
}
// dbg(dp);
ll ans = 0;
FOR(i, 0, n) {
ans = sum(ans, dp[i][m - 1]);
}
ans = sum(ans, 1);
pr(ans, '\n');
return 0;
}
