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QOJ
ID | Problem | Submitter | Result | Time | Memory | Language | File size | Submit time | Judge time |
---|---|---|---|---|---|---|---|---|---|
#765284 | #8049. Equal Sums | The_cosmos | ML | 1ms | 5776kb | C++14 | 2.2kb | 2024-11-20 13:35:39 | 2024-11-20 13:35:39 |
Judging History
answer
#include <bits/stdc++.h>
using namespace std;
using db = double;
using lb = long double;
using ull = unsigned long long;
using uint = unsigned int;
using ll = long long;
//using i128 = __int128;
//using ui128 = unsigned __int128;
#define REP(i, first, last) for(int i = (first); i <= (last); ++ i)
#define DOW(i, first, last) for(int i = (first); i >= (last); -- i)
#define int long long
#define pb emplace_back
#define ob pop_back
#define pii pair<int, int>
#define MmodR make_pair
#define fi first
#define se second
#define tpl tuple<int, int, int>
#define MTmod make_tuple
#define poly vector<int>
#define polyp vector<pii>
#define polyt vector<tpl>
#define all(x) x.begin(), x.end()
#define CVR(a, q) memset(a, q, sizeof a)
#define CLR(a) memset(a, 0, sizeof a)
#define CmodY(a, q) memcpy(a, q, sizeof a)
#define modCT __builtin_popcount
const int N = 500 + 5, M = 1e6 + 5;
const int mod1 = 1e9 + 7, mod2 = 998244353, mod = mod2;
const long long inf = 1e18;
const int dx[] = {-1, -1, -1, 0, 0, 0, 1, 1, 1};
const int dy[] = {-1, 0, 1, -1, 0, 1, -1, 0, 1};
mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
int n, m, f[N][N][N << 1], s[N << 1];
int a[N][2], b[N][2];
void Add(int& x, int y) {
return (x += y) >= mod ? x -= mod : x, void();
}
void Solve() {
cin >> n >> m;
REP(i, 1, n) cin >> a[i][0] >> a[i][1];
REP(i, 1, m) cin >> b[i][0] >> b[i][1];
f[0][0][N] = 1;
REP(i, 0, n) REP(j, 0, m) {
if (i > 0) {
REP(k, -500, 500) s[k + N] = f[i - 1][j][k + N], Add(s[k + N], s[k + N - 1]);
REP(k, 0, 500) Add(f[i][j][k + N], s[k + N - a[i][0]]), Add(f[i][j][k + N], mod - s[k + N - a[i][1] - 1]);
}
if (j > 0) {
DOW(k, 500, -500) s[k + N] = f[i][j - 1][k + N], Add(s[k + N], s[k + N + 1]);
REP(k, -500, -1) Add(f[i][j][k + N], s[k + N + b[j][0]]), Add(f[i][j][k + N], mod - s[k + N + b[j][1] + 1]);
}
}
REP(i, 1, n) REP(j, 1, m)
cout << f[i][j][N] << " \n"[j == m];
}
signed main() {
// freopen(".in", "r", stdin);
// freopen(".out", "w", stdout);
ios_base::sync_with_stdio(false);
cin.tie(nullptr), cout.tie(nullptr);
int T = 1;
// cin >> T;
while (T --) {
Solve();
}
return 0;
}
Details
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Test #1:
score: 100
Accepted
time: 1ms
memory: 5776kb
input:
2 3 1 2 2 3 1 4 2 2 1 3
output:
2 0 0 3 4 4
result:
ok 6 numbers
Test #2:
score: -100
Memory Limit Exceeded
input:
500 500 19 458 1 480 7 485 50 461 12 476 15 461 48 466 40 453 46 467 9 458 27 478 26 472 46 459 29 490 6 500 17 487 48 484 28 472 28 459 25 480 4 491 29 481 36 460 2 491 44 499 22 473 20 458 4 483 27 471 2 496 11 461 43 450 2 478 37 466 15 459 42 482 7 451 19 455 2 453 47 475 48 450 1 474 46 471 9 4...
output:
411 79401 9145270 673005095 180581065 984223118 586589234 293043270 404363796 865361724 665487988 118838806 926189944 226338288 521479857 808644951 786041288 340769021 177100 21 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 ...