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ID | 题目 | 提交者 | 结果 | 用时 | 内存 | 语言 | 文件大小 | 提交时间 | 测评时间 |
---|---|---|---|---|---|---|---|---|---|
#382311 | #7782. Ursa Minor | nguyentunglam | WA | 93ms | 19924kb | C++17 | 3.3kb | 2024-04-08 11:59:51 | 2024-04-08 11:59:52 |
Judging History
answer
#include<bits/stdc++.h>
#define all(v) v.begin(), v.end()
#define endl "\n"
//#define int long long
using namespace std;
const int N = 2e5 + 10, T1 = 0, T2 = 300;
const long long base = 177013, mod = 1e9 + 7;
long long pw[N], f[N], a[N], sum[T1 + 10][N / T2 + 10];
long long sum_block[N / T2 + 10], pref_block[N / T2 + 10], pref[N];
int lg[N], rmq[20][N];
int b[N];
int n, m, q;
void handle (int p) {
long long old = f[p];
f[p] = p > 1 ? a[p] - a[p - 1] : a[p] - a[n];
for(int c = 1; c <= T1; c++) {
sum[c][p / T2] += (f[p] - old) * pw[p % c];
}
sum_block[p / T2] += (f[p] - old) * pw[p];
sum_block[p / T2] %= mod;
pref_block[0] = sum_block[0];
for(int i = 1; i <= n / T2; i++) pref_block[i] = pref_block[i - 1] + sum_block[i];
for(int i = max(1, p / T2 * T2); i <= min(n, p / T2 * T2 + T2 - 1); i++) {
pref[i] = pref[i - 1] + f[i] * pw[i] % mod;
}
}
long long get (int l, int r) {
if (l / T2 == r / T2) return pref[r] - pref[l - 1];
return pref[r] - pref[r / T2 * T2] + pref[l / T2 * T2 + T2 - 1] - pref[l - 1] + pref_block[r / T2 - 1] - pref_block[l / T2];
}
void update(int p, int v) {
a[p] = v;
handle(p);
if (p + 1 <= n) handle(p + 1);
else handle(1);
}
tuple<int, int, int, int> query[N];
int get_gcd (int l, int r) {
int k = lg[r - l + 1];
return __gcd(rmq[k][l], rmq[k][r - (1 << k) + 1]);
}
int32_t main() {
#define task ""
cin.tie(0) -> sync_with_stdio(0);
if (fopen("task.inp", "r")) {
freopen("task.inp", "r", stdin);
freopen("task.out", "w", stdout);
}
if (fopen(task".inp", "r")) {
freopen (task".inp", "r", stdin);
freopen (task".out", "w", stdout);
}
cin >> n >> m >> q;
pw[0] = 1;
for(int i = 1; i <= n; i++) {
pw[i] = pw[i - 1] * base % mod;
}
for(int i = 1; i <= n; i++) {
cin >> a[i];
update(i, a[i]);
}
for(int i = 2; i <= m; i++) lg[i] = lg[i / 2] + 1;
for(int i = 1; i <= m; i++) cin >> rmq[0][i];
for(int j = 1; j <= lg[m]; j++) for(int i = 1; i + (1 << j) - 1 <= m; i++) {
rmq[j][i] = __gcd(rmq[j - 1][i], rmq[j - 1][i + (1 << j - 1)]);
}
while (q--) {
char type; cin >> type;
if (type == 'U') {
int p, v; cin >> p >> v;
update(p, v);
}
else {
int x, y, u, v; cin >> x >> y >> u >> v;
int c = __gcd(y - x + 1, get_gcd(u, v));
long long hs = 0;
if (c <= T1) {
int st = x / T2;
int ed = y / T2;
for(int i = st + 1; i < ed; i++) hs += sum[c][i];
if (st == ed) {
for(int i = x; i <= y; i++) hs += f[i] * pw[i % c];
}
else {
for(int i = x; i <= st * T2 + T2 - 1; i++) hs += f[i] * pw[i % c];
for(int i = y; i >= ed * T2; i--) hs += f[i] * pw[i % c];
}
hs -= f[x] * pw[x % c];
hs += (a[x] - a[y]) * pw[x % c];
// cout << hs << endl;
cout << (hs == 0 ? "Yes\n" : "No\n");
}
else {
for(int i = x; i <= y; i += c) {
hs += get(i, i + c - 1) % mod * pw[n - i] % mod;
}
hs -= f[x] * pw[n];
hs += (a[x] - a[y]) * pw[n];
hs %= mod;
// cout << hs << endl;
cout << (hs == 0 ? "Yes\n" : "No\n");
}
}
}
}
详细
Test #1:
score: 100
Accepted
time: 2ms
memory: 11800kb
input:
6 4 5 1 1 4 5 1 4 3 3 2 4 Q 1 5 1 2 Q 2 5 3 4 U 5 2 Q 1 6 1 2 Q 2 5 3 4
output:
Yes No No Yes
result:
ok 4 tokens
Test #2:
score: 0
Accepted
time: 0ms
memory: 11916kb
input:
1 1 1 0 1 Q 1 1 1 1
output:
Yes
result:
ok "Yes"
Test #3:
score: -100
Wrong Answer
time: 93ms
memory: 19924kb
input:
2000 2000 200000 1 1 2 0 0 2 0 2 0 0 0 0 0 2 2 1 2 0 0 2 2 2 1 0 1 2 1 2 0 0 1 1 1 2 0 0 2 2 2 2 0 2 0 0 2 1 2 0 0 1 2 2 1 0 2 0 0 0 1 2 2 1 2 2 0 0 1 1 1 0 0 2 0 0 1 1 0 2 2 2 1 0 0 1 0 1 2 2 2 1 1 2 2 1 2 1 0 2 2 3 1 3 2 3 1 0 1 2 0 1 1 1 0 2 2 3 2 0 3 2 3 3 1 2 3 1 2 0 1 0 3 1 0 0 2 0 1 2 1 3 2 2...
output:
Yes Yes No Yes Yes No No No Yes Yes Yes Yes Yes Yes Yes No Yes Yes Yes Yes Yes Yes Yes No Yes Yes Yes No Yes Yes No No No No No Yes No No No Yes Yes No Yes No Yes Yes Yes Yes Yes Yes Yes Yes Yes No Yes Yes No Yes Yes Yes No No Yes No Yes No No No Yes Yes Yes Yes Yes Yes Yes Yes Yes No Yes Yes Yes No...
result:
wrong answer 111th words differ - expected: 'Yes', found: 'No'