QOJ.ac
QOJ
ID | Problem | Submitter | Result | Time | Memory | Language | File size | Submit time | Judge time |
---|---|---|---|---|---|---|---|---|---|
#761324 | #9747. 字符串复制 | Symmetree | TL | 1ms | 8004kb | C++17 | 2.3kb | 2024-11-18 22:03:44 | 2024-11-18 22:03:44 |
Judging History
answer
#include<bits/stdc++.h>
const int N = 3e5+5, base = 233, mod = 998244353;
using u64 = unsigned long long;
using L = __uint128_t;
struct FastMod {
u64 b, m;
FastMod(u64 b): b(b), m(u64((L(1) << 64) / b)) {}
u64 reduce(u64 a) {
u64 q = (u64)((L(m) * a) >> 64);
u64 r = a - q * b;
return r >= b ? r - b: r ;
}
} F(mod);
int n, m, len, hash[N * 3], pw[N * 3], rk[N * 3]; char s[N * 3];
inline int ask(int l, int r) {
return F.reduce(hash[r] - F.reduce(1ll * hash[l - 1] * pw[r - l + 1]) + mod);
}
bool cmp(int a, int b) {
if(a > b) return !cmp(b, a);
if(a == b) return false;
if(s[a] != s[b]) return s[a] < s[b];
int l = 1, r = std::min(len - a + 1, len - b + 1);
while(l < r) {
int mid = (l + r) >> 1;
if(ask(a, a + mid - 1) ^ ask(b, b + mid - 1)) r = mid;
else l = mid + 1;
}
if(ask(a, a + l - 1) == ask(b, b + l - 1)) return true;
else return s[a + l - 1] < s[b + l - 1];
}
int lcp(int a, int b) {
if(a > b) return lcp(b, a);
if(s[a] != s[b]) return 0;
int l = 1, r = std::min(len - a + 1, len - b + 1);
while(l < r) {
int mid = (l + r) >> 1;
if(ask(a, a + mid - 1) ^ ask(b, b + mid - 1)) r = mid;
else l = mid + 1;
}
if(ask(a, a + l - 1) ^ ask(b, b + l - 1)) --l;
return l;
}
int solve() {
hash[1] = s[1] - 'a';
for(int i = 2; i <= len; ++i) hash[i] = F.reduce(1ll * hash[i - 1] * base + (s[i] - 'a'));
for(int i = 1; i <= len; ++i) rk[i] = i;
std::sort(rk + 1, rk + len + 1, cmp);
int ans = 1ll * len * (len + 1) / 2 % mod;
//for(int i = 1; i <= n; ++i) printf("%d ", rk[i]); puts("");
for(int i = 1; i < len; ++i) ans = F.reduce(ans + mod - lcp(rk[i], rk[i + 1]));
return ans;
}
signed main() {
scanf("%d%d%s", &n, &m, s + 1), pw[0] = 1;
for(int i = 1; i <= n * 3; ++i) pw[i] = F.reduce(1ll * pw[i - 1] * base);
for(int i = 1; i <= n; ++i) s[i + n] = s[i + 2 * n] = s[i];
if(m <= 3) {
len = n * m;
printf("%d\n", solve());
return 0;
}
int res2, res3;
len = 3 * n, res3 = solve();
len = 2 * n, res2 = solve();
// printf("%d %d\n", res2, res3);
printf("%lld\n", (res2 + 1ll * (res3 - res2 + mod) % mod * (m - 2) % mod) % mod);
return 0;
}
Details
Tip: Click on the bar to expand more detailed information
Test #1:
score: 100
Accepted
time: 1ms
memory: 7872kb
input:
6 2 mantle
output:
57
result:
ok single line: '57'
Test #2:
score: 0
Accepted
time: 0ms
memory: 7956kb
input:
12 1919810 ifamjlifamjl
output:
138226305
result:
ok single line: '138226305'
Test #3:
score: 0
Accepted
time: 0ms
memory: 8004kb
input:
13 935330878 aabbbbababbaa
output:
348310505
result:
ok single line: '348310505'
Test #4:
score: -100
Time Limit Exceeded
input:
300000 1000000000 rqwcfnskpxmdplyemxtntenuvjcbtggljuravmoaipbujlseqtuakblnqzbirxowfaeykxwldpeovyuuvjvskbllmtdlsfswoklvdpuuujryuvzbabieklgkcbqdocbmmwsjlfiqxvwocsexuldtiaucurmiewfrggbnbfmxqtaabxnbzcjnavvuuvowvyazzmgtdwvjlmwxpxontrusqgipfvfsjtckxdrloofzmjxxhlexhqcjchgryycbybcnmbjsaffaeaitkalawjuporbnzo...