QOJ.ac
QOJ
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 |
---|---|---|---|---|---|---|---|---|---|
#205783 | #7561. Digit DP | ucup-team052# | RE | 10ms | 29476kb | C++14 | 5.5kb | 2023-10-07 17:23:09 | 2023-10-07 17:23:10 |
Judging History
answer
#include <bits/stdc++.h>
#define rep(i, a, b) for (int i = a; i <= b; i++)
#define per(i, a, b) for (int i = a; i >= b; i--)
using namespace std;
typedef unsigned long long ull;
typedef pair <int, int> pii;
typedef long long ll;
template <typename _T>
inline void read(_T &f) {
f = 0; _T fu = 1; char c = getchar();
while (c < '0' || c > '9') { if (c == '-') { fu = -1; } c = getchar(); }
while (c >= '0' && c <= '9') { f = (f << 3) + (f << 1) + (c & 15); c = getchar(); }
f *= fu;
}
template <typename T>
void print(T x) {
if (x < 0) putchar('-'), x = -x;
if (x < 10) putchar(x + 48);
else print(x / 10), putchar(x % 10 + 48);
}
template <typename T>
void print(T x, char t) {
print(x); putchar(t);
}
typedef __int128 i128;
const int md = 998244353;
inline int add(int x, int y) {
if (x + y >= md) return x + y - md;
return x + y;
}
inline void addx(int &x, int y) {
x += y;
if (x >= md) x -= md;
}
inline int sub(int x, int y) {
if (x < y) return x - y + md;
return x - y;
}
inline void subx(int &x, int y) {
x -= y;
if (x < 0) x += md;
}
inline int mul(int x, int y) { return 1ull * x * y % md; }
inline int fpow(int x, int y) {
int ans = 1;
while (y) {
if (y & 1) ans = mul(ans, x);
y >>= 1; x = mul(x, x);
}
return ans;
}
const int N = 2e7 + 5;
struct atom {
int val[4], len;
};
atom w[N], base[100];
int lc[N], rc[N], tag[N];
int sum[5][1 << 20], pre[1 << 20], a[100]; char c[200];
int n, q, root, tot;
inline int lowbit(int x) { return x & -x; }
inline int getsum(i128 x) {
int ans = 0;
for (int i = 0; i < 5; i++) {
ans = add(ans, sum[i][x & 1048575]);
x >>= 20;
}
return ans;
}
int ifac[4];
int C(i128 n, int m) {
if (n < m) return 0;
int ans = ifac[m];
for (int i = 1; i <= m; i++) ans = mul(ans, (n - i + 1) % md);
return ans;
}
atom addatom(atom a, int b) {
if (!b) return a;
atom ans;
memset(ans.val, 0, sizeof(ans.val)); ans.len = a.len;
for (int i = 0; i <= 3; i++) {
for (int j = 0; j <= 3; j++) {
addx(ans.val[i + j], mul(a.val[i], mul(fpow(b, j), C(a.len - i, j))));
}
}
return ans;
}
atom getatom(int bit, i128 state) {
int tmp = getsum(state & (((i128)1 << 100) - ((i128)1 << bit)));
if (bit == 0) {
atom ans;
ans.val[0] = 1;
ans.val[1] = tmp;
ans.val[2] = ans.val[3] = 0;
return ans;
}
atom ans = base[bit - 1];
return addatom(ans, tmp);
}
atom merge(atom a, atom b) {
atom ans;
memset(ans.val, 0, sizeof(ans.val)); ans.len = add(a.len, b.len);
for (int i = 0; i <= 3; i++) {
for (int j = 0; j <= 3 - i; j++) {
addx(ans.val[i + j], mul(a.val[i], b.val[j]));
}
}
return ans;
}
void newNode(int &u, i128 l, i128 r, int bit) {
u = ++tot;
w[u] = getatom(bit, l);
}
void add(int u, i128 L, i128 R, i128 l, i128 r, int x, int bit) {
if (l <= L && R <= r) {
w[u] = addatom(w[u], x);
tag[u] = add(tag[u], x);
return;
}
i128 mid = (L + R) >> 1;
if (!lc[u]) newNode(lc[u], L, mid, bit - 1);
if (!rc[u]) newNode(rc[u], mid + 1, R, bit - 1);
if (mid >= l) add(lc[u], L, mid, l, r, x, bit - 1);
if (mid + 1 <= r) add(rc[u], mid + 1, R, l, r, x, bit - 1);
w[u] = addatom(merge(w[lc[u]], w[rc[u]]), tag[u]);
}
atom query(int u, i128 L, i128 R, i128 l, i128 r, int bit) {
if (l <= L && R <= r) return w[u];
i128 mid = (L + R) >> 1;
if (!lc[u]) newNode(lc[u], L, mid, bit - 1);
if (!rc[u]) newNode(rc[u], mid + 1, R, bit - 1);
atom ans; memset(ans.val, 0, sizeof(ans.val)); ans.val[0] = 1; ans.len = 0;
if (mid >= l) ans = query(lc[u], L, mid, l, r, bit - 1);
if (mid + 1 <= r) ans = merge(ans, query(rc[u], mid + 1, R, l, r, bit - 1));
return addatom(ans, tag[u]);
}
int main() {
ifac[0] = 1; ifac[1] = 1; ifac[2] = fpow(2, md - 2); ifac[3] = fpow(6, md - 2);
read(n); read(q);
for (int i = 0; i < n; i++) read(a[i]), a[i] %= md;
n = 100;
for (int i = 0; i < 20; i++) pre[1 << i] = i;
for (int i = 0; i < 5; i++) {
for (int j = 1; j < (1 << 20); j++) {
sum[i][j] = sum[i][j ^ lowbit(j)] + a[i * 20 + pre[lowbit(j)]];
}
}
base[0].val[0] = 1;
base[0].val[1] = a[0];
base[0].len = 2;
for (int i = 1; i < 100; i++) {
for (int l = 0; l <= 3; l++) {
for (int r = 0; r <= 3 - l; r++) {
for (int x = 0; x <= 3 - l - r; x++) {
addx(base[i].val[l + r + x], mul(base[i - 1].val[l], mul(base[i - 1].val[r], mul(fpow(a[i], x), C(((i128)1 << i) - r, x)))));
}
}
}
base[i].len = mul(base[i - 1].len, 2);
}
newNode(root, 0, 0, 100);
while (q--) {
int opt; read(opt);
i128 l = 0, r = 0;
scanf("%s", c); int len = strlen(c);
for (int i = 0; i < len; i++) {
if (c[i] == '1') l |= (i128)1 << i;
}
scanf("%s", c);
for (int i = 0; i < len; i++) {
if (c[i] == '1') r |= (i128)1 << i;
}
if (opt == 1) {
int x; read(x);
add(root, 0, ((i128)1 << 100) - 1, l, r, x, 100);
} else {
atom ans = query(root, 0, ((i128)1 << 100) - 1, l, r, 100);
print(ans.val[3], '\n');
}
}
return 0;
}
详细
Test #1:
score: 100
Accepted
time: 0ms
memory: 29476kb
input:
3 3 1 2 4 2 000 111 1 010 101 1 2 000 111
output:
1960 3040
result:
ok 2 number(s): "1960 3040"
Test #2:
score: 0
Accepted
time: 10ms
memory: 28308kb
input:
2 2 1 1 2 00 10 2 00 11
output:
0 2
result:
ok 2 number(s): "0 2"
Test #3:
score: -100
Runtime Error
input:
99 49952 470888 74578 802746 396295 386884 721198 628655 722503 207868 647942 87506 792718 761498 917727 843338 908043 952768 268783 375312 414369 319712 96230 277106 168102 263554 936674 246545 667941 198849 268921 191459 436316 134606 802932 515506 837311 465964 394766 17626 650419 984050 790137 4...