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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 |
---|---|---|---|---|---|---|---|---|---|
#205801 | #7561. Digit DP | ucup-team052# | RE | 10ms | 30220kb | C++14 | 5.5kb | 2023-10-07 17:27:22 | 2023-10-07 17:27:23 |
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++) {
l = l * 2 + (c[i] - '0');
}
scanf("%s", c);
for (int i = 0; i < len; i++) {
r = r * 2 + (c[i] - '0');
}
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: 10ms
memory: 30220kb
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: -100
Runtime Error
input:
2 2 1 1 2 00 10 2 00 11