結果
問題 | No.2215 Slide Subset Sum |
ユーザー |
👑 |
提出日時 | 2023-02-05 14:05:12 |
言語 | C (gcc 13.3.0) |
結果 |
AC
|
実行時間 | 2,810 ms / 3,000 ms |
コード長 | 4,964 bytes |
コンパイル時間 | 262 ms |
コンパイル使用メモリ | 34,536 KB |
実行使用メモリ | 82,344 KB |
最終ジャッジ日時 | 2024-07-04 06:52:58 |
合計ジャッジ時間 | 41,389 ms |
ジャッジサーバーID (参考情報) |
judge2 / judge1 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 2 |
other | AC * 45 |
ソースコード
#include <stdio.h>const int Mod = 998244353,bit[21] = {1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768, 65536, 131072, 262144, 524288, 1048576},bit_inv[21] = {1, 499122177, 748683265, 873463809, 935854081, 967049217, 982646785, 990445569, 994344961, 996294657, 997269505, 997756929,998000641, 998122497, 998183425, 998213889, 998229121, 998236737, 998240545, 998242449, 998243401},root[21] = {1, 998244352, 911660635, 372528824, 929031873, 452798380, 922799308, 781712469, 476477967, 166035806, 258648936, 584193783, 63912897,350007156, 666702199, 968855178, 629671588, 24514907, 996173970, 363395222, 565042129},root_inv[21] = {1, 998244352, 86583718, 509520358, 337190230, 87557064, 609441965, 135236158, 304459705, 685443576, 381598368, 335559352,129292727, 358024708, 814576206, 708402881, 283043518, 3707709, 121392023, 704923114, 950391366};void NTT_inline(int kk, int a[], int x[]){int h, hh, i, ii, j, jj, k, l, r = bit[kk], d = bit[kk-1], tmpp, cur, prev;int *pi, *pii, *pj, *pjj;static int y[2][256];long long tmp;for (i = 0; i < r; i++) y[0][i] = a[i];for (k = 1, kk--, cur = 1, prev = 0; kk >= 0; k++, kk--, cur ^= 1, prev ^= 1) {for (h = 0, tmp = 1; h << (kk + 1) < r; h++, tmp = tmp * root[k] % Mod) {for (hh = 0, pi = &(y[cur][h<<kk]), pii = pi + d, pj = &(y[prev][h<<(kk+1)]), pjj = pj + bit[kk]; hh < bit[kk]; hh++, pi++, pii++, pj++,pjj++) {tmpp = tmp * (*pjj) % Mod;*pi = *pj + tmpp;if (*pi >= Mod) *pi -= Mod;*pii = *pj - tmpp;if (*pii < 0) *pii += Mod;}}}for (i = 0; i < r; i++) x[i] = y[prev][i];}void NTT_reverse_inline(int kk, int a[], int x[]){int h, hh, i, ii, j, jj, k, l, r = bit[kk], d = bit[kk-1], tmpp, cur, prev;int *pi, *pii, *pj, *pjj;static int y[2][256];long long tmp;for (i = 0; i < r; i++) y[0][i] = a[i];for (k = 1, kk--, cur = 1, prev = 0; kk >= 0; k++, kk--, cur ^= 1, prev ^= 1) {for (h = 0, tmp = 1; h << (kk + 1) < r; h++, tmp = tmp * root_inv[k] % Mod) {for (hh = 0, pi = &(y[cur][h<<kk]), pii = pi + d, pj = &(y[prev][h<<(kk+1)]), pjj = pj + bit[kk]; hh < bit[kk]; hh++, pi++, pii++, pj++,pjj++) {tmpp = tmp * (*pjj) % Mod;*pi = *pj + tmpp;if (*pi >= Mod) *pi -= Mod;*pii = *pj - tmpp;if (*pii < 0) *pii += Mod;}}}for (i = 0; i < r; i++) x[i] = y[prev][i];}// Compute the product of two polynomials a[0-da] and b[0-db] using NTT in O(d * log d) timevoid prod_poly_NTT(int da, int db, int a[], int b[], int c[]){int i, k;static int aa[256], bb[256], cc[256];for (k = 0; bit[k] <= da + db; k++);for (i = 0; i <= da; i++) aa[i] = a[i];for (i = da + 1; i < bit[k]; i++) aa[i] = 0;for (i = 0; i <= db; i++) bb[i] = b[i];for (i = db + 1; i < bit[k]; i++) bb[i] = 0;static int x[256], y[256], z[256];NTT_inline(k, aa, x);if (db == da) {for (i = 0; i <= da; i++) if (a[i] != b[i]) break;if (i <= da) NTT_inline(k, bb, y);else for (i = 0; i < bit[k]; i++) y[i] = x[i];} else NTT_inline(k, bb, y);for (i = 0; i < bit[k]; i++) z[i] = (long long)x[i] * y[i] % Mod;NTT_reverse_inline(k, z, cc);for (i = 0; i <= da + db; i++) c[i] = (long long)cc[i] * bit_inv[k] % Mod;}// Compute the product of two polynomials a[0-da] and b[0-db] naively in O(da * db) timevoid prod_poly_naive(int da, int db, int a[], int b[], int c[]){int i, j;static long long tmp[256];for (i = 0; i <= da + db; i++) tmp[i] = 0;for (i = 0; i <= da; i++) for (j = 0; j <= db; j++) tmp[i+j] += (long long)a[i] * b[j] % Mod;for (i = 0; i <= da + db; i++) c[i] = tmp[i] % Mod;}// Compute the product of two polynomials a[0-da] and b[0-db] in an appropriate wayvoid prod_polynomial(int da, int db, int a[], int b[], int c[]){if (da <= 70 || db <= 70) prod_poly_naive(da, db, a, b, c);else prod_poly_NTT(da, db, a, b, c);}int main(){int i, N, M, K, A[200001];scanf("%d %d %d", &N, &M, &K);for (i = 1; i <= N; i++) scanf("%d", &(A[i]));int j, k, kk, l, ans[200001], tmp[101], tmpp[256];static int prod[200001][101];for (l = 1; l + M - 1 <= N; l += M + 1) {for (k = 1, prod[l+M][0] = 1; k < K; k++) prod[l+M][k] = 0;for (i = l + M - 1; i >= l; i--) {for (k = 0; k < K; k++) prod[i][k] = prod[i+1][k];for (k = 0, kk = A[i]; k < K; k++, kk++) {if (kk == K) kk = 0;prod[i][kk] += prod[i+1][k];if (prod[i][kk] >= Mod) prod[i][kk] -= Mod;}}ans[l] = prod[l][0] - 1;for (k = 1, tmp[0] = 1; k < K; k++) tmp[k] = 0;for (i = l + 1; i <= l + M && i + M - 1 <= N; i++) {for (k = 0; k < K; k++) tmpp[k] = tmp[k];for (k = 0, kk = A[i+M-1]; k < K; k++, kk++) {if (kk == K) kk = 0;tmp[kk] = tmpp[k] + tmpp[kk];if (tmp[kk] >= Mod) tmp[kk] -= Mod;}prod_polynomial(K - 1, K - 1, prod[i], tmp, tmpp);ans[i] = tmpp[0] + tmpp[K] - 1;if (ans[i] >= Mod) ans[i] -= Mod;}}for (i = 1; i <= N - M + 1; i++) printf("%d\n", ans[i]);fflush(stdout);return 0;}