結果

問題 No.1086 桁和の桁和2
コンテスト
ユーザー kriii
提出日時 2020-06-19 22:27:20
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
TLE  
実行時間 -
コード長 2,077 bytes
コンパイル時間 512 ms
コンパイル使用メモリ 46,080 KB
最終ジャッジ日時 2025-01-11 07:01:52
ジャッジサーバーID
（参考情報） 		judge5 / judge1
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 5
other AC * 11 WA * 1 TLE * 19
権限があれば一括ダウンロードができます
コンパイルメッセージ
main.cpp: In function ‘int main()’:
main.cpp:57:15: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result]
   57 |         scanf ("%d", &N);
      |         ~~~~~~^~~~~~~~~~
main.cpp:58:43: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result]
   58 |         for (int i = 0; i < N; i++) scanf ("%lld", &L[i]);
      |                                     ~~~~~~^~~~~~~~~~~~~~~
main.cpp:59:43: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result]
   59 |         for (int i = 0; i < N; i++) scanf ("%lld", &R[i]);
      |                                     ~~~~~~^~~~~~~~~~~~~~~
main.cpp:60:43: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result]
   60 |         for (int i = 0; i < N; i++) scanf ("%d", &D[i]);
      |                                     ~~~~~~^~~~~~~~~~~~~

ソースコード

diff # 

#include <stdio.h>
#include <algorithm>
using namespace std;

int N, D[100100];
long long L[100100], R[100100];

const long long mod = 1000000007;
struct matrix{
	matrix(){
		for (int i = 0; i < 9; i++) for (int j = 0; j < 9; j++) a[i][j] = 0;
	}
	long long a[9][9];

	matrix operator +(matrix t){
		matrix r;
		for (int i = 0; i < 9; i++) for (int j = 0; j < 9; j++){
			r.a[i][j] = (a[i][j] + t.a[i][j]) % mod;
		}
		return r;
	}
	matrix operator *(matrix t){
		matrix r;
		for (int i = 0; i < 9; i++) for (int j = 0; j < 9; j++){
			r.a[i][j] = 0;
			for (int k = 0; k < 9; k++){
				r.a[i][j] = (r.a[i][j] + a[i][k] * t.a[k][j]) % mod;
			}
		}
		return r;
	}
}iden, all, base[70];

pair<matrix, matrix> fpow(int i, long long n)
{
	if (n <= 0) return { iden, matrix() };
	if (n == 1) return { base[i], iden };

	auto half = fpow(i + 1, n / 2);
	half.second = half.second * (iden + base[i]);
	if (n % 2){
		half.second = half.second + half.first;
		half.first = half.first * base[i];
	}
	return half;
}

int main()
{
	for (int i = 0; i < 9; i++){
		iden.a[i][i] = 1;
		for (int j = 0; j < 9; j++) all.a[i][j] = 1;
	}
	for (int i = 0; i < 9; i++) for (int j = 0; j < 10; j++) base[0].a[i][(i + j) % 9]++;
	for (int i = 1; i < 70; i++) base[i] = base[i - 1] * base[i - 1];

	scanf ("%d", &N);
	for (int i = 0; i < N; i++) scanf ("%lld", &L[i]);
	for (int i = 0; i < N; i++) scanf ("%lld", &R[i]);
	for (int i = 0; i < N; i++) scanf ("%d", &D[i]);

	int s = 0;
	while (s < N && D[s] == 0) s++;

	for (int i = s; i < N; i++) if (D[i] == 0){
		puts("0");
		return 0;
	}

	for (int i = s; i < N; i++){
		L[i - s] = L[i];
		R[i - s] = R[i];
		D[i - s] = D[i];
	}
	N -= s;

	long long ans = 1; int ld = 0;
	for (int i = 0; i < N; i++){
		int nd = (D[i] + 9 - ld) % 9;

		auto low = fpow(0, L[i]);
		auto upp = fpow(0, R[i] - L[i]);

		matrix p = all * low.first * upp.second;
		ans = ans * (p.a[0][nd] + (nd == 0)) % mod;
		ld = nd;
	}

	bool g = N > 0;
	for (int i = 0; i < N; i++) if (D[i] != 9) g = 0;
	if (g) ans = (ans + mod - 1) % mod;
	
	printf ("%lld\n", ans);

	return 0;
}
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