結果

問題 No.194 フィボナッチ数列の理解(1)
ユーザー tottoripapertottoripaper
提出日時 2015-04-26 23:58:24
言語 C++11
(gcc 13.3.0)
結果
AC  
実行時間 43 ms / 5,000 ms
コード長 2,922 bytes
コンパイル時間 372 ms
コンパイル使用メモリ 51,820 KB
実行使用メモリ 9,388 KB
最終ジャッジ日時 2024-07-05 03:05:12
合計ジャッジ時間 1,961 ms
ジャッジサーバーID
(参考情報)
judge2 / judge5
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 * 3
other AC * 37
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:123:10: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result]
  123 |     scanf("%lld %lld", &N, &K);
      |     ~~~~~^~~~~~~~~~~~~~~~~~~~~
main.cpp:126:14: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result]
  126 |         scanf("%lld", A+i);
      |         ~~~~~^~~~~~~~~~~~~

ソースコード

diff #
プレゼンテーションモードにする

// Wrongri-La Shower
#include <cstdio>
#include <string>
#include <tuple>
typedef long long ll;
typedef std::tuple<ll,ll> P;
struct Matrix{
Matrix(){
for(int i=0;i<=60;i++){
for(int j=0;j<=60;j++){
element[i][j] = 0;
}
}
}
ll element[61][61];
};
const ll MOD = 1000000000ll + 7;
ll N, K;
ll A[10001], B[1000001];
P solve(){
ll sum = 0ll, sum2 = 0ll;
for(int i=1;i<=K;i++){
if(i <= N){
B[i] = A[i];
sum = (sum + B[i]) % MOD;
}else{
B[i] = sum;
sum = (sum + B[i] + (-B[i-N] % MOD + MOD) % MOD) % MOD;
}
sum2 = (sum2 + B[i]) % MOD;
}
return std::make_tuple(B[K], sum2);
}
Matrix multiply(const Matrix &matrix, const Matrix &matrix2, int n){
Matrix mat;
for(int i=1;i<=n;i++){
for(int j=1;j<=n;j++){
mat.element[i][j] = 0ll;
}
}
for(int i=1;i<=n;i++){
for(int k=1;k<=n;k++){
for(int j=1;j<=n;j++){
mat.element[i][j] = (mat.element[i][j] + matrix.element[i][k] * matrix2.element[k][j] % MOD) % MOD;
}
}
}
return mat;
}
P solve2(){
if(K <= N){
ll sum = 0ll;
for(int i=1;i<=K;i++){
sum = (sum + A[i]) % MOD;
}
return std::make_tuple(A[K] % MOD, sum);
}
Matrix matrix, matrix2, matrix3;
for(int i=1;i<=N;i++){
for(int j=1;j<=N;j++){
if(i == 1){
matrix.element[i][j] = 1;
}else if(j+1 == i){
matrix.element[i][j] = 1;
}
}
matrix.element[i+N][i] = 1;
matrix.element[i+N][i+N] = 1;
}
for(int i=1;i<=2*N;i++){
matrix2.element[i][i] = 1;
}
matrix3 = matrix;
ll k = K-1;
while(k > 0){
if(k & 1){
matrix2 = multiply(matrix2, matrix, 2*N);
}
k >>= 1;
matrix = multiply(matrix, matrix, 2*N);
}
// for(int i=1;i<=N;i++){
// for(int j=1;j<=N;j++){
// printf("%lld%c", matrix2.element[i][j], ",\n"[j==N]);
// }
// }
// puts(std::string(40, '-').c_str());
ll value = 0ll, sum = 0ll;
for(int i=1;i<=N;i++){
value = (value + matrix2.element[N][i] * A[N-i+1] % MOD) % MOD;
}
matrix2 = multiply(matrix2, matrix3, 2*N);
for(int i=1;i<=N;i++){
sum = (sum + matrix2.element[2*N][i] * A[N-i+1] % MOD) % MOD;
}
// for(int i=1;i<=N;i++){
// res = (res + matrix2.element[1][i] * A[N-i+1] % MOD) % MOD;
// }
return std::make_tuple(value, sum);
}
int main(){
scanf("%lld %lld", &N, &K);
for(int i=1;i<=N;i++){
scanf("%lld", A+i);
}
ll x, y;
if(K <= 1000000){
std::tie(x, y) = solve();
}else{
std::tie(x, y) = solve2();
}
printf("%lld %lld\n", x, y);
}
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