結果

問題 No.1086 桁和の桁和2
ユーザー kriii
提出日時 2020-06-19 23:05:15
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 131 ms / 3,000 ms
コード長 1,094 bytes
コンパイル時間 370 ms
コンパイル使用メモリ 40,704 KB
最終ジャッジ日時 2025-01-11 07:46:32
ジャッジサーバーID
(参考情報)
judge5 / judge3
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 * 31
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:18:15: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result]
   18 |         scanf ("%d", &N);
      |         ~~~~~~^~~~~~~~~~
main.cpp:19:43: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result]
   19 |         for (int i = 0; i < N; i++) scanf ("%lld", &L[i]);
      |                                     ~~~~~~^~~~~~~~~~~~~~~
main.cpp:20:43: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result]
   20 |         for (int i = 0; i < N; i++) scanf ("%lld", &R[i]);
      |                                     ~~~~~~^~~~~~~~~~~~~~~
main.cpp:21:43: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result]
   21 |         for (int i = 0; i < N; i++) scanf ("%d", &D[i]);
      |                                     ~~~~~~^~~~~~~~~~~~~

ソースコード

diff #
プレゼンテーションモードにする

#include <stdio.h>
#include <algorithm>
using namespace std;
int N, D[100100];
long long L[100100], R[100100], B[77], BB[77];
const int mod = 1000000007;
int main()
{
B[0] = 10; BB[0] = 1;
for (int i = 1; i < 70; i++){
B[i] = B[i - 1] * B[i - 1] % mod;
BB[i] = BB[i - 1] * (B[i - 1] + 1) % mod;
}
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;
for (int i = 0; i < N; i++){
int p = 0;
long long upp = R[i] - L[i];
for (int b = 0; upp; b++){
if (upp & 1) p = (B[b] * p + BB[b]) % mod;
upp /= 2;
}
long long low = L[i];
for (int b = 0; low; b++){
if (low & 1) p = B[b] * p % mod;
low /= 2;
}
ans = ans * (p + (i > 0 && D[i - 1] == D[i])) % mod;
}
printf ("%lld\n", ans);
return 0;
}
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