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 23:02:22 |
| 言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
| 結果 |
WA
|
| 実行時間 | - |
| コード長 | 1,198 bytes |
| コンパイル時間 | 438 ms |
| コンパイル使用メモリ | 41,216 KB |
| 最終ジャッジ日時 | 2025-01-11 07:42:53 |
|
ジャッジサーバーID (参考情報) |
judge5 / judge2 |
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| ファイルパターン | 結果 |
|---|---|
| sample | AC * 5 |
| other | AC * 30 WA * 1 |
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
#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;
}
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;
}
kriii