結果
問題 | No.1086 桁和の桁和2 |
ユーザー |
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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 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 5 |
other | AC * 31 |
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;}printf ("%lld\n", ans);return 0;}