結果
問題 |
No.1097 Remainder Operation
|
ユーザー |
|
提出日時 | 2020-06-26 21:38:35 |
言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 275 ms / 2,000 ms |
コード長 | 1,712 bytes |
コンパイル時間 | 1,170 ms |
コンパイル使用メモリ | 128,964 KB |
最終ジャッジ日時 | 2025-01-11 11:03:44 |
ジャッジサーバーID (参考情報) |
judge2 / judge1 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 2 |
other | AC * 21 |
コンパイルメッセージ
main.cpp: In function ‘int main()’: main.cpp:46:20: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result] 46 | ll n; scanf("%lld", &n); | ~~~~~^~~~~~~~~~~~ main.cpp:47:59: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result] 47 | vector<ll> a(n); for (int i = 0; i < n; i++) scanf("%lld", &a[i]); | ~~~~~^~~~~~~~~~~~~~~ main.cpp:55:23: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result] 55 | int kkt; scanf("%d", &kkt); | ~~~~~^~~~~~~~~~~~ main.cpp:57:28: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result] 57 | ll k; scanf("%lld", &k); | ~~~~~^~~~~~~~~~~~
ソースコード
#include <cstdio> #include <iostream> #include <string> #include <sstream> #include <stack> #include <algorithm> #include <cmath> #include <queue> #include <map> #include <set> #include <cstdlib> #include <bitset> #include <tuple> #include <assert.h> #include <deque> #include <bitset> #include <iomanip> #include <limits> #include <chrono> #include <random> #include <array> #include <unordered_map> #include <functional> #include <complex> #include <numeric> template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; } template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; } constexpr long long MAX = 5100000; constexpr long long INF = 1LL << 60; constexpr int inf = 1000000007; constexpr long long mod = 1000000007LL; //constexpr long long mod = 998244353LL; const long double PI = acos((long double)(-1)); using namespace std; typedef unsigned long long ull; typedef long long ll; typedef long double ld; int main() { /* cin.tie(nullptr); ios::sync_with_stdio(false); */ ll n; scanf("%lld", &n); vector<ll> a(n); for (int i = 0; i < n; i++) scanf("%lld", &a[i]); vector<vector<ll>> to(60, vector<ll>(n)), sum(60, vector<ll>(n)); for (int i = 0; i < n; i++) { ll nx = (i + a[i]) % n; to[0][i] = nx; sum[0][i] = a[i]; } for (int i = 1; i < 60; i++) for (int j = 0; j < n; j++) to[i][j] = to[i - 1][to[i - 1][j]], sum[i][j] = sum[i - 1][j] + sum[i - 1][to[i - 1][j]]; int kkt; scanf("%d", &kkt); while (kkt--) { ll k; scanf("%lld", &k); ll cur = 0; ll res = 0; for (int i = 59; i >= 0; i--) { if (k >> i & 1) { res += sum[i][cur]; cur = to[i][cur]; } } printf("%lld\n", res); } return 0; }