結果
| 問題 |
No.1201 お菓子配り-4
|
| コンテスト | |
| ユーザー |
|
| 提出日時 | 2020-09-22 21:22:33 |
| 言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
| 結果 |
AC
|
| 実行時間 | 480 ms / 4,000 ms |
| コード長 | 3,285 bytes |
| コンパイル時間 | 2,488 ms |
| コンパイル使用メモリ | 194,684 KB |
| 最終ジャッジ日時 | 2025-01-14 19:38:01 |
|
ジャッジサーバーID (参考情報) |
judge3 / judge1 |
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| ファイルパターン | 結果 |
|---|---|
| other | AC * 36 |
コンパイルメッセージ
main.cpp: In function ‘int main()’:
main.cpp:123:14: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result]
123 | scanf("%d%d", &n, &m);
| ~~~~~^~~~~~~~~~~~~~~~
main.cpp:124:24: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result]
124 | rep(i, n) scanf("%d", a+i);
| ~~~~~^~~~~~~~~~~
main.cpp:125:24: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result]
125 | rep(i, m) scanf("%d", b+i);
| ~~~~~^~~~~~~~~~~
ソースコード
//#define _GLIBCXX_DEBUG
#include <bits/stdc++.h>
#define rep(i, n) for(int i=0; i<n; ++i)
#define all(v) v.begin(), v.end()
#define rall(v) v.rbegin(), v.rend()
using namespace std;
using ll = int64_t;
using ld = long double;
using P = pair<int, int>;
using vs = vector<string>;
using vi = vector<int>;
using vvi = vector<vi>;
template<class T> using PQ = priority_queue<T>;
template<class T> using PQG = priority_queue<T, vector<T>, greater<T> >;
const int INF = 0xccccccc;
const ll LINF = 0xcccccccccccccccLL;
template<typename T1, typename T2>
inline bool chmax(T1 &a, T2 b) {return a < b && (a = b, true);}
template<typename T1, typename T2>
inline bool chmin(T1 &a, T2 b) {return a > b && (a = b, true);}
template<typename T1, typename T2>
istream &operator>>(istream &is, pair<T1, T2> &p) { return is >> p.first >> p.second;}
template<typename T1, typename T2>
ostream &operator<<(ostream &os, const pair<T1, T2> &p) { return os << p.first << ' ' << p.second;}
const unsigned int mod = 1000000007;
//const unsigned int mod = 998244353;
struct mint {
unsigned int x;
mint():x(0) {}
mint(int64_t x_) {
int64_t v = int64_t(x_ % mod);
if(v < 0) v += mod;
x = (unsigned int)v;
}
static mint row(int v) {
mint v_;
v_.x = v;
return v_;
}
mint operator-() const { return mint(-x);}
mint& operator+=(const mint a) {
if ((x += a.x) >= mod) x -= mod;
return *this;
}
mint& operator-=(const mint a) {
if ((x += mod-a.x) >= mod) x -= mod;
return *this;
}
mint& operator*=(const mint a) {
uint64_t z = x;
z *= a.x;
x = (unsigned int)(z % mod);
return *this;
}
mint operator+(const mint a) const { return mint(*this) += a;}
mint operator-(const mint a) const { return mint(*this) -= a;}
mint operator*(const mint a) const { return mint(*this) *= a;}
friend bool operator==(const mint &a, const mint &b) {return a.x == b.x;}
friend bool operator!=(const mint &a, const mint &b) {return a.x != b.x;}
mint &operator++() {
x++;
if(x == mod) x = 0;
return *this;
}
mint &operator--() {
if(x == 0) x = mod;
x--;
return *this;
}
mint operator++(int) {
mint result = *this;
++*this;
return result;
}
mint operator--(int) {
mint result = *this;
--*this;
return result;
}
mint pow(int64_t t) const {
mint x_ = *this, r = 1;
while(t) {
if(t&1) r *= x_;
x_ *= x_;
t >>= 1;
}
return r;
}
//for prime mod
mint inv() const { return pow(mod-2);}
mint& operator/=(const mint a) { return *this *= a.inv();}
mint operator/(const mint a) {return mint(*this) /= a;}
};
int64_t gcd(int64_t a, int64_t b) {return b?gcd(b, a%b):a;}
int64_t lcm(int64_t a, int64_t b) {return a/gcd(a,b)*b;}
int64_t extgcd(int64_t a, int64_t b, int64_t &x, int64_t &y) { // ax+by=gcd(a, b)の解を与える
int64_t d = a;
if(b) {
d = extgcd(b, a%b, y, x);
y -= (a/b)*x;
} else {
x = 1; y = 0;
}
return d;
}
istream& operator>>(istream& is, mint& a) { return is >> a.x;}
ostream& operator<<(ostream& os, const mint& a) { return os << a.x;}
#define N 2010
//head
int n, m;
int a[N], b[N];
mint ans;
int main() {
scanf("%d%d", &n, &m);
rep(i, n) scanf("%d", a+i);
rep(i, m) scanf("%d", b+i);
rep(i, n) rep(j, m) {
int u = gcd(a[i], b[j]);
ans += mint::row(b[j]+1)*a[i] - b[j] + u;
}
printf("%u\n", ans.x);
}