#684381 (C++17) No.1629 Sorting Integers (SUM of M)

• 問題: No.1629 Sorting Integers (SUM of M)
• ユーザー: 風月
• 提出日時: 2021-07-30 21:36:31
• 言語: C++17 (gcc 13.3.0 + boost 1.87.0)
• 結果: AC
• 実行時間: 8 ms / 2,000 ms
• コード長: 2,362 bytes
• コンパイル時間: 2,022 ms
• コンパイル使用メモリ: 192,236 KB
• 最終ジャッジ日時: 2025-01-23 11:18:39
• ジャッジサーバーID （参考情報）: judge3 / judge5

ファイルパターン	結果
sample	AC * 4
other	AC * 14

ソースコード

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#include<bits/stdc++.h>
#include<unordered_map>
#define for_int(i,a,b) for(int i=(a);i<=(b);++i)
#define rep_int(i,a,b) for(int i=(a);i>=(b);--i)
#define for_ll(i,a,b) for(ll i=(a);i<=(b);++i)
#define rep_ll(i,a,b) for(ll i=(a);i>=(b);--i)
#define Ios std::ios::sync_with_stdio(false),std::cin.tie(0)
#define ull unsigned long long
#define ll long long
#define db double
#define PII std::pair<int,int>
#define PLI std::pair<long long , int>
template<class T> void chkmax(T& a, T b) { a > b ? (a = a) : (a = b); }
template<class T> void chkmin(T& a, T b) { a > b ? (a = b) : (a = a); }
template<class T> T min(T a, T b) { return a > b ? b : a; }
template<class T> T max(T a, T b) { return a < b ? b : a; }
template<class T> T abs(T a) { return a < 0 ? -a : a; }
//using namespace std;
/*
    1.注意边界情况
    2.优先考虑时间复杂度
!!! 3.求最大值时，答案应至多初始化为INT_MIN；求最小值时，答案应至少初始化为INT_MAX
    4.遇事不决先暴力
    5.代码要写简洁
    6.计数题要开long long
*/
//快速IO
template<class T>
inline bool rd(T& ret) {
    char c; int sgn;
    if (c = getchar(), c == EOF) return 0;
    while (c != '-' && (c < '0' || c > '9')) c = getchar();
    sgn = (c == '-') ? -1 : 1;
    ret = (c == '-') ? 0 : (c - '0');
    while (c = getchar(), c >= '0' && c <= '9') ret = ret * 10 + (c - '0');
    ret *= sgn;
    return true;
}
template<class T>
inline void print(T x) {
    if (x > 9) print(x / 10);
    putchar(x % 10 + '0');
    return;
}
using std::vector;
using std::queue;
using std::string;
using std::map;
using std::unordered_map;
using std::priority_queue;
using std::cout;
using std::cin;
int mod = 1e9 + 7;
int quickPow(int a, int n) {
    int res = 1;
    while (n) {
        if (n & 1) res = 1ll * res * a % mod;
        n >>= 1;
        a = 1ll * a * a % mod;
    }
    return res;
}
int n, ct, a[10];
int main() {
    rd(n);
    for (int i = 1; i <= 9; ++i) rd(a[i]);
    int m = 1;
    for (int i = 1; i <= n; ++i) {
        m = 1ll * m * i % mod;
        ct = (1ll * ct * 10 + 1) % mod;
    }
    long long res = 0;
    for (int i = 1; i <= 9; ++i) {
        int tmp = 1;
        for (int j = 1; j <= a[i]; ++j)
            tmp = 1ll * tmp * j % mod;
        m = 1ll * m * quickPow(tmp, mod - 2) % mod;
    }
    int invn = quickPow(n, mod - 2);
    m = 1ll * m * invn % mod;
    for (int i = 1; i <= 9; ++i) {
        res = (res + ((((1ll * m * a[i])% mod) * i)%mod) * ct) % mod;
    }
    print(res); puts("");
    return 0;
}
 
 
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