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abc058d <公式化简>

时间:2023-06-23 17:12:54浏览次数:51  
标签:return abc058d int res LL sum mod

D - ###

原计算公式为:

\[\sum\limits_{1\le i<j\le n}\sum\limits_{1\le k<l\le m}(x_j-x_i)(y_l-y_k) \]

可将xy拆分:

\[\left(\sum\limits_{1\leq i<j\leq n}(x_j-x_i)\right)\left(\sum\limits_{1\leq k<l\leq m}(y_l-y_k)\right) \]

仅计算x侧可进一步化简:

\[\sum\limits_{1\le i<j\le n}(x_j-x_i)=\sum\limits_{1\le k\le n}((k-1)x_k-(n-k)x_k) \]

// https://atcoder.jp/contests/abc058/tasks/arc071_b
// 公式化简

#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
const int N = 1e5 + 10, mod = 1e9 + 7;
int n, m;
int x[N], y[N];

LL qpow(LL a, LL k)
{
    LL res = 1;
    a %= mod;
    while (k)
    {
        if (k & 1) res = res * a % mod;
        k >>= 1;
        a = a * a % mod;
    }
    return res;
}

LL C(LL n, LL m)
{
    if (m < 0 || m > n) return 0;
    if (m > n / 2) return C(n, n - m);
    LL res = 1;
    LL t = 1;
    for (int i = n, j = 1; j <= m; j ++, i --)
    {
        res = res * i % mod;
        t = t * j % mod;
    }
    return res * qpow(t, mod-2) % mod;
}

void solv()
{
    cin >> n >> m;
    for (int i = 1; i <= n; i ++) cin >> x[i];
    for (int i = 1; i <= m; i ++) cin >> y[i];
    sort(x+1, x+1+n); sort(y+1, y+1+m);
    LL xx = 0, yy = 0;
    for (LL i = 1; i <= n; i ++) xx = (xx + (i*2 - 1 - n) * x[i] % mod) % mod;
    for (LL i = 1; i <= m; i ++) yy = (yy + (i*2 - 1 - m) * y[i] % mod) % mod;
    LL ans = xx * yy % mod;
    cout << ans << endl;
}

int main()
{
    ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    int T = 1;
    // cin >> T;
    while (T --)
    {
        solv();
    }
    return 0;
}

标签:return,abc058d,int,res,LL,sum,mod
From: https://www.cnblogs.com/o2iginal/p/17499345.html

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