牛客练习赛63-C

思路：
大佬们->推推推
蒟蒻的我->打表猜猜猜

打表结果

每一列分别表示i,j,a[i]*[j]出现的次数
之后就很简单啦

维护sum[i]= a[i]*(n-i+1)
枚举第一列的 那么每次乘上的数字就是sum[i-n]*i。
所以
注意负数取余即可

#pragma GCC optimize(3,"Ofast","inline")      //G++
#include<bits/stdc++.h>
#define mem(a,x) memset(a,x,sizeof(a))
#define debug(x) cout << #x << ": " << x << endl;
#define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fcout cout<<setprecision(4)<<fixed
using namespace std;
typedef long long ll;
//======================================
namespace FastIO{
char print_f[105];void read() {}void print() {putchar('\n');}
template <typename T, typename... T2>
inline void read(T &x, T2 &... oth){x = 0;char ch = getchar();ll f = 1;while (!isdigit(ch)){if (ch == '-')f *= -1;ch = getchar();}while (isdigit(ch)){x = x * 10 + ch - 48;ch = getchar();}x *= f;read(oth...);}
template <typename T, typename... T2>
inline void print(T x, T2... oth){ll p3=-1;if(x<0) putchar('-'),x=-x;do{print_f[++p3] = x%10 + 48;}while(x/=10);while(p3>=0) putchar(print_f[p3--]);putchar(' ');print(oth...);}} // namespace FastIO
using FastIO::print;
using FastIO::read;
//======================================
typedef pair<ll,ll> pii;
const ll inf=0x3f3f3f3f;
const ll mod=1e9+7;
const ll maxn = 1e6+5;

ll a[maxn],sum[maxn];
main() {
#ifndef ONLINE_JUDGE
//    freopen("H:\\code\\in.in", "r", stdin);
//    freopen("H:\\code\\bf_out.out", "w", stdout);
    clock_t c1 = clock();
#endif
//**************************************
    ll n;
    read(n);
    ll ans=0;
    for(ll i=1;i<=n;i++) read(a[i]);
    for(ll i=1;i<=n;i++) sum[i]=(sum[i-1]+(a[i]*(n-i+1))%mod)%mod;
    for(ll i=1;i<=n;i++){
        ll s=((sum[n]-sum[i-1])%mod+mod)%mod;
        s=s*i%mod;
        ans=(ans+(a[i]*s)%mod)%mod;
    }
    ans=((ans%mod)+mod)%mod;
    print(ans);
//**************************************

#ifndef ONLINE_JUDGE
    cerr << "Time:" << clock() - c1 << "ms" << endl;
#endif
    return 0;
}