NC19798 区间权值
区间权值
https://ac.nowcoder.com/acm/problem/19798
题意:
题解:
AC代码
/*
Author : zzugzx
Lang : C++
Blog : blog.csdn.net/qq_43756519
*/
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(), (x).end()
#define endl '\n'
#define SZ(x) (int)x.size()
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
const int mod = 1e9 + 7;
const int MOD = 998244353;
const double eps = 1e-6;
const double pi = acos(-1.0);
const int maxn = 1e6 + 10;
const int N = 4e3 + 10;
const ll inf = 0x3f3f3f3f;
const int dir[][2]={{0, 1}, {1, 0}, {0, -1}, {-1, 0}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1}};
ll a[maxn], w[maxn], s[maxn];
int main()
{
ios::sync_with_stdio(false);
cin.tie(0); cout.tie(0);
// freopen("in.txt", "r", stdin);
// freopen("out.txt", "w", stdout);
int n;
cin >> n;
for (int i = 1; i <= n; i++) {
cin >> a[i];
s[i] = (s[i - 1] + a[i]) % mod;
}
ll sum = 0, ans = 0;
for (int i = 1, w; i <= n; i++) {
cin >> w;
sum = (sum + s[n - i + 1] - s[i - 1] + mod) % mod;
ans = (ans + w * sum) % mod;
}
cout << ans;
return 0;
}
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