hdu 5496 Beauty of Sequence(高效)

题目链接：hdu 5496 Beauty of Sequence

解题思路

考虑每个位置的贡献度。

代码

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
const int maxn = 1e5 + 5;
const int mod = 1e9 + 7;

struct Element {
    int val, pos;
    bool operator < (const Element& u) const { if (val != u.val) return val < u.val; return pos < u.pos; }
}E[maxn];

int N, P[maxn];

int solve () {
    int ret = 0, del;
    sort(E, E + N);
    for (int i = 0; i < N; i++) {
        if (i == 0 || E[i].val != E[i-1].val) del = 0;

        int l = E[i].pos, r = N-E[i].pos-1;
        ret = (ret + 1LL * (P[l] - del + mod) * P[r] % mod * E[i].val % mod) % mod;
        del = (del + P[l]) % mod;
    }
    return ret;
}

int main () {
    P[0] = 1;
    for (int i = 1; i <= (int)1e5; i++) P[i] = P[i-1] * 2 % mod;

    int cas;
    scanf("%d", &cas);
    while (cas--) {
        scanf("%d", &N);
        for (int i = 0; i < N; i++) {
            scanf("%d", &E[i].val);
            E[i].pos = i;
        }

        printf("%d\n", solve());
    }
    return 0;
}