hdu 5496 Beauty of Sequence(高效)

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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;}

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