429c Leha and Function

题目

解题报告

F(n, k)是在集合{1, 2, 3, ..., n}中所有的具有k个元素的子集中分别取最小值，相加后的期望。
例如：要求F(4, 2)，根据定义有{1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {3, 4}，则F(4, 2)=(1+1+1+2+2+3)/6=1.6666666666666...

对于F(n, k)，我们有这么一个结论，
$$ F(n, k) > F(m, k), n > m $$
$$F(n, k) > F(n, q), k < q $$

因此，原问题变为将A按照由大到小排序后，求B数组每个元素在排序后的编号，在此位置输出排序后的Ai

#pragma comment(linker, "/STACK:1024000000,1024000000")#include <bits/stdc++.h>using namespace std;typedef long long ll;typedef unsigned long long ull;#define ms(s) memset(s, 0, sizeof(s))#define REP(i, k, n) for (int i = k; i < n; i++)#define REPP(i, k, n) for (int i = k; i <= n; i++)const int inf = 0x3f3f3f3f;#define LOCALint a[200005], h[200005];pair<int, int> b[200005];bool cmp(int a, int b){    return a > b;}bool cmp1(pair<int, int> a, pair<int, int> b){    return a.first < b.first;}int main(int argc, char * argv[]) {    #ifdef LOCAL    freopen("/Users/huangjiaming/Documents/Algorithm/oj/data.in", "r", stdin);    //freopen("/Users/huangjiaming/Documents/Algorithm/oj/data.out", "w", stdout);    #endif    int n;    while (~scanf("%d", &n))    {        REPP(i, 1, n)        scanf("%d", a+i);        REPP(i, 1, n)        {            scanf("%d", &b[i].first);            b[i].second = i;        }        sort(a+1, a+n+1, cmp);        sort(b+1, b+n+1, cmp1);             REPP(i, 1, n)        h[b[i].second] = i;        REPP(i, 1, n)        printf("%d ", a[h[i]]);                printf("\n");    }    return 0;}