D. Mike and distribution?構造法

798D - Mike and distribution

In the beginning, it's quite easy to notice that the condition "?2·(ap1?+?...?+?apk)?is greater than the sum of all elements in?A?" is equivalent to "?ap1?+?...?+?apk?is greater than the sum of the remaining elements in?A?".

Now, let's store an array of indices?C?with?Ci?=?i?and then sort it in decreasing order according to array?A, that is we must have?ACi?≥?ACi?+?1.

Our answer will always have size?. First suppose that?N?is odd. Add the first index to our set, that is make?p1?=?C1. Now, for the remaining elements, we will consider them consecutively in pairs. Suppose we are at the moment inspecting?AC2k?and?AC2k?+?1. If?BC2k?≥?BC2k?+?1?we make?pk?+?1?=?C2k, else we make?pk?+?1?=?C2k?+?1.

Why does this subset work? Well, it satisfies the condition for?B?because each time for consecutive non-intersecting pairs of elements we select the bigger one, and we also add?BC1?to the set, so in the end the sum of the selected elements will be bigger than the sum of the remaining ones.

It also satisfies the condition for?A, because?Ap1?is equal or greater than the complement element of?p2?(that is — the index which we could've selected instead of?p2?from the above procedure — if we selected?C2k?then it would be?C2k?+?1?and vice-versa). Similarly?Ap2?is greater than the complement of?p3?and so on. In the end we also add the last element from the last pair and this makes the sum of the chosen subset strictly bigger than the sum of the remaining elements.

The case when?N?is even can be done exactly the same as when?N?is odd, we just pick the last remaining index in the end.

The complexity is?.

#include <cstdio> #include <iostream> #include <algorithm> using namespace std; const int N = 1e5 + 7; int a[N], b[N], c[N], p[N/2]; bool cmp(int i, int j) {return a[i] > a[j]; } int main() {//ios::sync_with_stdio(0);int n;while(~scanf("%d", &n)) {for (int i = 1; i <= n; ++i) {scanf("%d", a +i);c[i] = i;}for (int i = 1; i <= n; ++i)scanf("%d", b + i);sort(c + 1, c + 1 + n, cmp);int k = n + 1 >> 1, cur = 0;p[++cur] = c[1];b[n + 1] = 0;//最邊界小值 for (int i = 2; i <= n; i += 2) //cur = k ?p[++cur] = b[c[i]] > b[c[i + 1]] ? c[i] : c[i+1];printf("%d\n%d", cur, p[1]); for (int i = 2; i <= cur; ++i)printf(" %d", p[i]);puts("");}return 0; }

?

轉載于:https://www.cnblogs.com/qinwenjie/p/7271339.html

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