CF798D:Mike and distribution(思维 )^
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题意:给两个N个元素的数组,最多选择K=N/2+1个不同下标,使得各自两个数组中这些下标的元素之和*2>各自数组的元素总和。
思路:即求取K个元素之和大于剩下的元素之和,按A排序,两个两个地按B选择。
# include <bits/stdc++.h>using namespace std;int a[1<<20], b[1<<20], c[1<<20], d[1<<20];int main(){ int n, cnt=0; scanf("%d",&n); for(int i=0; i<n; ++i) scanf("%d",&a[i]); for(int i=0; i<n; ++i) scanf("%d",&b[i]); for(int i=0; i<n; ++i) c[i] = i; sort(c, c+n, [&](int x, int y){return a[x] > a[y];}); d[cnt++] = c[0]; for(int i=1; i<n; i+=2) { int t = c[i]; if(i+1<n && b[c[i+1]] > b[t]) t = c[i+1]; d[cnt++] = t; } printf("%d\n",cnt); for(int i=0; i<cnt; ++i) printf("%d ",d[i]+1); return 0;}
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