Codeforces Round #410 (Div. 2) D. Mike and distribution 思维
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题意是给两个序列,让你选几个下标,使得对于这两个序列都满足所选的数和大于未选的数和
思维水题,先按a值从大到小排序,扫一遍,先取最大的那个值,然后两个两个的选,优先选b值较大的,如果总数为偶数个,则最后还会剩一个,选上就好了,这样不难想ab都会满足那个条件。#include<stdio.h>#include<string.h>#include<math.h>#include<string>#include<algorithm>#include<queue>#include<vector>#include<map>#include<set>#define eps 1e-9#define PI 3.141592653589793#define bs 1000000007#define bsize 256#define MEM(a) memset(a,0,sizeof(a))typedef long long ll;using namespace std;struct node{int a,b,index;}num[100005];int cmp(node a,node b){return a.a>b.a;}int main(){int n,i,j;scanf("%d",&n);for(i=1;i<=n;i++)scanf("%d",&num[i].a),num[i].index=i;for(i=1;i<=n;i++)scanf("%d",&num[i].b);sort(num+1,num+n+1,cmp);vector<int>ans;ans.push_back(num[1].index);for(i=2;i<=n;i+=2){if(i+1<=n&&num[i+1].b>num[i].b)ans.push_back(num[i+1].index);elseans.push_back(num[i].index);}printf("%d\n",ans.size());for(i=0;i<ans.size();i++)printf("%d%c",ans[i]," \n"[i==n]);return 0; }
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