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[NOIP2009 普及组] 分数线划定
2022-07-06 04:42:00 【Recurss】
题目描述
世博会志愿者的选拔工作正在 A 市如火如荼的进行。为了选拔最合适的人才, A A A市对所有报名的选手进行了笔试,笔试分数达到面试分数线的选手方可进入面试。面试分数线根据计划录取人数的 150 % 150\% 150%划定,即如果计划录取 m m m名志愿者,则面试分数线为排名第 m × 150 % m \times 150\% m×150%(向下取整)名的选手的分数,而最终进入面试的选手为笔试成绩不低于面试分数线的所有选手。
现在就请你编写程序划定面试分数线,并输出所有进入面试的选手的报名号和笔试成绩。
输入格式
第一行,两个整数 n , m ( 5 ≤ n ≤ 5000 , 3 ≤ m ≤ n ) n,m(5 ≤ n ≤ 5000,3 ≤ m ≤ n) n,m(5≤n≤5000,3≤m≤n),中间用一个空格隔开,其中 n n n表示报名参加笔试的选手总数, m m m表示计划录取的志愿者人数。输入数据保证 m × 150 % m \times 150\% m×150%向下取整后小于等于 n n n。
第二行到第 n + 1 n+1 n+1 行,每行包括两个整数,中间用一个空格隔开,分别是选手的报名号 k ( 1000 ≤ k ≤ 9999 ) k(1000 ≤ k ≤ 9999) k(1000≤k≤9999)和该选手的笔试成绩$ s(1 ≤ s ≤ 100)$。数据保证选手的报名号各不相同。
输出格式
第一行,有 2 2 2个整数,用一个空格隔开,第一个整数表示面试分数线;第二个整数为进入面试的选手的实际人数。
从第二行开始,每行包含 2 2 2个整数,中间用一个空格隔开,分别表示进入面试的选手的报名号和笔试成绩,按照笔试成绩从高到低输出,如果成绩相同,则按报名号由小到大的顺序输出。
样例 #1
样例输入 #1
6 3
1000 90
3239 88
2390 95
7231 84
1005 95
1001 88
样例输出 #1
88 5
1005 95
2390 95
1000 90
1001 88
3239 88
提示
【样例说明】
m × 150 % = 3 × 150 % = 4.5 m \times 150\% = 3 \times150\% = 4.5 m×150%=3×150%=4.5,向下取整后为 4 4 4。保证 4 4 4个人进入面试的分数线为 88 88 88,但因为 88 88 88有重分,所以所有成绩大于等于 88 88 88 的选手都可以进入面试,故最终有 5 5 5个人进入面试。
NOIP 2009 普及组 第二题
/* * @Description: To iterate is human, to recurse divine. * @Autor: Recursion * @Date: 2022-07-04 18:46:06 * @LastEditTime: 2022-07-05 03:18:13 */
#include <bits/stdc++.h>
#define LL long long
using namespace std;
const int maxn = 1e6 + 10;
const int mod = 1e9 + 7;
const int INF = 1e9 + 10;
const int N = 1e6;
int n;
double m;
int a[N];
struct node{
int k;
int s;
}t[N];
bool cmp(node x,node y){
if(x.s > y.s)
return true;
if(x.s == y.s){
if(x.k < y.k )
return true;
}
return false;
}
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin >> n >> m;
int x;
x = m * 1.5;
for(int i = 1;i <= n;i ++){
cin >> t[i].k >> t[i].s;
}
sort(t + 1, t + 1 + n,cmp);
int num = x;
for(int i = x + 1;i <= n;i ++){
if(t[x].s == t[i].s)
num++;
}
cout << t[x].s << " " << num << endl;
for(int i = 1;i <= x;i ++){
cout << t[i].k <<" "<<t[i].s << endl;
}
for(int i = x + 1;i <= n;i ++){
if(t[x].s == t[i].s)
cout << t[i].k << " " << t[i].s << endl;
}
return 0;
}
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