HDU6040 排列问题 element运用_c语言计算6040×3953表格-优快云博客

本文链接：https://blog.youkuaiyun.com/LinzhiQQQ/article/details/76596149

本文解析了一道关于ACM竞赛的编程题目，介绍了如何通过特定的算法和数据结构技巧来解决涉及选手评级和比赛策略的问题。

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Hints of sd0061

Time Limit: 5000/2500 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 2359 Accepted Submission(s): 713

Problem Description

sd0061, the legend of Beihang University ACM-ICPC Team, retired last year leaving a group of noobs. Noobs have no idea how to deal with

m coming contests. sd0061 has left a set of hints for them.

There are

n noobs in the team, the

i-th of which has a rating

ai. sd0061 prepares one hint for each contest. The hint for the

j-th contest is a number

bj, which means that the noob with the

(bj+1)-th lowest rating is ordained by sd0061 for the

j-th contest.

The coach asks constroy to make a list of contestants. constroy looks into these hints and finds out:

bi+bj≤bk is satisfied if

bi≠bj,

bi<bk and

bj<bk.

Now, you are in charge of making the list for constroy.

Input

There are multiple test cases (about

10).

For each test case:

The first line contains five integers

n,m,A,B,C.

(1≤n≤107,1≤m≤100)

The second line contains

m integers, the

i-th of which is the number

bi of the

i-th hint.

(0≤bi<n)

The

n noobs' ratings are obtained by calling following function

n times, the

i-th result of which is

ai.

unsigned x = A, y = B, z = C;
unsigned rng61() {
  unsigned t;
  x ^= x << 16;
  x ^= x >> 5;
  x ^= x << 1;
  t = x;
  x = y;
  y = z;
  z = t ^ x ^ y;
  return z;
}

Output

For each test case, output "Case #

⋯

ym" in one line (without quotes), where

x indicates the case number starting from

1 and

(1≤i≤m) denotes the rating of noob for the

i-th contest of corresponding case.

Sample Input

题意：题目给出了一个函数，直接copy就可以了，根据那个函数直接计算出数组a，然后给出了数组b，要你根据数组b对a进行排序 ai 必须是第bi + 1 大的数

我们对b的下标进行排序得到一个下标数组然后从大到小去扫一遍这个数组使用 nth_element 函数（将a组第b[pos[i]]小的数放到数组左边，比他大的放到数组右边）

然后再拿一个答案数组去根据下标数组记录答案

#include<bits/stdc++.h>
using namespace std;
#define N 10001000
unsigned a[N],b[110],ans[110],under[110],x,y,z;
unsigned rng61() {
	unsigned t;
	x ^= x << 16;
	x ^= x >> 5;
	x ^= x << 1;
	t = x;
	x = y;
	y = z;
	z = t ^ x ^ y;
	return z;
}
unsigned cmp(unsigned x,unsigned y){
	return b[x] < b[y];
}
int main(){
	unsigned n,m,A,B,C,Case = 1;
	while(scanf("%u %u %u %u %u",&n,&m,&A,&B,&C) != EOF){
		x = A;
		y = B;
		z = C;
		for(int i = 0;i < m;i++){
			scanf("%u",&b[i]);
			under[i] = i;
		}
		for(int i = 0;i < n;i++){
			a[i] = rng61();
		}
		sort(under,under + m,cmp);
		under[m] = m;
		b[m] = n;
		for(int i = m - 1;i >= 0;i--){
			nth_element(a,a + b[under[i]],a + b[under[i + 1]]);
			ans[under[i]] = a[b[under[i]]];
		}//nth_element(a + x,a + z,a + y) 在 x 到 y之间找第z小的元素放到z位置上 
		printf("Case #%u:",Case++);
		for(int i = 0;i < m;i++){
			printf(" %u",ans[i]);
		}
		printf("\n");
	}
	return 0;
}