L2-009 抢红包（模拟）

Posted 2022-03-30 MangataTS

题目链接

https://pintia.cn/problem-sets/994805046380707840/problems/994805066890854400

思路

我们开一个结构体Node去存储编号id、收到红包数量num 以及获得的总红包金额money，然后按照每一行的人收到的红包我们进行相应的处理，将发红包的人的money金额减少，然后将收红包的人的money增加，并且将收红包的人的红包数量增加，最后排序输出就好啦

代码

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define mod 1000000007
#define endl "\\n"
#define PII pair<int,int>
#define INF 0x3f3f3f3f

const int N = 1e4+10;

int n;

struct Node
	int id,num;
	double money;
people[N];

bool cmp(Node a,Node b) 
	if(fabs(a.money-b.money) <= 0.0001)
		if(a.num == b.num) 
			return a.id < b.id;
	else
		return a.num > b.num;
	return a.money > b.money;


map<int,bool> vis;

int main()

	ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
	cin>>n;
	int k,a;
	double b;
	for(int i = 1;i <= n; ++i)
		cin>>k;
		people[i].id = i;
		double sum = 0;
		vis.clear();
		for(int j = 1;j <= k; ++j) 
			cin>>a>>b;
			if(vis[a]) continue;
			vis[a] = true;
			people[i].money -= b;
			people[a].num++;
			people[a].money += b;
		
	
	
	sort(people+1,people+1+n,cmp);
	cout<<fixed<<setprecision(2);
	for(int i = 1;i <= n; ++i) 
		cout<<people[i].id<<" "<<people[i].money/100.0<<endl;
	
	return 0;