PAT甲1063Set的使用

1063

1063 Set Similarity (25分)

Given two sets of integers, the similarity of the sets is defined to be N
​c
​​ /N
​t
​​ ×100%, where N
​c
​​ is the number of distinct common numbers shared by the two sets, and N
​t
​​ is the total number of distinct numbers in the two sets. Your job is to calculate the similarity of any given pair of sets.

Input Specification:
Each input file contains one test case. Each case first gives a positive integer N (≤50) which is the total number of sets. Then N lines follow, each gives a set with a positive M (≤10
​4
​​ ) and followed by M integers in the range [0,10
​9
​​ ]. After the input of sets, a positive integer K (≤2000) is given, followed by K lines of queries. Each query gives a pair of set numbers (the sets are numbered from 1 to N). All the numbers in a line are separated by a space.

Output Specification:
For each query, print in one line the similarity of the sets, in the percentage form accurate up to 1 decimal place.

Sample Input:
3
3 99 87 101
4 87 101 5 87
7 99 101 18 5 135 18 99
2
1 2
1 3
Sample Output:
50.0%
33.3%

题意：给你n个集合，然后输入q个查询，查询两个集合内，相同元素，和除相同元素外的和为多少

题解：用set的find，find如果没找到则返回set的end，然后计算即可

代码：

#include<iostream>
#include<set>
using namespace std;
set<int> st[100];
int main()
{
	int n,k,t,q,x1,x2;
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	{
		scanf("%d",&k);
		for(int j=1;j<=k;j++)
		{
			scanf("%d",&t);
			st[i].insert(t);
		}
	}
	scanf("%d",&q);
	for(int i=1;i<=q;i++)
	{
		cin>>x1>>x2;
		int cnt=0;
		for(set<int>::iterator it=st[x1].begin();it!=st[x1].end();it++)
		{
			if(st[x2].find(*it)!=st[x2].end())
				cnt++;
		}
		printf("%.1f%\n",100.0*cnt/(st[x1].size()+st[x2].size()-cnt));
	}
	return 0;
}