问题 A: Set Similarity (25)

问题 A: Set Similarity (25)

时间限制: 1 Sec  内存限制: 32 MB
提交: 387  解决: 185
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题目描述

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

输入

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 (<=104) and followed by M integers in the range [0, 109]. 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.

输出

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

样例输入

3
3 99 87 101
4 87 101 5 87
7 99 101 18 5 135 18 99
2
1 2
1 3

样例输出

50.0%
33.3%

这一题用set容器数组存储不重复的元素

将set容器a的元素依次在set容器b里进行查找找到相同元素数量Nc就加一。

而不同元素的个数就是两个容器的size相加再减去相同元素的个数Nc。

 

#include
#include 
#include 
#include 
using namespace std;
const int maxn=60;
set Array[maxn];

int main(){
	int N,M,K,group=0;
	int Nc,Nt;
	scanf("%d",&N);
	for(int i=1;i<=N;i++){
		scanf("%d",&M);
		for(int j=0;j::iterator it=Array[a].begin();it!=Array[a].end();it++){
			if(Array[b].find(*it) != Array[b].end()) //在b中找a的各个元素是否存在 
			Nc++;
		}
		Nt=Array[a].size()+Array[b].size()-Nc;
		printf("%.1f%%\n", (double)Nc / Nt * 100);
	}
	return 0;
}