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?Nlines 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%

解題思路：

題目大意：給出n個(gè)序列和其中的數(shù)，要求給出指定兩個(gè)集合的(相同個(gè)數(shù))/(不同個(gè)數(shù))的比率

使用STL中的set容器，因?yàn)閟et容器自動(dòng)去掉重復(fù)的數(shù)并且按升序排序集合里的數(shù)。代碼分為2段：首先按照要求輸入數(shù)據(jù)，即輸入“要輸入的集合數(shù)n”和“n個(gè)集合，每個(gè)集合要輸入的數(shù)m，以及輸入m個(gè)數(shù)”，用set的insert()函數(shù)可以過濾掉重復(fù)的數(shù)據(jù)；下一步是整理數(shù)據(jù)，例如統(tǒng)一確定st[set2]是被查找的集合，循環(huán)st[set1]，如果找到st[set2].find(*it)!=st[set2].end()，意思是非空，也就是找到了，就相同數(shù)加一，反之就是st[set1]中有的數(shù)在st[set2]中是沒有的，那么把st[set2].size()++，每一次輸出即可。

總結(jié)：在st[set2].find(*it)!=st[set2].end()，這里需要思路清晰，大腦不能犯糊涂，對(duì)照1查2,2中查不到就++，不是對(duì)照1查1

代碼：?

#include <stdio.h> #include <set> using namespace std; set<int> st[55];int main(){int n,m,x,set1,set2,inquiry,same,count=0;scanf("%d",&n); //集合的數(shù)量n//輸入數(shù)據(jù) for(int i=1;i<=n;i++){scanf("%d",&m); //集合中的數(shù)量mfor(int j=0;j<m;j++){scanf("%d",&x); //集合中的每一個(gè)數(shù)，賦值為xst[i].insert(x); //將數(shù)插入到第i個(gè)集合中 } }//整理數(shù)據(jù)scanf("%d",&inquiry); //需要查詢的次數(shù)for(int i=0;i<inquiry;i++){same=0;scanf("%d%d",&set1,&set2); //比較的集合編號(hào)int count=st[set2].size();for(set<int>::iterator it=st[set1].begin();it!=st[set1].end();it++){//從第一個(gè)集合中的每一個(gè)數(shù)，看是否有數(shù)和st[set2]中的數(shù)一致，一致就same++,不一致則count++if(st[set2].find(*it)!=st[set2].end()){//一開始弄錯(cuò)的地方same++;//printf("%d\n",same);}else{count++;//printf("c%d\n",count);}}printf("%.1lf%%\n",(double)(same*100.0/count));}return 0; }

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