1063 Set Similarity (25 分)
2021/9/14 6:08:22
本文主要是介绍1063 Set Similarity (25 分),对大家解决编程问题具有一定的参考价值,需要的程序猿们随着小编来一起学习吧!
Given two sets of integers, the similarity of the sets is defined to be N c / N t × 100 N_c/N_t×100 Nc/Nt×100%, where N c N_c Nc is the number of distinct common numbers shared by the two sets, and N t N_t 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.
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 ( ≤ 1 0 4 ≤10^4 ≤104) and followed by M integers in the range [0, 1 0 9 10^9 109]. 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.
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%
解释:
输入集合总数N
然后接下来N行输入集合,包括集合包元素数,和集合内各个元素
然后输入M个查询,每次输入两个集合序号(1-N)
输出集合的交并比=交集元素个数/并集元素个数*100%,每个元素要去重
#include<cstdio> #include<set> using namespace std; const int N=51; //定义一个集合的数组 set<int> st[N]; void compare_set(int x,int y) { int totaNum=st[y].size(),samenum=0; for(set<int> ::iterator it=st[x].begin();it!=st[x].end();it++)//迭代器访问set中各个元素 { //找到一样的,交集元素个数 if(st[y].find(*it)!=st[y].end()) samenum++; //并集元素个数++ else totaNum++; } printf("%.1f%\n",samenum*100.0/totaNum); } int main() { int n,k,v,q; int st1,st2; //输入·set总数 scanf("%d",&n); for(int i=1;i<=n;i++) { scanf("%d",&k);//输入当前set元素总数 for(int j=0;j<k;j++) { scanf("%d",&v); st[i].insert(v);//输入到set[i]中 } } //输入查询总数 scanf("%d",&q); for(int i=0;i<q;i++) { scanf("%d %d",&st1,&st2); compare_set(st1,st2); } return 0; } /* 3 3 99 87 101 4 87 101 5 87 7 99 101 18 5 135 18 99 2 1 2 1 3 */
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