PAT 甲级 1063 Set Similarity (25 分) （新学，set的使用，printf 输出%，要%%）

1063 Set Similarity (25 分)

 

Given two sets of integers, the similarity of the sets is defined to be /, 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 (≤) which is the total number of sets. Then N lines follow, each gives a set with a positive M (≤) and followed by M integers in the range [0]. After the input of sets, a positive integer K (≤) is given, followed by Klines 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%

作者: CHEN, Yue

单位: 浙江大学

时间限制: 500 ms

内存限制: 64 MB

 

题意：

本题定义集合相似度Nc/Nt*100%,其中Nc是两个集合中共有的不相等元素个数，Nt是两个集合中不相等元素的个数。你的任务是计算给定的集合的相似度。

Nc是两个集合的交集元素数量。
Nt是两个集合的并集元素数量。

题解:

本来用数组，明显内存超限，速学了set，很方便啊！！！一下子过了。

printf 输出%，要%%

AC代码：

 

#include<iostream>
#include<set>
#include<algorithm>
#include<string>
#include<cstring>
using namespace std;
set<int>s[55];
int n,m,k;
int main(){
    cin>>n;
    for(int i=1;i<=n;i++){
        cin>>m;
        for(int j=1;j<=m;j++){
            int x;
            cin>>x;
            s[i].insert(x);
        }
    }
    cin>>k;
    for(int i=1;i<=k;i++){
        int a,b;
        cin>>a>>b;
        set<int>::iterator it;
        int count=0;
        for(it=s[a].begin();it!=s[a].end();it++){
            if(s[b].find(*it)!=s[b].end()) count++;
        }    
        printf("%.1f%%\n",count*1.0/(s[a].size()+s[b].size()-count)*100);
    }
    return 0;
}