1063. Set Similarity

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 (<=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.

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%

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<set>
using namespace std;
set<int> num[51];
int main(){
    int N, M, temp, K;
    scanf("%d", &N);
    for(int i = 1; i <= N; i++){
        scanf("%d", &M);
        for(int j = 0; j < M; j++){
            scanf("%d", &temp);
            num[i].insert(temp);
        }
    }
    scanf("%d", &K);
    int s1, s2;
    for(int i = 0; i < K; i++){
        scanf("%d %d", &s1, &s2);
        int same = 0, diff = num[s2].size();
        for(set<int> ::iterator it = num[s1].begin(); it != num[s1].end(); it++){
            if(num[s2].find(*it) != num[s2].end())
                same++;
            else diff++;
        }
        double rate = 1.0 * same / diff * 100.0;
        printf("%.1lf%%\n", rate);
    }
    cin >> N;
    return 0;
}

总结：

1、set可以用于hash表解决不了的情况，比如键值不是int的情况。

2、printf输出%时，需要输出%%。