PAT A1063

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.

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%

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#include "stdio.h"
//#include "math.h"
#include "string.h"
//#include "iostream"
//#include "stdlib.h"
#include "algorithm"
//#include "vector"
#include "set"
using namespace std;
//typedef long long LL;
const int N = 51;
set<int> st[N];//N个集合

void compare(int x, int y){//比较集合st[x]与集合st[y]
int totalNum = st[y].size(), sameNum = 0;//不同数的个数,相同数的个数
//遍历集合st[x]
for (set<int>::iterator it = st[x].begin(); it != st[x].end(); it++) {
if (st[y].find(*it) != st[y].end()) sameNum++;//在st[y]中能找到该元素
else totalNum++;//在st[y]中不能找到该元素
}
printf("%.1f%\n", sameNum * 100.0 / totalNum);//输出比率
}
int main(){
int n, k, q, v, st1, st2;
scanf("%d", &n);//集合个数
for (int i = 1; i <= n; i++) {
scanf("%d", &k);//集合i中的元素个数
for (int j = 0; j < k; j++) {
scanf("%d", &v);//集合i中的元素v
st[i].insert(v);
}
}
scanf("%d", &q); //q个查询
for (int i = 0; i < q; i++) {
scanf("%d%d", &st1, &st2);
compare(st1, st2);
}
return 0;
}

set.find(*it) 找不到会到set.end()位置处