题目
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%
题目大意
给出两个集合数据，Nc表示两个集合中相同的元素，Nt表示两个集合总的不相同元素的个数，求Nc/Nt。
思路
首先题目要对集合去重，这里可以直接用set实现，并且set内部自动实现排序，后面统计数据会更方便，统计Nc,Nt可以用两个指针实现，具体见代码。
代码
#include <iostream>
#include <cstdio>
#include <set>
using namespace std;
bool cmp(int a, int b){
return a < b;
}
int main(){
int n, k;
set<int> sets[51];
scanf("%d", &n);
for(int i=1, t; i<=n; i++){
scanf("%d", &t);
for(int j=0, c; j<t; j++){
scanf("%d", &c);
sets[i].insert(c);
}
}
scanf("%d", &k);
while(k--){
int s, t, nc = 0, nt = 0;
scanf("%d%d", &s, &t);
set<int>::iterator it1 = sets[s].begin(), it2 = sets[t].begin();
while(it1 != sets[s].end() && it2 != sets[t].end()){
if(*it1 > *it2)
nt++, it2++;
else if(*it1 == *it2)
nt++, nc++, it1++, it2++;
else
nt++, it1++;
}
while(it1 != sets[s].end())
nt++, it1++;
while(it2 != sets[t].end())
nt++, it2++;
printf("%.1f%%\n", (nc*1.0)/nt*100);
}
return 0;
}