In Berland each high school student is characterized by academic performance — integer value between 1 and 5.
In high school 0xFF there are two groups of pupils: the group A and the group B. Each group consists of exactly n students. An academic performance of each student is known — integer value between 1 and 5.
The school director wants to redistribute students between groups so that each of the two groups has the same number of students whose academic performance is equal to 1, the same number of students whose academic performance is 2 and so on. In other words, the purpose of the school director is to change the composition of groups, so that for each value of academic performance the numbers of students in both groups are equal.
To achieve this, there is a plan to produce a series of exchanges of students between groups. During the single exchange the director selects one student from the class A and one student of class B. After that, they both change their groups.
Print the least number of exchanges, in order to achieve the desired equal numbers of students for each academic performance.
The first line of the input contains integer number n (1 ≤ n ≤ 100) — number of students in both groups.
The second line contains sequence of integer numbers a1, a2, ..., an (1 ≤ ai ≤ 5), where ai is academic performance of the i-th student of the group A.
The third line contains sequence of integer numbers b1, b2, ..., bn (1 ≤ bi ≤ 5), where bi is academic performance of the i-th student of the group B.
Print the required minimum number of exchanges or -1, if the desired distribution of students can not be obtained.
4
5 4 4 4
5 5 4 5
1
6
1 1 1 1 1 1
5 5 5 5 5 5
3
1
5
3
-1
9
3 2 5 5 2 3 3 3 2
4 1 4 1 1 2 4 4 1
4
题意:两组长度为n的数字 每个数字的范围为[1,5] 现在要求两个组内 [1,5]各个数字出现的次数相同 问最少要交换两个组内的数字多少次.
题解:若某一个数字在两组内出现的次数之和为奇数 则无论怎么交换都无法满足要求,输出“-1” 其他情况下题目都会有解
1 #include <iostream> 2 #include <cstdio> 3 #include <cstdlib> 4 #include <cstring> 5 #include <algorithm> 6 #include <stack> 7 #include <queue> 8 #include <cmath> 9 #include <map> 10 #define ll __int64 11 #define mod 1000000007 12 #define dazhi 2147483647 13 using namespace std; 14 int n; 15 int a[10]; 16 int b[10]; 17 int c[10]; 18 int x; 19 int main() 20 { 21 int ans1=0; 22 scanf("%d",&n); 23 for(int i=1;i<=n;i++) 24 { 25 scanf("%d",&x); 26 a[x]++; 27 c[x]++; 28 } 29 for(int i=1;i<=n;i++) 30 { 31 scanf("%d",&x); 32 b[x]++; 33 c[x]++; 34 } 35 for(int i=1;i<=5;i++) 36 { 37 if(c[i]%2) 38 { 39 printf("-1\n"); 40 return 0; 41 } 42 else{ 43 if(a[i]>b[i]) 44 { 45 ans1+=a[i]-c[i]/2; 46 } 47 } 48 } 49 printf("%d\n",ans1); 50 return 0; 51 }