Replacing

Problem Statement

You have a sequence A composed of N positive integers: 1,2,⋯,A1​,A2​,⋯,AN​.

You will now successively do the following Q operations:

• In the i-th operation, you replace every element whose value is Bi​ withCi​.

For each i (1≤i≤Q), find Si​: the sum of all elements in A just after the i-th operation.

Constraints

• All values in input are integers.
• 1≤N,Q,Ai​,Bi​,Ci​≤105
• Bi​!=Ci​

Input

Input is given from Standard Input in the following format:

N
1A1​ 2A2​ ⋯⋯ AN​
Q
B1​ C1​
B2​ C2​
⋮⋮
BQ​ CQ​

Output

Print Q integers Si​ to Standard Output in the following format:

S1​
S2​
⋮⋮
SQ​

Note that Si​ may not fit into a 3232-bit integer.

Sample 1

Input	Output
4 1 2 3 4 3 1 2 3 4 2 4	11 12 16

Initially, the sequence A is 1,2,3,41,2,3,4.

After each operation, it becomes the following:

• 2,2,3,4
• 2,2,4,4
• 4,4,4,4

Sample 2

Input	Output
4 1 1 1 1 3 1 2 2 1 3 5	8 4 4

Note that the sequenceA may not contain an element whose value is Bi​.

Sample 3

Inputcopy	Outputcopy
2 1 2 3 1 100 2 100 100 1000	102 200 2000

题目分析&解题思路:

每次输入a,b把数组当中所有的a全部换成b,那么对于所有数的和来说就是加上（b-a）*a的个数，然后再更新b的个数和a的个数就行

代码:

#include<bits/stdc++.h>
#define _for(a,b) for(int i=a;i<=b;i++)
#define for_(a,b) for(int i=a;i>=b;i--)
#define pii pair<int,int>
#define use int T;cin>>T;while(T--)
using namespace std;
const int N =1e5+6;
int ar[N];
int cnt[N];
int main()
{
   ios::sync_with_stdio(0);
   cin.tie(0);cout.tie(0);
 
 int n;
 cin>>n;
 long long sum=0;
 _for(1,n)
 	{
 	cin>>ar[i];
	 cnt[ar[i]]++;
	 	sum+=ar[i];
}
 	int q;cin>>q;
 	while(q--)
 {
 	int a,b;cin>>a>>b;
 	if(cnt[a]!=0)
 	{
 		
 		sum+=(b-a)*cnt[a];
 		cnt[b]+=cnt[a];
 		cnt[a]=0;
	}
	cout<<sum<<endl;
 }

return 0;
}