D - Replacing

Time Limit: 2 sec / Memory Limit: 1024 MB

Score :$400400$points

Problem Statement

You have a sequence$\mathrm{AA}$composed of$\mathrm{NN}$positive integers:${\mathrm{A1,A2,\cdots ,ANA1,A2,\cdots ,AN}}^{}$.

You will now successively do the following$\mathrm{QQ}$operations:

• In the$\mathrm{ii}$-th operation, you replace every element whose value is${\mathrm{BiBi}}^{}$with${\mathrm{CiCi}}^{}$.

For each

Constraints

• All values in input are integers.
• $\mathrm{1\le N,Q,Ai,Bi,Ci\le 1051\le N,Q,Ai,Bi,Ci\le 105}$
• ${\mathrm{Bi\ne CiBi\ne Ci}}^{}$

Input

Input is given from Standard Input in the following format:

$\mathrm{NN}$
${\mathrm{A1A1}}^{}$ ${\mathrm{A2A2}}^{}$ $\cdots \cdots$ ${\mathrm{ANAN}}^{}$
$\mathrm{QQ}$
${\mathrm{B1B1}}^{}$ ${\mathrm{C1C1}}^{}$
${\mathrm{B2B2}}^{}$ ${\mathrm{C2C2}}^{}$
$⋮⋮$
${\mathrm{BQBQ}}^{}$ ${\mathrm{CQCQ}}^{}$

Output

Print$\mathrm{QQ}$integers${\mathrm{SiSi}}^{}$to Standard Output in the following format:

${\mathrm{S1S1}}^{}$
${\mathrm{S2S2}}^{}$
$⋮⋮$
${\mathrm{SQSQ}}^{}$

Note that${\mathrm{SiSi}}^{}$may not fit into a$3232$-bit integer.

Sample Input 1Copy

4
1 2 3 4
3
1 2
3 4
2 4

Sample Output 1Copy

11
12
16

Initially, the sequence$\mathrm{AA}$is$\mathrm{1,2,3,41,2,3,4}$

After each operation, it becomes the following:

• $\mathrm{2,2,3,42,2,3,4}$
• $\mathrm{2,2,4,42,2,4,4}$
• $\mathrm{4,4,4,44,4,4,4}$

Sample Input 2Copy

4
1 1 1 1
3
1 2
2 1
3 5

Sample Output 2Copy

8
4
4

Note that the sequence$\mathrm{AA}$may not contain an element whose value is${\mathrm{BiBi}}^{}$.

Sample Input 3Copy

2
1 2
3
1 100
2 100
100 1000

Sample Output 3Copy

102
200
2000

題意：輸入一個n，然後輸入n個數字，再輸入一個Q，接下來Q行每行輸入一個B和C，然後將陣列中所有的B換成C，然後把新陣列的和輸出
題解：開一個數組b，用來儲存陣列元素中出現的次數，我們在輸入陣列的元素的時候，先用sum把每個元素加起來，然後陣列b中a【i】元素出現的次數++(這樣能讓我們在後面O(1)的時間算出新陣列的總和)
每次B,C輸入的時候其實直接用公式sum=sum-(C-B)*b[B];求出sum，然後將陣列b更新一下。
程式碼如下：

#include<cstdio>
#include<iostream>
#define maxn 100005
#define ll long long 
using namespace std;
ll n,q,a[maxn],sum;
int tm[maxn];
int main(void)
{
    while(~scanf("%lld",&n))
    {
        sum=0;
        for(int i=0;i<n;++i)
        {
            scanf("%lld",&a[i]);
            sum+=a[i];//求和
            tm[a[i]]++;//a[i]出現的次數自增
        }
        scanf("%lld",&q);
        ll b,c;
        for(int i=0;i<q;++i)
        {
            scanf("%lld %lld",&b,&c);
            sum+=(c-b)*tm[b];
            tm[c]+=tm[b];//更新tm[c]
            tm[b]=0;//更新tm[b]
            printf("%lld\n",sum);
        }
    }
    return 0;
}