zoj The 12th Zhejiang Provincial Collegiate Programming Contest Beauty of Array

http://acm.zju.edu.cn/onlinejudge/showContestProblem.do?problemId=5496

The 12th Zhejiang Provincial Collegiate Programming Contest - D

Beauty of Array

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Time Limit: 2 Seconds       Memory Limit: 65536 KB

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Edward has an array A with N integers. He defines the beauty of an array as the summation of all distinct integers in the array. Now Edward wants to know the summation of the beauty of all contiguous subarray of the array A.

Input

There are multiple test cases. The first line of input contains an integer T indicating the number of test cases. For each test case:

The first line contains an integer N (1 <= N <= 100000), which indicates the size of the array. The next line contains N positive integers separated by spaces. Every integer is no larger than 1000000.

Output

For each case, print the answer in one line.

Sample Input

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

Sample Output

105
21
38

分析：

水题、

 AC代码：

 

 1 #include <iostream>
 2 #include <stdio.h>
 3 #include <string.h>
 4 #include <stack>
 5 #include <queue>
 6 #include <map>
 7 #include <set>
 8 #include <vector>
 9 #include <math.h>
10 #include <algorithm>
11 using namespace std;
12 #define ls 2*i
13 #define rs 2*i+1
14 #define up(i,x,y) for(i=x;i<=y;i++)
15 #define down(i,x,y) for(i=x;i>=y;i--)
16 #define mem(a,x) memset(a,x,sizeof(a))
17 #define w(a) while(a)
18 #define LL long long
19 const double pi = acos(-1.0);
20 #define Len 1000005
21 #define mod 786433
22 #define exp 1e-5
23 const int INF = 0x3f3f3f3f;
24 
25 LL dp[Len];
26 int hsh[Len];
27 
28 int main()
29 {
30     int t,n,a;
31     scanf("%d",&t);
32     w(t--)
33     {
34         memset(hsh,0,sizeof(hsh));
35         scanf("%d",&n);
36         dp[0]=0;
37         for(int i=1;i<=n;i++)
38         {
39             scanf("%d",&a);
40             dp[i]=dp[i-1]+a+(i-1-hsh[a])*a;
41             hsh[a]=i;
42         }
43         LL ans=0;
44         for(int i=1;i<=n;i++)
45             ans+=dp[i];
46         printf("%lld\n",ans);
47     }
48 }

View Code

 

 

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <iostream>
 4 #include <algorithm>
 5 #include <cmath>
 6 #include <queue>
 7 #include <stack>
 8 #include <vector>
 9 #include <map>
10 #include <set>
11 #include <deque>
12 #include <cctype>
13 #define LL long long
14 #define INF 0x7fffffff
15 using namespace std;
16 long long int dp[100005];
17 int a[100005];
18 int main() {
19     int t;
20     scanf("%d",&t);
21     while(t--){
22         int n;
23                 
24         scanf("%d",&n);
25         map<int,int> num; //num记录前一个a[i]的位置 
26         
27         for(int i = 0;i < n;i++){
28             scanf("%d",&a[i]);
29             num[a[i]] = 0;     
30         }
31         dp[0] = 0;
32         dp[1] = a[0];
33         num[a[0]] = 1;
34         
35         for(int i = 1;i < n;i++){
36             /*
37                 dp[i] 为前i个美丽的数组的和
38                 前i个和包括前i-1个的和 dp[i-1] 还有由第i个数组成的和 dp[i] - dp[i-1] + a[i] * (i+1 - num[a[i]]);
39                 其中的i+1 - num[a[i]] 是用于去除重复的a[i],重复的只计算一次 
40             */ 
41             dp[i+1] = dp[i] + dp[i] - dp[i-1] + a[i] * (i+1 - num[a[i]]);
42             num[a[i]] = i+1; //将新的位置记录下来 
43         }
44         cout << dp[n] << endl;
45     }
46     return 0;
47 }

View Code

 

另外附上本次比赛的终榜：（1-104）