Beauty of Array ZOJ - 3872(思維題)
阿新 • • 發佈:2018-11-29
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
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
題意就是給你一個數組,讓你求1~1,1~2,1~3,1~4.... 1~n,
2~2,2~3, 2~4.... 2~n
....................
n-1~n-1 n-1~n
n~n
這些區間內不同數字的和,然後求總和。
思路:就是從右往左遍歷,找某個數右邊第一次出現這個數的位置,如果是第一次出現那麼這個位置就是n
然後加加乘乘就行了
程式碼如下:
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 using namespace std; 5 const int N=1e6+10; 6 const int NN=1e5+100; 7 int pos[N],a[NN]; 8 int main() 9 { 10 int t; 11 int n; 12 scanf("%d",&t); 13 long long ans,tans; 14 while(t--) 15 { 16 scanf("%d",&n); 17 for(int i=0;i<n;i++) 18 { 19 scanf("%d",&a[i]); 20 pos[a[i]]=-1;//標記每個數都沒出現過 21 } 22 ans=0,tans=0; 23 for(int i=n-1;i>=0;i--) 24 { 25 if(pos[a[i]]==-1) 26 tans+=a[i]*(n-i); 27 else 28 tans+=a[i]*(pos[a[i]]-i); 29 pos[a[i]]=i; 30 ans+=tans; 31 } 32 printf("%lld\n",ans); 33 } 34 return 0; 35 }