ZOJ 3872 Beauty of Array
2015-09-06 12:38
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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
Sample Output
定义一个序列的美丽值为序列中不同数的和,求一个序列的所有子序列的美丽值的和
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
定义一个序列的美丽值为序列中不同数的和,求一个序列的所有子序列的美丽值的和
#include <stdio.h> #include <string.h> int main() { long long int dp[100005],pre[100005]; int n,cas,temp; scanf("%d",&cas); while(cas--) { scanf("%d",&n); memset(pre,0,sizeof pre); memset(dp,0,sizeof dp); for(int i=1;i<=n;i++) { scanf("%d",&temp); dp[i]=dp[i-1]+(i-pre[temp])*temp;//之前的加上 i*temp(每项都多了一个temp)但因为重复了 要减去pre[temp]的temp次 pre[temp]=i; } long long int ans=0; for(int i=1;i<=n;i++) ans+=dp[i]; printf("%lld\n",ans); } return 0; }
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