求一个序列所有区间（区间内不同数的个数）的和

求一个序列所有区间（区间内不同数的个数）的和

链接：https://ac.nowcoder.com/acm/contest/888/B
来源：牛客网

Gromah and LZR have entered the second level. There is a sequence a1,a2,⋯ ,an on the wall.

There is also a note board saying "the beauty value of a sequence is the number of different elements in the sequence".

LZR soon comes up with the password of this level, which is the sum of the beauty values of all successive subintervals of the sequence on the wall.

Please help them determine the password!

输入描述:

The first line contains one positive integer nn_{}n, denoting the length of the sequence.

The second line contains nn_{}n positive integers a1,a2,⋯ ,ana_1, a_2, \cdots, a_na1,a2,⋯,an, denoting the sequence.

1 ≤ \(a_i\) ≤ n ≤ \(10^5\)

输出描述:

Print a non-negative integer in a single line, denoting the answer.示例1

输入

4
1 2 1 3

输出

18

说明
The beauty values of subintervals [1,1],[2,2],[3,3],[4,4] are all 1.
The beauty values of subintervals [1,2],[1,3],[2,3],[3,4] are all 2.
The beauty values of subintervals [1,4],[2,4] are all 3.
As a result, the sum of all beauty values are 1×4+2×4+3×2=18

题意

求一个序列所有区间（区间内不同数的个数）的和

题解

递推，记录\(a_i\)出现的位置，如果\(a_i\)以前出现的位置是cnt[\(a_i\)],那么区间[L, i]和[L, i-1]（0<L<=cnt[\(a_i\)])的结果是一样的，[L, i]比[L,i-1]（cnt[\(a_i\)] < L <= i)多1,因为在这些区间，\(a_i\)属于不同的数。

#include <cstdio>
int cnt[100010];
int main() {
    int n, a;
    scanf("%d", &n);
    long long sum[100010], ans = 0;
    sum[0] = 0;
    for(int i = 1; i <= n; i++) {
        scanf("%d", &a);
        sum[i] = sum[i-1] + (i - cnt[a]);
        cnt[a] = i;
        ans += sum[i];
    }
    printf("%lld\n", ans);
    return 0;
}