51nod 子序列个数 dp

看上去很裸的题,依旧是dp.
晚点来补.
dp[i]前i个的所有子序列,sum[j],以j结尾的子序列数量.
那么dp[i] = dp[i-1] (不加第i个) +(dp[i-1]-sum[a[i]]) (分三块,剪去还没以a[i]结尾的所有子序列数量.)

#include <algorithm>
#include <iostream>
#include <vector>
using namespace std;
#define debug(x) std::cerr << #x << " = " << (x) << std::endl
typedef long long LL;
const int MAXN = 1e5 + 17;
const int MOD = 1e9 + 7;
int a[MAXN], sum[MAXN], dp[MAXN];
int main(int argc, char const* argv[])
{
#ifdef noob
    freopen("Input.txt", "r", stdin);
    freopen("Output.txt", "w", stdout);
#endif
    int n;
    cin >> n;
    for (int i = 1; i <= n; ++i) {
        cin >> a[i];
    }
    dp[0] = 1;
    for (int i = 1; i <= n; ++i) {
        dp[i] = (dp[i] + dp[i - 1] - sum[a[i]]) % MOD;
        dp[i] = (dp[i] + dp[i - 1]) % MOD;
        sum[a[i]] = (sum[a[i]] + dp[i - 1] - sum[a[i]]);
    }
    cout << ((dp[n] - 1) % MOD + MOD) % MOD << endl;
    return 0;
}