51nod1202【DP-树状数组维护】

思路：
DP[i]代表从1 到 i 以 a[i] 为末尾的子序列个数，dp[i]=dp[i]+dp[j](a[i]!=a[j]) +1

利用树状数组维护以值 a[i] 结尾的子序列个数。

#include<bits/stdc++.h>using namespace std;typedef long long LL;const LL mod=1e9+7;const int N=1e5+10;LL c[N];int lowbit(int x){return x&(-x);}void add(int d,LL v){while(d<N){c[d]=(c[d]+v)%mod;d+=lowbit(d);}}LL Sum(int x){LL ans=0;while(x){ans=(ans+c[x])%mod;x-=lowbit(x);}return ans;}int main(){int n,x;memset(c,0,sizeof(c));scanf("%d",&n);for(int i=1;i<=n;i++){LL sum=0;scanf("%d",&x);sum=(Sum(100000)+1)%mod;sum=(sum-Sum(x)+Sum(x-1)+mod)%mod;add(x,sum);}printf("%lld\n",Sum(100000));return 0;}