Light oj 1085 - All Possible Increasing Subsequences (简单dp + 离散化 + BIT)

题目链接：http://www.lightoj.com/volume_showproblem.php?problem=1085

题意：

问你有多少个上升子序列。

思路：

dp[i]表示以第i个数结尾的上升序列数量。

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <map>
 using namespace std;
 typedef long long LL;
 const int N = 1e5 + ;
 int a[N];
 int b[N], n;
 LL bit[N], mod = 1e9 + ;
 LL dp[N];

 void add(int i, LL val) {
     for( ; i <= n; i += (i&-i))
         bit[i] += val;
 }

 LL sum(int i) {
     LL s = ;
     for( ; i >= ; i -= (i&-i))
         s += bit[i];
     return s;
 }

 int main()
 {
     int t;
     scanf("%d", &t);
     for(int ca = ; ca <= t; ++ca) {
         scanf("%d", &n);
         for(int i = ; i <= n; ++i) {
             scanf("%d", a + i);
             b[i] = a[i];
         }
         sort(b + , b + n + );
         memset(bit, , sizeof(bit));
         memset(dp, , sizeof(dp));
         for(int i = ; i <= n; ++i) {
             a[i] = lower_bound(b + , b + n + , a[i]) - b;
             dp[i]++;
         }
         for(int i = ; i <= n; ++i) {
             dp[i] += sum(a[i] - );
             dp[i] %= mod;
             add(a[i], dp[i]);
         }
         LL ans = ;
         for(int i = ; i <= n; ++i) {
             ans += dp[i];
             ans %= mod;
         }
         printf("Case %d: %lld\n", ca, ans);
     }
     return ;
 }