Light oj 1085

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 + 5;
int a[N];
int b[N], n;
LL bit[N], mod = 1e9 + 7;
LL dp[N];

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

LL sum(int i) {
    LL s = 0;
    for( ; i >= 1; i -= (i&-i))
        s += bit[i];
    return s;
}

int main()
{
    int t;
    scanf("%d", &t);
    for(int ca = 1; ca <= t; ++ca) {
        scanf("%d", &n);
        for(int i = 1; i <= n; ++i) {
            scanf("%d", a + i);
            b[i] = a[i];
        }
        sort(b + 1, b + n + 1);
        memset(bit, 0, sizeof(bit));
        memset(dp, 0, sizeof(dp));
        for(int i = 1; i <= n; ++i) {
            a[i] = lower_bound(b + 1, b + n + 1, a[i]) - b;
            dp[i]++;
        }
        for(int i = 1; i <= n; ++i) {
            dp[i] += sum(a[i] - 1);
            dp[i] %= mod;
            add(a[i], dp[i]);
        }
        LL ans = 0;
        for(int i = 1; i <= n; ++i) {
            ans += dp[i];
            ans %= mod;
        }
        printf("Case %d: %lld
", ca, ans);
    }
    return 0;
}