lightoj 1085

#include <set>
#include <map>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <cassert>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#define pi acos(-1.0)
#define maxn (100000 + 50)
#define mod 1000000007
#define Lowbit(x) (x & (-x))
using namespace std;
typedef long long int LLI;

int a[maxn];
int acopy[maxn];
int c[maxn];

void Add(int x,int add,int n) {
while(x <= n) {
c[x] = (c[x] + add) % mod;
x = x + Lowbit(x);
}
}

int Sum(int n) {
int sum = 0;
while(n > 0) {
sum = (sum + c
) % mod;
n = n - Lowbit(n);
}
return sum % mod;
}

int main() {
//    freopen("in.txt", "r", stdin);
//    freopen("out.txt", "w", stdout);
int t,n;
scanf("%d",&t);
for(int Case = 1; Case <= t; Case ++) {
scanf("%d",&n);
int cnt = 1;
memset(c,0,sizeof(c));
for(int i = 1; i <= n; i ++)  {
scanf("%d",&a[i]);
acopy[i] = a[i];
}
sort(acopy + 1,acopy + n + 1);
int n2 = unique(acopy + 1,acopy + n + 1) - acopy;//unique删除重复元素(实际上是把重复的元素都挪到数组后面，返回不重复数组最后一个元素的下标 + 1)
for(int i = 1; i <= n; i ++) {
int pos = lower_bound(acopy + 1,acopy + n2,a[i]) - acopy;//离散化
Add(pos,Sum(pos - 1) + 1,n2 - 1);
}
printf("Case %d: %d\n",Case,Sum(n2 - 1));//减1是因为访问不到n2的位置
}
return 0;
}
/*
c[i]表示以第i个元素a[i]结尾的符合条件的子序列有c[i]个,假设上一个比a[i]小的元素下标为k,显然c[i]等于c[k] + 1;
那么离散化之后a[i]在这个数组中的位置是pos,那么上一个比它小的元素的下标一定是 pos - 1
其实相当于一个一个的把a[i]放进序列
附数据(input)：
10
2
7 -7
3
16 4 -10
12
-14 3 5 16 8 -11 7 0 5 8 8 -8
12
15 3 -6 -13 1 -9 6 15 -9 14 16 13
6
-14 12 10 -8 8 11
10
-4 10 1 0 0 16 -11 -9 7 -4
2
5 14
以及离散化后a[i]的坐标以及a[i] 的值：
Case 1:
pos =   2   a[i] =   7
pos =   1   a[i] =  -7
Case 2:
pos =   3   a[i] =  16
pos =   2   a[i] =   4
pos =   1   a[i] = -10
Case 3:
pos =   1   a[i] = -14
pos =   5   a[i] =   3
pos =   6   a[i] =   5
pos =  12   a[i] =  16
pos =   8   a[i] =   8
pos =   2   a[i] = -11
pos =   7   a[i] =   7
pos =   4   a[i] =   0
pos =   6   a[i] =   5
pos =   8   a[i] =   8
pos =   8   a[i] =   8
pos =   3   a[i] =  -8
Case 4:
pos =   9   a[i] =  15
pos =   5   a[i] =   3
pos =   3   a[i] =  -6
pos =   1   a[i] = -13
pos =   4   a[i] =   1
pos =   2   a[i] =  -9
pos =   6   a[i] =   6
pos =   9   a[i] =  15
pos =   2   a[i] =  -9
pos =   8   a[i] =  14
pos =  10   a[i] =  16
pos =   7   a[i] =  13
Case 5:
pos =   1   a[i] = -14
pos =   6   a[i] =  12
pos =   4   a[i] =  10
pos =   2   a[i] =  -8
pos =   3   a[i] =   8
pos =   5   a[i] =  11
Case 6:
pos =   3   a[i] =  -4
pos =   7   a[i] =  10
pos =   5   a[i] =   1
pos =   4   a[i] =   0
pos =   4   a[i] =   0
pos =  10   a[i] =  16
pos =   1   a[i] = -11
pos =   2   a[i] =  -9
pos =   6   a[i] =   7
pos =   3   a[i] =  -4
Case 7:
pos =   1   a[i] =   5
pos =   2   a[i] =  14
Case 8:
pos =   1   a[i] =   5
pos =   2   a[i] =  14
Case 9:
pos =   1   a[i] =   5
pos =   2   a[i] =  14
Case 10:
pos =   1   a[i] =   5
pos =   2   a[i] =  14
运行结果(output)：
Case 1: 2
Case 2: 3
Case 3: 121
Case 4: 137
Case 5: 21
Case 6: 37
Case 7: 3
Case 8: 3
Case 9: 3
Case 10: 3
*/