Light OJ 1085 - All Possible Increasing Subsequences

题目

分析

Code

#include <bits/stdc++.h>

using namespace std;
typedef long long LL;
const int maxn = 1e5 + 131;
const LL MOD = 1e9 + 7;

LL Num[maxn], A[maxn], B[maxn];
int n;
int lowebit(int x) { return x&(-x); }
LL Sum(int x) {
LL ret = 0;
while(x > 0)
{
ret = (ret + Num[x]) % MOD;
x -= lowebit(x);
}
return ret;
}
void Update(int x, LL Val) {
while(x <= n)
{
Num[x] = (Num[x] + Val) % MOD;
x += lowebit(x);
}
}

int main()
{
int t;
scanf("%d",&t);
for(int kase = 1; kase <= t; ++kase)
{
scanf("%d",&n);
memset(Num, 0, sizeof(Num));
for(int i = 0; i < n; ++i)
scanf("%lld",A+i), B[i] = A[i];
sort(A, A+n);
int Max = unique(A, A+n)-A;
for(int i = 0; i < n; ++i)
{
B[i] = lower_bound(A, A+Max, B[i])-A + 1;
}
//cout << "1" <<endl;
for(int i = 0; i < n; ++i)
{
LL s = Sum(B[i]-1) + 1;
//cout << "2" << "  " <<B[i] <<endl;
s %= MOD;
Update(B[i], s);
//cout << "3" <<endl;
}
LL Ans = Sum(n);
printf("Case %d: %lld\n",kase, Ans);
}
return 0;
}