Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2048 Accepted Submission(s): 805
Problem Description
Recently, dobby is addicted in the Fruit Ninja. As you know, dobby is a free elf, so unlike other elves, he could do whatever he wants.But the hands of the elves are somehow strange, so when he cuts the fruit, he can only make specific move of his hands. Moreover, he can only start his hand in point A, and then move to point B,then move to point C,and he must make sure that point A is the lowest, point B is the highest, and point C is in the middle. Another elf, Kreacher, is not interested in cutting fruits, but he is very interested in numbers.Now, he wonders, give you a permutation of 1 to N, how many triples that makes such a relationship can you find ? That is , how many (x,y,z) can you find such that x < z < y ?
Input
The first line contains a positive integer T(T <= 10), indicates the number of test cases.For each test case, the first line of input is a positive integer N(N <= 100,000), and the second line is a permutation of 1 to N.
Output
For each test case, ouput the number of triples as the sample below, you just need to output the result mod 100000007.
Sample Input
2
6
1 3 2 6 5 4
5
3 5 2 4 1
Sample Output
Case #1: 10
Case #2: 1
Source
题意：给出1～N的一种排列，问存在多少个三元组(x,y,z) x < z < y 其中x y z 的位置递增
可先求出满足 x < y < z 和x < z < y的总数tot， 总数为：对于每一个数 x， 从x后面的位置中比x大的num个数中选择任意两个， 即 tot += comb[ num ][2]
再从总数tot中减去 x < y < z的数量即为答案， x < y < z 的个数为： 对于一个数y， 在y的前面比 y小的个数为 low， 在y的后面比y大的个数为 high， 根据组合原理，在low中选一个x， 在high中选一个z，共有low × high中情况
如何求 每个数的low值和high值， 就用树状数组来统计
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <string>
#include <queue>
#include <map>
#define Lowbit(x) ((x) & (-x))
using namespace std;
typedef long long LL;
const int N = 100005;
const int M = 100000007;
int c[N], a[N];
LL low[N], high[N], comb[N][5];
int n;
void pre_c()
{
for(int i = 0; i < N; ++i)
for(int j = 0; j <= min(i,2); ++j)
comb[i][j] = (i == 0 || j == 0) ? 1:((comb[i - 1][j] % M + comb[i - 1][j - 1]) % M);
}
void update(int pos)
{
while(pos <= n) {
c[pos]++;
pos += Lowbit(pos);
}
}
LL sum(int pos)
{
LL res = 0;
while(pos) {
res += c[pos];
pos -= Lowbit(pos);
}
return res;
}
int main()
{
int _, v, cas = 1;
pre_c();
scanf("%d", &_);
while(_ --)
{
scanf("%d", &n);
memset(c, 0, sizeof c);
for(int i = 1; i <= n; ++i) {
scanf("%d", &v);
a[i] = v;
low[i] = sum(v); update(v);
high[i] = (n - v) - (i - 1 - low[i]);
} // for(int i = 1; i <= n; ++i) printf("%lld %lld\n", low[i], high[i]);
LL tot = 0;
for(int i = 1; i <= n; ++i) {
tot = tot % M + comb[ high[i] ][2];
tot = (tot - (low[i] % M * high[i] % M) + M) % M;
}
printf("Case #%d: %lld\n",cas++, tot % M ); }
}