Codeforces Beta Round #57 (Div. 2) Enemy is weak

 

E. Enemy is weak

time limit per test
5 seconds

memory limit per test
256 megabytes

The Romans have attacked again. This time they are much more than the Persians but Shapur is ready to defeat them. He says: "A lion is never afraid of a hundred sheep".

Nevertheless Shapur has to find weaknesses in the Roman army to defeat them. So he gives the army a weakness number.

In Shapur's opinion the weakness of an army is equal to the number of triplets i, j, k such that i < j < k and ai > aj > ak where ax is the power of man standing at position x. The Roman army has one special trait — powers of all the people in it are distinct.

Help Shapur find out how weak the Romans are.

Input
The first line of input contains a single number n (3 ≤ n ≤ 106) — the number of men in Roman army. Next line contains n different positive integers ai (1 ≤ i ≤ n, 1 ≤ ai ≤ 109) — powers of men in the Roman army.

Output
A single integer number, the weakness of the Roman army.

Please, do not use %lld specificator to read or write 64-bit integers in C++. It is preffered to use cout (also you may use %I64d).

Sample test(s)

input

3
3 2 1

output

1

input

3
2 3 1

output

0

input

4
10 8 3 1

output

4

input

4
1 5 4 3

output

1

题意是找到i > j > k且满足ai < aj < ak的数据的组数。

我们不妨按照中间的数进行考虑，就可以确定x是j的顺序的数的个数，y是j的逆序数的个数，这样的数的组数是x * y

求逆序对数的方法之前已经说过，这里稍作修改从前往后遍历，思想相同。代码如下：

/*************************************************************************
> File Name: Enemy_is_weak.cpp
> Author: Zhanghaoran
> Mail: chilumanxi@xiyoulinux.org
> Created Time: 2016年02月04日 星期四 22时18分58秒
************************************************************************/

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cstdlib>

using namespace std;

long long n;
struct node{
long long num;
long long pos;
}nn[1000010];

long long uu[1000010];
long long tree[1000010];
long long ans;
long long aa[1000010];
long long bb[1000010];
long long cc[1000010];
bool cmp(node a, node b){
return a.num < b.num;
}

void add(long x){
while(x < 1000010){
tree[x] ++;
x += x & -x;
}
}

long long check(long x){
long long sum = 0;
while(x){
sum += tree[x];
x -= x & -x;
}
return sum;
}

int main(void){
cin >> n;
for(long long i = 1; i <= n; i ++){
scanf("%I64d", &nn[i].num);
nn[i].pos = i;
}
sort(nn + 1, nn + 1 + n, cmp);
for(long long i = 1; i <= n; i ++){
uu[nn[i].pos] = i;
}

for(long long i = 1; i <= n; i ++){
aa[uu[i]] = check(uu[i]);
bb[uu[i]] = i - aa[uu[i]] - 1;
cc[uu[i]] = uu[i] - 1 - aa[uu[i]];
ans += cc[uu[i]] * bb[uu[i]];
add(uu[i]);
}
cout << ans << endl;

}

 

查看原文：http://chilumanxi.org/2016/02/04/codeforces-beta-round-57-div-2-enemy-is-weak/