Description
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 Input
3 3 2 1
1
3 2 3 1
0
4 10 8 3 1
4
4 1 5 4 3
1
题意：求满足三个数是下标i < j < k, 值 a[i] > a[j] > a[k] 的个数
思路：对于每一个数我们向前找比它大的数，向后找比它小的数，相乘就得到这个数的结果了。然后统计全部数的可能，基于数量太大，我们用归并算法。在处理的时候计算
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
const int maxn = 1e6 + 10;
struct Node {
ll val, front, rear;
} a[maxn], b[maxn];
ll ans;
void merge_sort(Node *A, int x, int y, Node *T) {
if (y - x > 1) {
int m = x + (y -x) / 2;
int p = x, q = m, i = x;
merge_sort(A, x, m, T);
merge_sort(A, m, y, T);
while (p < m || q < y) {
if (q >= y || (p < m && A[p].val <= A[q].val)) {
ans += A[p].front * (q - m);
A[p].rear += (q - m);
T[i++] = A[p++];
}
else {
ans += A[q].rear * (m - p);
A[q].front += (m - p);
T[i++] = A[q++];
}
}
for (i = x; i < y; i++)
A[i] = T[i];
}
}
int main() {
int n;
scanf("%d", &n);
for (int i = 0; i < n; i++)
scanf("%lld", &a[i].val);
ans = 0;
merge_sort(a, 0, n, b);
cout << ans << endl;
return 0;
}CodeForces - 61E Enemy is weak
原文：http://www.cnblogs.com/mengfanrong/p/5211994.html