标签：des style blog os io for 2014 ar
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，然后统计所有数的可能，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,布布扣,bubuko.com
CodeForces - 61E Enemy is weak
标签：des style blog os io for 2014 ar
原文地址：http://blog.csdn.net/u011345136/article/details/38415023
