hoj 2275 Number Sequence

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知道两点即可：

一：计算出每个Ai，之前以及之后比他小的个数做乘法运算，然后求和，即是最后结果

二：找出Ai之前比Ai小的数，可以用树状数组，同理求之后的数也一样；

#include <iostream>#include <cstring>#include <cstdio>#define MAXN 50010using namespace std;int N;int c[MAXN], a[MAXN], b[MAXN], d[MAXN];int lowbit( int x ){return x&(-x);}void UFset(int i, int data){while(i <= 32768){c[i] += data;i += lowbit(i);}}int Querry( int x ){int sum = 0;while(x > 0){sum += c[x];x -= lowbit(x);}return sum;}int main(){int i;long long ans;while(scanf("%d", &N) != EOF){memset(a, 0, sizeof(a));memset(b, 0, sizeof(b));memset(c, 0, sizeof(c));for(i = 0; i < N; ++i){scanf("%d", &d[i]);a[i] += Querry( d[i] );//在输入的时候就计算i之前的值比它小的值有多少个UFset( d[i] + 1, 1 );//然后再更新}memset(c, 0, sizeof(c));for(i = N-1; i >= 0; --i){b[i] = Querry(d[i]);UFset(d[i] + 1, 1);}ans = 0;for(i = 0; i < N; ++i)ans += (long long)a[i] * (long long)b[i];printf("%lld\n", ans);}return 0;}