hdu5147 (sequence 2) 树状数组

#include <iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<vector>
#include<map>
using namespace std;
#define INF 0x7fffffff
#define MOD 1000000007


#define N 100010
#define MAX 1000100

int C[MAX], S[MAX], b[N],d[N];
long long f[N];
long long total[N], ans;
int num[N], T, s, t, i, j;

int Lowbit(int x){
    return x&(x^(x-1));
}

void add(int pos,int num,int *P) {
    while(pos <= MAX) {
        P[pos] += num;
        pos += Lowbit(pos);
    }
}

int Sum(int end,int *P) {
    int cnt = 0;
    while(end > 0) {
        cnt += P[end];
        end -= Lowbit(end);
    }
    return cnt;
}

void init(){
    total[0] = 0;
    for(i = 1; i < N; ++i){
        total[i] = total[i-1] + i;
    }
}

int main() {
    //init();
    int t;
    cin>>t;

    while(t--) {

        scanf("%d",&T);

        memset(C,0,sizeof(C));
        memset(S,0,sizeof(S));
        memset(b,0,sizeof b);
        memset( d,0,sizeof d);
        memset(f,0,sizeof f);
        memset(num,0,sizeof num);
        //memset(num,0,sizeof(num));
        //memset(b,0,sizeof(b));
        //ans = 0;
        for(j = 0; j < T; j ++) {//因为第一个数前面比它小的数没有，所以j要从0开始
            scanf("%d",&num[j]);
            add(num[j]+1,1,C);
            b[j] = Sum(num[j], C);//Sum(num[j],C)求的就是小于s的个数，j - Sum(num[j],C)就是前j个数中大于num[j]的个数
            //printf("%d ",b[j]);
        }
        //printf("
");
        //ans = 0;
        for(j = T-1; j > -1; --j){//反过来求第j个数右边中大于它的数的个数。
            add(num[j]+1 ,1, S);
            d[j] =T-1-j-Sum(num[j] ,S);//Sum(num[j],S)求的就是小于num[j]的个数
            //b[j] -= Sum(num[j]+1,S) - Sum(num[j],S)-1;
            //printf("%d ",b[j]);
            //ans += total[b[j]];
        }
        f[T-1]=0;
       for(int j=T-2;j>=0;j--)
        f[j]=d[j+1]+f[j+1];
       long long int res=0;
       for(int j=0;j<T;j++)
        res+=b[j]*f[j];
       cout<<res<<endl;

    }
    return 0;
}

代码君

理解有问题的话，先看树状数组求逆序数。