hdu 2688 Rotate(樹狀數組)

有個序列找出所有正序數的對數，R是將S,E之間的數rotate。

Q是求當前的正序數。

樹狀數組+類比

這道題坑的啊，用G++交的話TLE了，C++就540ms

/*Problem ID:meaning:Analyzing:*/#include <iostream>#include <algorithm>#include<cstdio>#include<cmath>#include<cstdlib>#include<cstring>#include<vector>using namespace std;typedef struct even{int y1,y2,x;}even;#define clr(A,k) memset(A,k,sizeof(A))#define FOR(i,s,t) for(int i=(s); i<(t); i++)#define LL long long#define BUG puts("here!!!")#define print(x) printf("%d\n",x)#define STOP system("pause")#define eps 1e-8#define PI acos(-1.0)#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define maxn 3000006#define maxm 10005int lowbit(int x){    return x&(-x);}LL gcd(LL a,LL b) {return a?gcd(b%a,a):b;}int A[maxn],C[maxm],n;void update(int pos,int val){    while(pos<=maxm-1){        C[pos]+=val;        pos+=lowbit(pos);    }}int query(int pos){    int ret=0;    while(pos>0){        ret+=C[pos];        pos-=lowbit(pos);    }    return ret;}int main(){    int m;    char op[5];    while(~scanf("%d",&n)){        clr(C,0);        LL sum=0;        for(int i=0;i<n;i++){            scanf("%d",&A[i]);            update(A[i],1);            sum+=query(A[i]-1);        }        scanf("%d",&m);        int S,E;        while(m--){            scanf("%s",op);            if(op[0]=='Q'){                printf("%I64d\n",sum);            }else {                scanf("%d%d",&S,&E);                int temp=A[S];                for(int i=S;i<E;i++){                    A[i]=A[i+1];                    if(A[i]>temp) sum--;                    else if(A[i]<temp) sum++;                }                A[E]=temp;            }        }    }return 0;}