hdu 6609 + interval half the prefix and conditions

Topic Portal // res tp hdu

purpose

When n elements gradually inserted at the end, requires the insertion of the i-th element, [. 1, the i) the number of elements omitted, and the prefix can [. 1, i] is not greater than m

Sets of Q [1,15]
n-[1,2e5]
m [1,1e9]
Each element W is _I [. 1, m] (i∈ [. 1, n-]);

data structure

Fenwick tree

analysis

Tree maintains two arrays, respectively, the number prefix and storage elements. First of all elements read, followed by the program while recording after a good index for each element arrangements. Press header inserting elements into the array tree. After two minutes to enumerate
the time complexity of O (Qnlognlogn)

#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
typedef long long ll;
const int L = 200010;
ll val[L],sor[L];
struct E{
    ll v;
    int pos;
}sorted[L];
int Q,POS[L];
ll n,m;
ll BIT[L],bit[L];
int Len;
int lowbit(int x){return x&-x;}
bool cmp(E a,E b){return a.v<b.v;}
void change(int x,int y){
    sor[x] = y;
    for(int i = x;i<=L;i+=lowbit(i)){
        BIT[i] += y;bit[i]++;
    }
}
ll query(int k){//对前缀和的询问
    ll ans = 0;
    for(int i = k;i > 0; i -= lowbit(i))
        ans += BIT[i];
    return ans;
}
int query1(int k){//对前缀内元素个数的询问
    int ans = 0;
    for(int i = k;i>0;i-=lowbit(i))
        ans += bit[i];
    return ans;
}
int getans(int x,int M){ //前x个元素的小序前缀之和不超过M
    int lo = 1,hi = n+1;
    int mi;
    ll sum;
    while(lo < hi){
        mi = (lo + hi)>>1;
        sum = query(mi);
        if(M <sum)  hi = mi;
        else lo = mi + 1;
    }
    --lo;
    return x-query1(lo);
}

int main(){
    scanf(" %d",&Q);
    while(Q--){
        scanf(" %lld %lld",&n,&m);
        for(int i = 1;i<=n;++i) BIT[i] = bit[i] = 0;
        for(int i = 1;i<=n;++i) sor[i] = 0;
        for(int i = 1;i<=n;++i) {
            scanf(" %lld",&val[i]);
            sorted[i].v = val[i];
            sorted[i].pos = i;
        }
        sort(sorted+1,sorted+1+n,cmp);
        for(int i = 1;i<=n;++i) POS[sorted[i].pos] = i;
        int ans;
        for(int i = 1;i<=n;++i){
            ans = getans(i-1,m-val[i]);
            change(POS[i],val[i]);
            printf("%d ",ans);
        }
        printf("\n");
    }
}