Find the answer

题目传送门

解题思路

要想变0的个数最少，显然是优先把大的变成0。所以离散化，建立一颗权值线段树，维护区间和与区间元素数量，假设至少减去k才能满足条件，查询大于等于k的最少数量即可。

代码如下

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
using namespace std;
typedef long long ll;

inline int read(){
    int res = 0, w = 0; char ch = 0;
    while(!isdigit(ch)){
        w |= ch == '-', ch = getchar();
    }
    while(isdigit(ch)){
        res = (res << 3) + (res << 1) + (ch ^ 48);
        ch = getchar();
    }
    return w ? -res : res;
}

const int N = 200005;

struct T{
    int l, r;
    int c;
    ll sum;
}tree[N<<2];
ll a[N], b[N];
void build(int k, int l, int r)
{
    tree[k].l = l, tree[k].r = r;
    tree[k].c = tree[k].sum = 0;
    if(tree[k].l == tree[k].r)
        return;
    int mid = (tree[k].l + tree[k].r) / 2;
    build(2*k, l, mid);
    build(2*k+1, mid + 1, r);
}

void insert(int k, int x)
{
    if(tree[k].l == tree[k].r){
        tree[k].c ++;
        tree[k].sum += b[x];
        return;
    }
    int mid = (tree[k].l + tree[k].r) / 2;
    if(x <= mid)
        insert(2*k, x);
    else
        insert(2*k+1, x);
    tree[k].c = tree[2*k].c + tree[2*k+1].c;
    tree[k].sum = tree[2*k].sum + tree[2*k+1].sum;
}

ll query(int k, ll x)
{
    if(tree[k].l == tree[k].r)
        return (x - 1) / b[tree[k].l] + 1;
    if(tree[2*k+1].sum > x)
        return query(2*k+1, x);
    else if(tree[2*k+1].sum < x){
        x -= tree[2*k+1].sum;
        return query(2*k, x) + tree[2*k+1].c;
    }
    else
        return tree[2*k+1].c;
}

int main()
{
    int q;
    scanf("%d", &q);
    while(q --){
        int n;
        ll m;
        scanf("%d%lld", &n, &m);
        for(int i = 1; i <= n; i ++){
            scanf("%lld", &a[i]);
            b[i] = a[i];
        }
        sort(b + 1, b + n + 1);
        int k = unique(b + 1, b + n + 1) - b - 1;
        build(1, 1, k);
        ll sum = 0;
        for(int i = 1; i <= n; i ++){
            sum += a[i];
            if(sum - m > 0)
                printf("%lld ", query(1, sum - m));
            else
                printf("0 ");
            int x = lower_bound(b + 1, b + k + 1, a[i]) - b;
            insert(1, x);
        }
        printf("\n");
    }
    return 0;
}

原文地址：https://www.cnblogs.com/whisperlzw/p/11272351.html