Codeforces1343D

Constant Palindrome Sum

题目链接：https://codeforces.com/problemset/problem/1343/D

题意：

　　给你一个全是偶数的数组(每个数的值都不超过K)

　　现在每次你可以将任意 Ai 改变为[1 , K]的值

　　要求改变后的数组必须满足Ai + A(n - i + 1) = X (i∈[1 , n / 2] , x∈[2 , 2 * K])

　　问你最少需要改变几次

分析：

　　计算 Ai + A(n - i + 1) = [2 , 2 * K] 分别需要改变的次数

　　定义 ma = max(Ai , A(n - i + 1)) , mi = min(Ai , A(n - i + 1))

　　①、Ai + A(n - i + 1) = ma + mi 需要改变的次数为 0 

　　②、Ai + A(n - i + 1) = [2 , mi] 需要改变的次数为 2

　　③、Ai + A(n - i + 1) = [mi + 1 , mi + ma - 1] 需要改变的次数为 1　

　　④、Ai + A(n - i + 1) = [mi + ma + 1 , ma + K] 需要改变的次数为 1

　　⑤、Ai + A(n - i + 1) = [ma + K + 1 , 2 * K] 需要改变的次数为 2

　　等我们更新完[2 , 2 * K]内所有点需要改变的次数后，输出最小的即可。

　　至于更新我们可以用差分 or 线段树来操作

差分:

#include<bits/stdc++.h>
#define rep(i,a,n) for (int i=a;i<=n;i++)
#define per(i,n,a) for (int i=n;i>=a;i--)
#define int long long
using namespace std;
const int N = 4e5 + 10;
int a[N] , sum[N];
signed main()
{
    int t; 
    cin >> t; 
    while(t --)
    {
        int n , k ;
        cin >> n >> k ; 
        rep(i , 1 , k << 1) sum[i] = 0; 
        rep(i , 1 , n) cin >> a[i] ;
        rep(i , 1 , n / 2)
        {
            int ma = max(a[i] , a[n - i + 1]);
            int mi = min(a[i] , a[n - i + 1]);
            sum[2] += 2 ;
            sum[mi + 1] -- ;
            sum[mi + ma] -- ;
            sum[mi + ma + 1] ++ ;
            sum[ma + k + 1] ++ ; 
        }
        int ans = n / 2;
        rep(i , 2 , k << 1)
        {
            sum[i] += sum[i - 1] ;
            ans = min(ans , sum[i]);
        }
        cout << ans << '\n';
    }
    return 0;
}

线段树:

#include <bits/stdc++.h>
#define rep(i, a, n) for (int i = a; i <= n; i++)
#define per(i, n, a) for (int i = n; i >= a; i--)
#define int long long
#define ll long long
#define il inline  
using namespace std;
struct Tree
{
    ll l, r, sum, lazy, maxn, minn;
} tree[2000000];
il void push_up(ll rt)
{
    tree[rt].sum = tree[rt << 1].sum + tree[rt << 1 | 1].sum;
    tree[rt].maxn = max(tree[rt << 1].maxn, tree[rt << 1 | 1].maxn);
    tree[rt].minn = min(tree[rt << 1].minn, tree[rt << 1 | 1].minn);
}
il void push_down(ll rt, ll length)
{
    if (tree[rt].lazy)
    {
        tree[rt << 1].lazy += tree[rt].lazy;
        tree[rt << 1 | 1].lazy += tree[rt].lazy;
        tree[rt << 1].sum += (length - (length >> 1)) * tree[rt].lazy;
        tree[rt << 1 | 1].sum += (length >> 1) * tree[rt].lazy;
        tree[rt << 1].minn += tree[rt].lazy;
        tree[rt << 1 | 1].minn += tree[rt].lazy;
        tree[rt << 1].maxn += tree[rt].lazy;
        tree[rt << 1 | 1].maxn += tree[rt].lazy;
        tree[rt].lazy = 0;
    }
}
il void build(ll l, ll r, ll rt, ll *aa)
{
    tree[rt].lazy = 0;
    tree[rt].l = l;
    tree[rt].r = r;
    if (l == r)
    {
        tree[rt].sum = aa[l];
        tree[rt].minn = tree[rt].sum;
        tree[rt].maxn = tree[rt].sum;
        return;
    }
    ll mid = (l + r) >> 1;
    build(l, mid, rt << 1, aa);
    build(mid + 1, r, rt << 1 | 1, aa);
    push_up(rt);
}
il void update_range(ll L, ll R, ll key, ll rt)
{
    if (tree[rt].r < L || tree[rt].l > R)
        return;
    if (L <= tree[rt].l && R >= tree[rt].r)
    {
        tree[rt].sum += (tree[rt].r - tree[rt].l + 1) * key;
        tree[rt].minn += key;
        tree[rt].maxn += key;
        tree[rt].lazy += key;
        return;
    }
    push_down(rt, tree[rt].r - tree[rt].l + 1);
    ll mid = (tree[rt].r + tree[rt].l) >> 1;
    if (L <= mid)
        update_range(L, R, key, rt << 1);
    if (R > mid)
        update_range(L, R, key, rt << 1 | 1);
    push_up(rt);
}
il ll query_min(ll L, ll R, ll rt)
{
    if (L <= tree[rt].l && R >= tree[rt].r)
    {
        return tree[rt].minn;
    }
    push_down(rt, tree[rt].r - tree[rt].l + 1);
    ll mid = (tree[rt].r + tree[rt].l) >> 1;
    ll ans = (0x3f3f3f3f3f3f3f3fll);
    if (L <= mid)
        ans = min(ans, query_min(L, R, rt << 1));
    if (R > mid)
        ans = min(ans, query_min(L, R, rt << 1 | 1));
    return ans;
}
const int N = 4e5 + 10;
int a[N] , zero[N];
signed main()
{
    int t ;
    cin >> t ;
    while(t --)
    {
        int n , k ;
        cin >> n >> k ;
        build(1 , 2 * k , 1 , zero);
        rep(i , 1 , n) cin >> a[i];
        rep(i , 1 , n / 2)
        {
            int mi = min(a[i] , a[n - i + 1]);
            int ma = max(a[i] , a[n - i + 1]);
            update_range(2 , mi , 2 , 1);
            update_range(mi + 1 , mi + ma - 1 , 1 , 1);
            update_range(mi + ma + 1 , ma + k , 1 , 1);
            update_range(ma + k + 1 , 2 * k , 2 , 1);
        }
        cout << query_min(2 , 2 * k , 1) << '\n';
    }
    return 0;
}