codeforces 1401 C. Mere Array

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题目链接:codeforces 1401 C. Mere Array

题意:

   给出一个数组,当只有 a[i]  和 a[j] 的最大公倍数 等于 数组中最小值的时候,两个数才可以交换。问最终能否交换成一个递增序列。

 

解题思路:

   设最小元素为 m ,对于所有的 m 的倍数, m 可以与这些元素互换

而这些元素之间也可以借助 m 完成互换

所有对与不是 m 的倍数的数,看排个序后是否还是一样

#include <bits/stdc++.h>
using namespace std;
const int maxn = 100005;
int a[maxn], b[maxn];
int main(){
    int t;
    scanf("%d", &t);
    while(t--) {
        int n, m = 1e9+1;
        scanf("%d", &n);
        for(int i = 1; i <= n; i++){
            scanf("%d", &a[i]);
            b[i] = a[i];
            m = min(m, a[i]);
        }
        if(n == 1) {
            printf("YES\n"); continue;
        }
        sort(a+1, a+1+n);
        bool flag = true;
        for(int i = 1; i <= n; i++) {
            if(a[i] != b[i] && a[i] % m != 0) { // 如果相等,或者能通过m进行交换
                flag = false;
                break;
            }
        }
        if(flag) {
            printf("YES\n");
        } else {
            printf("NO\n");
        }
    }
    return 0;
}
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传送门 就先不说了,咱先上代码,明早再来写思路吧。 #include<bits/stdc++.h> #define endl '\n' #define null NULL #define ll long long #define int long long #define pii pair<int, int> #define lowbit(x) (x &(-x)) #define ls(x) x<<1 #define rs(x) (x<<1+1) #d

Satur9的博客 269

codeforces C. Mere Array

题目 题意: 给你一个序列,如果这个序列中的两个数满足gcd(ai,aj)=mini=1nagcd(a_i,a_j)=min^n_{i=1}agcd(ai​,aj​)=mini=1n​a,那么这两个数字就可以进行置换,问最后能不能通过置换将这个序列变成一个不递减的序列。 思路: 如果要满足gcd(ai,aj)=mini=1nagcd(a_i,a_j)=min^n_{i=1}agcd(ai​,aj​)=mini=1n​a,那么肯定有ai%Min=aj%Min=0a_i\%Min=a_j\%Min=0ai​%.

忘梦心的博客 609

Codeforces1401D题解

题目描述 D. Maximum Distributed Tree time limit per test2 seconds memory limit per test256 megabytes inputstandard input outputstandard output You are given a tree that consists of n nodes. You should label each of its n−1 edges with an integer in such way tha

李田所的博客 571

Codeforces1401 E. Divide Square(差分)

There is a square of size 106×106 on the coordinate plane with four points (0,0), (0,106), (106,0), and (106,106) as its vertices. You are going to draw segments on the plane. All segments are either horizontal or vertical and intersect with at least one s

tomjobs的博客 457

codeforces1401D dfs+贪心

1401D 题意:给你一棵树n个节点,n-1条边,然后给你m个值且都为素数,m和n之间的大小未知,让你对这n-1条边赋值使得这n-1条边的乘积为m个值的乘积,现让你在用1最少的情况下(即为边的值),使得 最大,简单的说就是节点1到节点2,3,4…n所经过的路程加上节点2到3,4,5…n所经过的路经加上…n-1到n所经过的路经。 思路:简单的说就是如果n>m那么就意味这有n-1-m条边是要被赋值1的,如果n<m,就意味着最大的那条边要多乘上m-n+1条边,这个是对与边的贪心,之后就是难点了,每条

Cichard 316

Codeforces 1401 F. Reverse and Swap —— 线段树,区间翻转

This way 题意: 给你一个长度为2n2^n2n的数组,有4种操作: 嗝太长了,自己看一下吧 题解: 我一开始还想着用FHQ Treap,没仔细考虑时间复杂度,如果2,3操作每次k都是0的话,就是n2logn^2logn2log的时间复杂度。 然后的话,由于是简单翻转没有插入删除,于是可以用线段树,区间打上翻转标记就行了,第二种和第三种有一些区别。 操作二的话,就需要在深度为k以及之后的所有深度都打上左右互换标记,这样的话,就能表示区间翻转了 操作三的话,需要在深度是k+1的位置打上左右互换标记就

天翼之城的博客 337
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