Codeforces Round #599 (Div. 2)

A.Maximum Square
B1.Character Swap (Easy Version)
B2.Character Swap (Hard Version)
C.Tile Painting
D.0-1 MST

A.Maximum Square

题目类型：数学【？】

题目链接：A题链接

题目大意：给你n条长度为a[i]的木棍，宽度为1的木棍，询问你如何排列能获得一个边最大的正方形

通过图我们能很好的理解题目的意思

解题思路：只需要排一下序，按从高到低去排序，每拼上一条宽度是增加1的，如果你的最小高度无法匹配你的总宽度的话就要跳出了，以此来寻找最大边长

/**
 *    This code has been written by YueGuang, feel free to ask me question. Blog: http://www.yx.telstudy.xyz
 *    created:
 */
#include <cstdio>
#include <iostream>
#include <set>
#include <map>
#include <algorithm>
#include <cstring>
#include <string>
#include <cmath>
#define REP(i, a, b) for(int i = a; i < b; i++)
#define REP_(i, a, b) for(int i = a; i <= b; i++)
#define sl(n) scanf("%lld", &n);
#define si(n) scanf("%d", &n);
#define RepAll(a) for(auto x: a)
#define cout(ans) cout << ans << endl;
typedef long long ll;

using namespace std;
bool cmp(int a, int b){return a > b;}
const int maxn = 2000;
int a[maxn];
int main(){
    //freopen("in.txt", "r", stdin);
    int k, n; scanf("%d", &k);
    while(k--){
        scanf("%d", &n);
        for(int i = 0; i < n; i++){scanf("%d", &a[i]);}
        sort(a, a+n, cmp);
        int j = 1;
        for(int i = 0; i < n; j++, i++){
            if(a[i] < j){break;}
        }
        printf("%d
", j-1);
    }
}

B1.Character Swap (Easy Version)

题目类型：模拟

题目链接：B1题链接

题目大意：给你两个长度为n的字符串，分别为s和t，让你交换一次ti和sj使得两个字符串相同，问是否能做到，如果可以就输出YES不行就输出NO

这句话听关键的，特意复制下来

Since Ujan is lazy, he will perform the following operation exactly once: he takes two positions i and j (1≤i,j≤n, the values i and j can be equal or different), and swaps the characters si and tj.

解题思路：寻找s中第一个与t不一样的字符，然后再寻找t中第二个与s不一样的字符，交换后比较

/**
 *    This code has been written by YueGuang, feel free to ask me question. Blog: http://www.yx.telstudy.xyz
 *    created:
 */
#include <cstdio>
#include <iostream>
#include <set>
#include <map>
#include <algorithm>
#include <cstring>
#include <string>
#include <cmath>
#define REP(i, a, b) for(int i = a; i < b; i++)
#define REP_(i, a, b) for(int i = a; i <= b; i++)
#define sl(n) scanf("%lld", &n);
#define si(n) scanf("%d", &n);
#define RepAll(a) for(auto x: a)
#define cout(ans) cout << ans << endl;
typedef long long ll;

using namespace std;
string s, t;
//char s[1010], t[1010];
//map<char, int> f;
int main(){
    //freopen("in.txt", "r", stdin);
    int q, n; scanf("%d", &q);
    while(q--){
        int cnt = 0;
        scanf("%d", &n);
        cin >> s >> t;
        //char a[];
        int flag = -1;
        for(int i = 0; i < n; i++){
            if(s[i]!=t[i] && flag == -1){
                //cout << "now : "<<flag << endl;
                 flag = i; continue;}
            if(s[i]!=t[i] && flag >= 0 ){
                //cout<< flag << '
';
                char tmp = t[i];
                t[i] = s[flag];
                s[flag] = tmp;
                break;
            }
        }
        //cout << s << " " << t << '
';
    if (s==t) printf("Yes
");
    else printf("No
");
    }
}

B2.Character Swap (Hard Version)

题目类型：

题目链接：

下次一定补

C. Tile Painting

题目类型：数学

题目链接：C题链接

题目大意：给你 n块边长为1的方块，对方快进行涂色，如果满足要求 |i - j| > 1 且 n mod |i - j| == 0 则第i个方块和第j个方块要求涂上相同的颜色，求解需要多少种颜色

左图：输入的n为4

i = 4, j = 2 满足所以2,4相同颜色
同理1,3满足， 所以1,3同色，而1，2或3,4无法满足所以1,2，3,4不能同色，综上需要使用两种颜色

解题思路：
（1）当看到n mod |i-j| = 0的时候，思路应该向公因数靠拢，因为当i 和 j 能够满足这条件的时候i和j的差值一定是n的公因数
（2）也可以考虑一下素数，因为输入的n是素数的话，n的因子只有本身和1，那就只能输出n了，每种色块都要是不相同的

/*
 *这个是想暴力找规律的一段代码
 */
#include <cstdio>
#include <iostream>
using namespace std;
int main(){
    int n; scanf("%d", &n);

    for(int i = 1; i <= n; i++){
        printf("%d
", i);
        for(int j = 1; j <= n; j++){
            if (abs(i-j) > 1 && n % (i-j) == 0){
                printf("%d ", j);
            }
        }
    cout << '
';
    }
}

能够验证，素数无可选因子

那么非素数，我们拿样例 n = 4测试一下

i = 1
j = 3
i = 2
j = 4
i = 3
j = 1
i = 4
j = 2

1. 可以的到这么几组数据，没错就是1、3共色，2、4共色
2. 循环不需要完全遍历n，只需要sqrt（n）就可以了，可以转化为数 i * i <= n 就可以了

/**
 *    This code has been written by YueGuang, feel free to ask me question. Blog: http://www.yx.telstudy.xyz
 *    created:
 */
#include <cstdio>
#include <iostream>
#include <set>
#include <map>
#include <algorithm>
#include <cstring>
#include <string>
#include <cmath>
#define REP(i, a, b) for(int i = a; i < b; i++)
#define REP_(i, a, b) for(int i = a; i <= b; i++)
#define sl(n) scanf("%lld", &n);
#define si(n) scanf("%d", &n);
#define RepAll(a) for(auto x: a)
#define cout(ans) cout << ans << endl;
typedef long long ll;

using namespace std;
inline ll gcd(ll a, ll b){return b == 0 ? a : gcd(b, a%b);}
int main(){
    ll n; scanf("%lld", &n);
    ll d = n;
    for(ll i = 2; i * i <= n; i++){
        if (n % i  == 0){
            d = gcd(i ,d);
            d = gcd(d, n/i);
        }
    }
    printf("%lld
", d);
}