A题纯属签到题目：
https://codeforces.com/contest/1622/problem/A
简述题意给定三个木棒让我们折断一个将其变成长方形问是否可以？(木棒只能折断成整数)
1.折断最大值：使得其和小的两根一致pd c=a+b：–当然看到tourist佬判断三个也可以
2.可能出现两个相等的这样有可能出现第三个不能被2整除的情况判断一下即可。
AC代码：

#include <iostream>
#include <algorithm>
#include <unordered_map>
#include <cmath>
#include <queue>
#include <sstream>
#include <cstring>
#include <set>
#include <map>
#include <deque>
#include <stack>
#include <unordered_set>

using namespace std;
typedef long long LL;
typedef unsigned long long ull;
typedef pair<int, int> PII;
typedef pair<int, PII> PIII;
typedef pair<double, double> PDD;
const int N = 1e6 + 10, M = 2 * 100010, inf = 0x3f3f3f3f, mod = 1e9 + 7;
const double eps = 1e-5;
#define x first
#define y second
#define gcd __gcd
#define sr string
#define sn scanf("%d%d",&n,&m);
#define mh memset(h,-1,sizeof h),idx=0;
#define lcm(a, b) (a/gcd(a,b))*b;

int dir[4][2] = {{-1, 0},
                 {0,  1},
                 {1,  0},
                 {0,  -1}};
int e[M], ne[M], h[N], idx, w[M], z[M];
int n, m, p[N], k, t, res, dfn[N], low[N], tot, id[N], T;
bool st[210][210], flag;
int root;
unordered_map<sr, int> q;
PII cp[3005];
bool cmp(PII W,PII C)
{
    if (W.x==C.x) return W.y<C.y;
    return W.x<C.x;
}
int main() {
  cin>>T;
    while (T--)
    {
        int a,b,c;
        cin>>a>>b>>c;
        n= min(a, min(b,c));
        m= max(a, max(b,c));
        k=a+b+c-n-m;
        if (k+n==m||a==b&&c%2==0||a==c&&b%2==0||b==c&&a%2==0)
            puts("YES");
        else
            puts("NO");
    }
        return 0;
}

B题读懂应该也是签到题目https://codeforces.com/contest/1622/problem/B
简述题意：本来给定一个打分表，让我们取修改这个打分表使得它满足一些性质
first：使得每个s串中的0要比1的打分要低
second：他还要是一个排列
使得|pi-qi]最小
我们需要从题目索取一些性质来用：
1.0的打分将是1.....m：m指的是0的个数
2.为了使得差值绝对值最小可以对pi进行排序小的剪小的差的绝对值相对来说较小
只需先处理0的从1.。m然后在处理1的依次排序输出即可
AC代码：

#include <iostream>
#include <algorithm>
#include <unordered_map>
#include <cmath>
#include <queue>
#include <sstream>
#include <cstring>
#include <set>
#include <map>
#include <deque>
#include <stack>
#include <unordered_set>

using namespace std;
typedef long long LL;
typedef unsigned long long ull;
typedef pair<int, int> PII;
typedef pair<int, PII> PIII;
typedef pair<double, double> PDD;
const int N = 2e5 + 10, M = 2 * 100010, inf = 0x3f3f3f3f, mod = 1e9 + 7;
const double eps = 1e-5;
#define x first
#define y second
#define gcd __gcd
#define sr string
#define sn scanf("%d%d",&n,&m);
#define mh memset(h,-1,sizeof h),idx=0;
#define lcm(a, b) (a/gcd(a,b))*b;

int dir[4][2] = {{-1, 0},
                 {0,  1},
                 {1,  0},
                 {0,  -1}};
int e[M], ne[M], h[N], idx, w[M], z[M];
int n, m, p[N], k, t, res, dfn[N], low[N], tot, id[N], T;
bool st[N], flag;
int root;
unordered_map<sr, int> q;
PII cp[3005];


int main() {
    cin>>T;
    while (T--)
    {m=1;
        cin>>n;
        vector<int> res(n,0);
        sr s;
        for(int i=0;i<n;i++)
            cin>>h[i];
        cin>>s;

   for(int x=0;x<=1;x++)
   {   vector<PII> a;
       for(int i=0;i<s.size();i++)
           if (s[i]-'0'==x)
             a.emplace_back(h[i],i);
        sort(a.begin(),a.end());
        for(auto &x:a)
            res[x.y]=m++;
   }
       for(int i=0;i<n;i++)
            cout<<res[i]<<" ";
        cout<<endl;
    }
        return 0;
}

C题我觉得是数学题
只推个公式枚举y即可

#include <iostream>
#include <algorithm>
#include <unordered_map>
#include <cmath>
#include <queue>
#include <sstream>
#include <cstring>
#include <set>
#include <map>
#include <deque>
#include <stack>
#include <unordered_set>

using namespace std;
typedef long long LL;
typedef unsigned long long ull;
typedef pair<int, int> PII;
typedef pair<int, PII> PIII;
typedef pair<double, double> PDD;
const int N = 2e5 + 10, M = 2 * 100010, inf = 0x3f3f3f3f, mod = 1e9 + 7;
const double eps = 1e-5;
#define x first
#define y second
#define gcd __gcd
#define sr string
#define sn scanf("%d%d",&n,&m);
#define mh memset(h,-1,sizeof h),idx=0;
#define lcm(a, b) (a/gcd(a,b))*b;

int dir[4][2] = {{-1, 0},
                 {0,  1},
                 {1,  0},
                 {0,  -1}};
int e[M], ne[M], h[N], idx, w[M], z[M];
int n, m, p[N], k, t, res, dfn[N], low[N], tot, id[N], T;
bool st[N], flag;
int root;
unordered_map<sr, int> q;
LL a[N],s[N];
LL solve(LL a, LL b) {
    LL c = (a) / b;
    while (c * b > a)
        c--;
    return c;
}

int main() {
    cin>>T;
    while (T--)
    {
        LL k;
       cin>>n>>k;
       LL res=1e18;
        memset(s,0,sizeof s);
        for(int i=1;i<=n;i++)
            cin>>a[i];
        sort(a+1,a+n+1);
        for(int i=1;i<=n;i++)
            s[i]=s[i-1]+a[i];//前缀和
    //枚举y的位置
    for(int y=1;y<=n;y++)
    {
        LL x=a[1] - solve(k - s[y] + a[1], n-y+1 );
        res= min(res, max(0LL,x)+n-y);
    }
    cout<<res<<endl;
    }
        return 0;
}