A題
答案為p^n mod b = 1的個數，就是d(p-1)

#include <iostream>
#include <bits/stdc++.h>
#define ll long long
using namespace std;

int main()
{
    int T;
    scanf("%d",&T);
    ll n;
    while (T--)
    {
       scanf("%lld",&n);
       n--;
       ll res=0;
       for (int i=1;i*i<=n;i++)
       if
 
 (n % i == 0 )
       {
           res++;
           if (i * i !=n) res++;
       }
       printf("%lld\n",res);
    }
    return 0;
} 

E題
隨便暴一下

#include <iostream>
#include <bits/stdc++.h>
#define ll long long
using namespace std;


int calc(int y, int m,int d)
{
    if (!( ((y % 4== 0) && (y
 
 % 100!=0)) | y % 400==0 ) && (m==2 && d==29)) return -1;
    if(m==1||m==2) m+=12,y--;
    int iWeek = (d+2*m+3*(m+1)/5+y+y/4-y/100+y/400)%7;
    return iWeek;
}
int y,m,d;
void solve()
{
    scanf("%d-%d-%d",&y,&m,&d);
    int wk=calc(y,m,d);
    int res=y+1;
    while ( calc(res,m,d) !=wk ) res++;
    printf
 
("%d\n",res);
}
int main()
{
    int T;
    scanf("%d",&T);
    while (T--)
        solve();
    return 0;
}

F題
先搜1，再從（0，0）開始把外圍的0填充為1，再搜一下0有幾塊

#include <iostream>
#include <bits/stdc++.h>
#include <queue>
#define clr(x,y) memset(x,y,sizeof x)
using namespace std;

int mp[110][110],vis[110][110];
char s[123];
int n,m;
struct Node
{
    int x,y;
    Node (int _x,int _y)
    {
        x=_x;
        y=_y;
    }
};
const int dir[4][2]= {0,1,0,-1,1,0,-1,0};
void bfs(Node st,int type)
{
    queue<Node> q;
    q.push(st);
    vis[st.x][st.y]=1;
    while (!q.empty())
    {
        Node now=q.front();
        q.pop();
        for (int i=0; i<4; i++)
        {
            int tx=now.x+dir[i][0];
            int ty=now.y+dir[i][1];
            if (tx<0 || ty<0 || tx>n+1 || ty>m+1 ) continue;
            if (mp[tx][ty] == type && vis[tx][ty]==0)
            {
                vis[tx][ty] = 1;
                if (type == 0) mp[tx][ty] = 1;
                q.push(Node(tx,ty));
            }
        }
    }
}

int main()
{
    while (scanf("%d%d",&n,&m)!=EOF)
    {
        clr(mp,0);
        clr(vis,0);
        int tot=0;
        for (int i=1; i<=n; i++)
        {
            scanf("%s",s+1);
            for (int j=1; j<=m; j++)
                mp[i][j]=s[j] - '0',tot+=(s[j]=='1');
        }
        tot = (n+1) * (m+1) - tot;
        int one=0;
        for (int i=1; i<=n; i++)
            for (int j=1; j<=m; j++)
                if (mp[i][j] == 1 && vis[i][j]==0)
                    bfs(Node(i,j),1),one++;
        bfs(Node(0,0),0);
        if (one!=1) printf("-1\n");
        else
        {
            int zero=0;
            for (int i=1; i<=n; i++)
                for(int j=1; j<=m; j++)
                    if (mp[i][j] == 0) bfs(Node(i,j),0), zero++; 
            if (zero==1) printf("0\n");
            if (zero==0) printf("1\n");
            if (zero>=2) printf("-1\n");
        }
    }
    return 0;
}