T1，不多说，pascal自带的pos轻松过，枚举第一个删AC或CA，水过~~

T2，meet in the middle，不过在用扩欧的时候要注意如果ex_gcd(a,b,x,y）中的a或b=0，直接退出。炸了85分~~

T3，一个暴力，靠着没有正确性的暴力居然有50分，感人。

总结：T2这种数学题对边界如0这种还是要好好思考一下的，比如说如果用费马小定理求逆元的话更加要注意了p=2的情况（YHYHYH3提醒）

写一下T3的题。

这道题我写了一个广搜，其实正解也是广搜，一般我们广搜都是对(x,y)的二元组惊醒搜索的，但是这道题我们发现一条蛇可以多次走一个点（不多于两次吧），所以我们要加入另一个状态(x,y,body)，表示其蛇头在这个点时的姿态？因为len<=8（哦照片没拍出来),所以我们可以用压位来操作。

 

#include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          using namespace std;typedef long long ll;int n, m, l, x, y, x2, y2;int cx[4] = {
   1, 0, -1, 0};int cy[4] = {
   0, 1, 0, -1};int head, tail, i, j, k, num, dk, ans;bool flag;struct Node{    int x, y, z;} d[8000000];int b[29][29][24010], f[29][29][24010], snackx[10], snacky[10], F[2100][2100], T[10];bool abletogo[2100][2100];int main(){    freopen("snake.in", "r", stdin);    freopen("snake.out", "w", stdout);    int t;    scanf("%d", &t);    while (t--)    {        scanf("%d%d%d", &n, &m, &l);        if (l == 1)        {            for (int i = 1; i <= n; i++)                for (int j = 1; j <= m; j++)                    abletogo[i][j] = 0, F[i][j] = -1;            scanf("%d%d", &x, &y);            abletogo[x][y] = 1;            head = tail = 1;            d[tail].x = x;            d[tail].y = y;            F[x][y] = 0;            scanf("%d", &k);            for (int i = 1; i <= k; i++)            {                scanf("%d%d", &x, &y);                abletogo[x][y] = 1;            }            while (head <= tail)            {                for (int i = 0; i < 4; i++)                {                    x2 = d[head].x + cx[i];                    y2 = d[head].y + cy[i];                    if (x2 < 1 || x2 > n || y2 < 1 || y2 > m || abletogo[x2][y2])                        continue;                    abletogo[x2][y2] = 1;                    F[x2][y2] = F[d[head].x][d[head].y] + 1;                    tail++;                    d[tail].x = x2;                    d[tail].y = y2;                }                head++;            }            printf("%d\n", F[1][1]);        }        else        {            for (int i = 1; i <= l; i++)                scanf("%d%d", &snackx[i], &snacky[i]);            num = 0;            for (int i = 1; i < l; i++)            {                for (int j = 0; j < 4; j++)                    if (snackx[i] + cx[j] == snackx[i + 1] && snacky[i] + cy[j] == snacky[i + 1])                    {                        num = num * 4 + j;                        break;                    }            }            for (int i = 1; i <= n; i++)                for (int j = 1; j <= m; j++)                    for (int dk = 0; dk <= 1 << 14; dk++)                        b[i][j][dk] = 0, f[i][j][dk] = -1;            b[snackx[1]][snacky[1]][num] = 1;            f[snackx[1]][snacky[1]][num] = 0;            head = tail = 1;            d[tail].x = snackx[1];            d[tail].y = snacky[1];            d[tail].z = num;            scanf("%d", &k);            for (int i = 1; i <= k; i++)            {                scanf("%d%d", &x, &y);                for (int j = 0; j <= 1 << 14; j++)                    b[x][y][j] = 1;            }            while (head <= tail)            {                num = d[head].z;                for (int i = l - 1; i; i--)                {                    T[i] = num % 4;                    num /= 4;                }                snackx[1] = d[head].x, snacky[1] = d[head].y;                for (int i = 1; i <= l - 1; i++)                {                    snackx[i + 1] = snackx[i] + cx[T[i]];                    snacky[i + 1] = snacky[i] + cy[T[i]];                }                for (int i = 0; i < 4; i++)                {                    x2 = snackx[1] + cx[i];                    y2 = snacky[1] + cy[i];                    if (x2 < 1 || x2 > n || y2 < 1 || y2 > m)                        continue;                    flag = false;                    for (int j = 1; j <= l - 1; j++)                        if (x2 == snackx[j] && y2 == snacky[j])                        {                            flag = true;                            break;                        }                    if (flag)                        continue;                    num = (i + 2) % 4;                    for (int j = 1; j < l - 1; j++)                        num = num * 4 + T[j];                    if (b[x2][y2][num])                        continue;                    b[x2][y2][num] = 1;                    f[x2][y2][num] = f[d[head].x][d[head].y][d[head].z] + 1;                    tail++;                    d[tail].x = x2;                    d[tail].y = y2;                    d[tail].z = num;                }                head++;            }            ans = INT_MAX;            for (int i = 0; i <= 1 << 14; i++)                if (f[1][1][i] != -1)                    ans = min(ans, f[1][1][i]);            if (ans == INT_MAX)                printf("-1\n");            else                printf("%d\n", ans);        }    }}
         
        
       
      
     

 

#include
     
      #include
      
       #include
       
        #include
        
         #include
         
          using namespace std;#define ll long longint n,m,z,L,body,step=-1;int dx[4]={
   0,1,0,-1},dy[4]={
   1,0,-1,0},Pow[10];int change[20000][4];bool f[20000],flag[21][21][20000],duang[2011][2011],bo[100][100];struct snake{    int x[8],y[8],z,body;}s,news;queue
          
            Q;int calc_body(snake s){ int id=0; for(int i=1;i
           
            n||y<=0||y>m||flag[x][y][body]||duang[x][y]||(!f[body])) continue; if(x==1&&y==1) { step=z; break; } news.z=z;news.x[0]=x,news.y[0]=y;news.body=body; Q.push(news); flag[x][y][body]=1; } } cout<
            
             <
             
              n||y<=0||y>m||duang[x][y])continue; if(x==1&&y==1) { step=z;break;} news.x[0]=x,news.y[0]=y,news.z=z; Q.push(news); duang[x][y]=1; } } cout<
              
               <
               
                =1;j--) { int t=i%Pow[j]/Pow[j-1]; x+=dx[t];y+=dy[t]; if(bo[x][y]) { f[i]=false; break; } bo[x][y]=1; } } }int main(){ int T; scanf("%d",&T); while(T--) { step=-1; memset(flag,0,sizeof(flag)); memset(duang,0,sizeof(duang)); scanf("%d %d %d",&n,&m,&L); prepare(); for(int i=0;i