既然bzoj上有这道题了就把这个坑填了吧。。。
题解见：http://blog.csdn.net/popoqqq/article/details/40984859
话说这个解法如果当时想到冲突的概率很小的话应该就能想出来233
代码：
1 #include<cstdio> 2 3 #include<cstdlib> 4 5 #include<cmath> 6 7 #include<cstring> 8 9 #include<algorithm> 10 11 #include<iostream> 12 13 #include<vector> 14 15 #include<map> 16 17 #include<set> 18 19 #include<queue> 20 21 #include<string> 22 23 #define inf 1000000000 24 25 #define maxn 100000 26 27 #define maxm 500+100 28 29 #define eps 1e-10 30 31 #define ll long long 32 33 #define pa pair<int,int> 34 35 #define for0(i,n) for(int i=0;i<=(n);i++) 36 37 #define for1(i,n) for(int i=1;i<=(n);i++) 38 39 #define for2(i,x,y) for(int i=(x);i<=(y);i++) 40 41 #define for3(i,x,y) for(int i=(x);i>=(y);i--) 42 43 #define mod 1000000007 44 45 using namespace std; 46 47 inline int read() 48 49 { 50 51 int x=0,f=1;char ch=getchar(); 52 53 while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} 54 55 while(ch>='0'&&ch<='9'){x=10*x+ch-'0';ch=getchar();} 56 57 return x*f; 58 59 } 60 int n,m,tot,ans[1005],a[1005][10],f[1000000][10]; 61 const int p[5]={10007,11261,14843,19997,21893}; 62 char s[10000+5]; 63 64 int main() 65 66 { 67 68 freopen("input.txt","r",stdin); 69 70 freopen("output.txt","w",stdout); 71 72 n=read();m=read(); 73 for0(i,n) 74 { 75 scanf("%s",s+1);int len=strlen(s+1);bool flag=0; 76 for1(j,len) 77 { 78 if(s[j]=='-')flag=1; 79 else for0(k,4)a[i][k]=(a[i][k]*10+s[j]-'0')%p[k]; 80 } 81 if(flag)for0(k,4)a[i][k]=p[k]-a[i][k]; 82 } 83 //for0(i,n)for0(j,4)cout<<i<<' '<<j<<' '<<a[i][j]<<endl; 84 for0(j,4) 85 for0(i,p[j]-1) 86 { 87 f[i][j]=0; 88 for3(k,n,0)f[i][j]=(f[i][j]*i+a[k][j])%p[j]; 89 } 90 for1(i,m) 91 { 92 int j; 93 for(j=0;j<5;j++) 94 if(f[i%p[j]][j])break; 95 if(j==5)ans[++tot]=i; 96 } 97 printf("%d\n",tot); 98 for1(i,tot)printf("%d\n",ans[i]); 99 100 return 0; 101 102 }