魔道研究

来源：互联网发布：阴阳师挂机软件ios 编辑：程序博客网时间：2024/06/09 14:32

题目大意

有3*10^5个多重集合，许多次操作，每次操作往某个集合里插入一个数。对于所有i，把第i个集合里前i大的元素加入集合S中。每次操作后你需要输出集合S中前N大元素的和。

模拟

模拟题意，用数据结构维护即可。
这里使用spaly。

#include<cstdio>#include<algorithm>#define fo(i,a,b) for(i=a;i<=b;i++)using namespace std;typedef long long ll;const ll mo=1000000007;ll flag[10][10]={    {1 ,0 ,0 ,0 ,0 ,0 ,0 ,0 ,0 ,0 },    {1 ,1 ,0 ,0 ,0 ,0 ,0 ,0 ,0 ,0 },    {1 ,0 ,1 ,0 ,0 ,0 ,0 ,0 ,0 ,0 },    {1 ,2 ,0 ,1 ,0 ,0 ,0 ,0 ,0 ,0 },    {1 ,2 ,0 ,0 ,1 ,0 ,0 ,0 ,0 ,0 },    {1 ,1 ,1 ,0 ,0 ,1 ,0 ,0 ,0 ,0 },    {1 ,0 ,2 ,0 ,0 ,0 ,1 ,0 ,0 ,0 },    {1 ,3 ,0 ,2 ,1 ,0 ,0 ,1 ,0 ,0 },    {1 ,2 ,1 ,1 ,0 ,2 ,0 ,0 ,1 ,0 },    {1 ,4 ,0 ,4 ,2 ,0 ,0 ,4 ,0 ,1 }},dt1[10]={12,9,6,6,6,3,0,3,0,0} ,dt2[10]={4 ,2,1,1,0,0,0,0,0,0};ll c[30][30],f[250][250][20];ll i,j,k,l,r,t,n,m,ans;int main(){    freopen("magic.in","r",stdin);freopen("magic.out","w",stdout);    scanf("%lld%lld",&n,&m);    f[0][0][0]=1;    f[0][1][1]=4;f[0][1][2]=4;    f[0][2][3]=4;f[0][2][4]=2;f[0][2][5]=8;f[0][2][6]=2;    f[0][3][7]=4;f[0][3][8]=4;    f[0][4][9]=1;    c[0][0]=1;    fo(i,1,14){        c[i][0]=1;        fo(j,1,i)            c[i][j]=(c[i-1][j]+c[i-1][j-1])%mo;    }    fo(i,0,n-1)        fo(j,0,m)            fo(k,0,9)                if (f[i][j][k])                    fo(l,0,9)                        if (flag[k][l])                            fo(t,0,dt2[l])                                if (j+t<=m)                                    fo(r,0,dt1[l]-2*t)                                        if (j+t+r<=m){                                            f[i+1][j+t+r][l]+=f[i][j][k]*flag[k][l]*c[dt2[l]][t]*c[dt1[l]-2*t][r];                                            f[i+1][j+t+r][l]%=mo;                                        }    fo(j,0,9) ans=(ans+f[n][m][j])%mo;    fo(i,1,m) ans=ans*i%mo;    printf("%lld\n",ans);    fclose(stdin);fclose(stdout);    return 0;}

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