CodeForces 140E
2016-11-30 22:51
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sum[i]=∑j=1l[i]dp[i][j];q[i]=∏j=1i(m−i+1)p[i]=∏j=1ijT[i][j]表示有i个位置,使用j个不同的颜色染色,相邻不同的情况总数T1[i][j]表示有i个位置,使用j个不同的颜色染色,相邻不同,本质不同的情况总数T[i][j]=T[i−1][j]∗(j−1)+T[i−1][j−1]∗jT1[i][j]=T1[i−1][j−1]+T1[i−1][j]∗(j−1);dp[i][j]=sum[i−1]∗T[l[i]][j]∗C[m][j]−dp[i−1][j]∗T[l[i]][j]=sum[i−1]∗T[l[i]][j]∗q[j]/p[j]−dp[i−1][j]∗T[l[i]][j]=(sum[i−1]∗q[j]−dp[i−1][j]∗p[j])∗T[l[i]][j]/p[j];T1[l[i]][j]=T[l[i]][j]/p[j];
#include <cstdio> #include <algorithm> using namespace std; long long P[5001],Q[5001]; int T[5001][5001],ans[2][5001],a[1000001]; int main(){ #ifdef DouBi freopen("in.cpp","r",stdin); #endif // DouBi int n,m,p; while(scanf("%d%d%d",&n,&m,&p)!=EOF){ **T=*P=*Q=1; for(int i=1;i<=5000;i++){ for(int j=1;j<=i;j++) T[i][j]=(T[i-1][j-1]+T[i-1][j]*(j-1ll))%p; P[i]=P[i-1]*(m-i+1)%p; Q[i]=Q[i-1]*i%p; } int S=1; for(int i=1;i<=n;i++){ scanf("%d",a+i); for(int j=1;j<=min(a[i],m);j++) ans[i&1][j]=((P[j]*S-ans[~i&1][j]*Q[j])%p+p)*T[a[i]][j]%p; //S=accumulate(ans[i&1],ans[i&1]+a[i]+1,0ll)%p; S=0; for(int j=0;j<a[i]+1;j++){ S=(S+ans[i&1][j])%p; } for(int j=0;j<a[i-1]+1;j++){ ans[~i&1][j]=0; } //fill_n(ans[~i&1],a[i-1]+1,0); } printf("%d\n",S); } return 0; }
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