一卡通大冒險
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1388 Accepted Submission(s): 915
{{A},{B},{C}} , {{A,B},{C}}, {{B,C},{A}}, {{A,C},{B}} ,{{A,B,C}} 於是,
這個邪惡計劃的組織者wf希望瞭解,如果ACM訓練對裏有n位帥哥(即有N張一卡通),那麼要把這些一卡通夾到書裏有多少種不同的方法。
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int dp[2005][2005],ans[2005];
void ini()
{
int i,j,sum;
memset(dp,0,sizeof(dp));
memset(ans,0,sizeof(ans));
for(i=1;i<=2000;i++)
dp[i][1]=dp[i][i]=1;
ans[1]=1;
for(i=2;i<=2000;i++)
{
sum=0;
for(j=2;j<i;j++)
{
dp[i][j]=(dp[i-1][j-1]+dp[i-1][j]*j)%1000;
sum+=dp[i][j];
}
ans[i]=sum%1000;
ans[i]+=2;
}
}
int main()
{
int t,n;
scanf("%d",&t);
ini();
while(t--)
{
scanf("%d",&n);
printf("%d\n",ans[n]);
}
return 0;
}