DP: dp[i][j]前i堆排出j长度的序列有多少种排法，
dp[i][j]=dp[i-1][j] (不用第i堆),
dp[i][j]+=dp[i-1][j-k]*C[j][k] (用第i堆的k个石头)
A Famous Stone Collector
Time Limit: 30000/15000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 845 Accepted Submission(s): 322
Problem Description
Mr. B loves to play with colorful stones. There are n colors of stones in his collection. Two stones with the same color are indistinguishable. Mr. B would like to
select some stones and arrange them in line to form a beautiful pattern. After several arrangements he finds it very hard for him to enumerate all the patterns. So he asks you to write a program to count the number of different possible patterns. Two patterns
are considered different, if and only if they have different number of stones or have different colors on at least one position.
select some stones and arrange them in line to form a beautiful pattern. After several arrangements he finds it very hard for him to enumerate all the patterns. So he asks you to write a program to count the number of different possible patterns. Two patterns
are considered different, if and only if they have different number of stones or have different colors on at least one position.
Input
Each test case starts with a line containing an integer n indicating the kinds of stones Mr. B have. Following this is a line containing n integers - the number of
available stones of each color respectively. All the input numbers will be nonnegative and no more than 100.
available stones of each color respectively. All the input numbers will be nonnegative and no more than 100.
Output
For each test case, display a single line containing the case number and the number of different patterns Mr. B can make with these stones, modulo 1,000,000,007,
which is a prime number.
which is a prime number.
Sample Input
3 1 1 1 2 1 2
Sample Output
Case 1: 15 Case 2: 8HintIn the first case, suppose the colors of the stones Mr. B has are B, G and M, the different patterns Mr. B can form are: B; G; M; BG; BM; GM; GB; MB; MG; BGM; BMG; GBM; GMB; MBG; MGB.
Source
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long int LL;
const LL MOD = 1000000007LL;
LL C[11000][210];
void getC()
{
for(int i=0;i<11000;i++) C[i][0]=C[i][i]=1;
for(int i=2;i<11000;i++)
for(int j=1;j<i&&j<=200;j++)
C[i][j]=(C[i-1][j]+C[i-1][j-1])%MOD;
}
LL dp[110][11000];
int n,a[110];
int main()
{
getC();
int cas=1;
while(scanf("%d",&n)!=EOF)
{
for(int i=1;i<=n;i++)
scanf("%d",a+i);
memset(dp,0,sizeof(dp));
int len=0;
dp[0][0]=1;
for(int i=1;i<=n;i++)
{
len+=a[i];
for(int j=0;j<=len;j++)
{
dp[i][j]=(dp[i][j]+dp[i-1][j])%MOD;
for(int k=1;k<=a[i]&&j-k>=0;k++)
{
dp[i][j]=(dp[i][j]+dp[i-1][j-k]*C[j][k])%MOD;
}
}
}
LL ans=0;
for(int i=1;i<=len;i++)
ans=(ans+dp[n][i])%MOD;
printf("Case %d: %lld\n",cas++,ans%MOD);
}
return 0;
}