HDOJ 题目4284 A Famous Stone Collector（组合数学）

A Famous Stone Collector

Time Limit: 30000/15000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1122    Accepted Submission(s): 421

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.

 

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.

 

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.

 

Sample Input

31 1 121 2

 

Sample Output

Case 1: 15Case 2: 8
Hint
In 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

Fudan Local Programming Contest 2012

 

Recommend

We have carefully selected several similar problems for you:  4247 4255 4252 4246 4249 

 题目大意：n用颜色的球，给出每种球的个数，问总共有多少种排法，石头可以不用上

#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>#define LL long long#define mod 1000000007using namespace std;LL a[110];LL c[11010][110];int dp[110][11010];void fun(){    int i,j;    for(i=0;i<10010;i++)        c[i][0]=1;    for(i=1;i<10010;i++)        for(j=1;j<=i&&j<110;j++)        c[i][j]=(c[i-1][j-1]+c[i-1][j])%mod;}int main(){    LL n;    int cas=0;    fun();    while(scanf("%lld",&n)!=EOF)    {        LL i,j,k;        LL sum=0;        int m;       // scanf("%d",&m);       // printf("++++%lld\n",c[n][m]);        for(i=1;i<=n;i++)        {            scanf("%lld",&a[i]);            sum+=a[i];        }        dp[0][0]=1;        for(i=1;i<=n;i++)        {            for(j=0;j<=sum;j++)            {                dp[i][j]=dp[i-1][j];                for(k=1;k<=a[i];k++)                {                    if(j-k<0)                        break;                    dp[i][j]=(dp[i][j]+dp[i-1][j-k]*c[j][k]%mod)%mod;                }            }        }        LL ans=0;        for(i=1;i<=sum;i++)        {            ans=(ans+dp[n][i])%mod;        }        printf("Case %d: %lld\n",++cas,ans);    }}