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这个题，比赛的时候一直在往dp的方向想，但是总有一个组合数学的部分没办法求，
纯粹组合数学撸，也想不到办法……
其实，很显然。。
从后往前推，把第k种颜色放在最后一个，剩下的k球，还有C(剩余的位置,k球的总数目-1)种放法
然后讨论第k-1种。。。推下去就好了
但是当时没想到……
这里要求组合数，由于比较大，用乘法逆元。。。
当然直接套lucas也是可以的。。。。
Kyoya Ootori has a bag with n colored balls that are colored with k different colors. The colors are labeled from 1 to k. Balls of the same color are indistinguishable. He draws balls from the bag one by one until the bag is empty. He noticed that he drew the last ball of color i before drawing the last ball of color i?+?1 for all i from 1 to k?-?1. Now he wonders how many different ways this can happen.
The first line of input will have one integer k (1?≤?k?≤?1000) the number of colors.
Then, k lines will follow. The i-th line will contain ci, the number of balls of the i-th color (1?≤?ci?≤?1000).
The total number of balls doesn‘t exceed 1000.
A single integer, the number of ways that Kyoya can draw the balls from the bag as described in the statement, modulo 1?000?000?007.
3 2 2 1
3
4 1 2 3 4
1680
In the first sample, we have 2 balls of color 1, 2 balls of color 2, and 1 ball of color 3. The three ways for Kyoya are:
1 2 1 2 3 1 1 2 2 3 2 1 1 2 3
#include<iostream>
using namespace std;
typedef long long ll;
const ll mod=1000000007;
ll qpow(ll a,ll b)
{
ll ans=1,c=a;
while(b)
{
if(b&1)
ans=ans*c%mod;
b>>=1;
c=c*c%mod;
}
return ans;
}
ll fac[1000010];
ll work(int a,int b)
{
return fac[a]*qpow(fac[b]*fac[a-b]%mod,mod-2)%mod;
}
int a[1010];
int main()
{
fac[0]=1;
for(int i=1;i<=1000000;i++)
fac[i]=fac[i-1]*i%mod;
int n;
cin>>n;
int sum=0;
for(int i=0;i<n;i++)
{
cin>>a[i];
sum+=a[i];
}
ll ans=1;
for(int i=n-1;i>-1;i--)
{
ans=ans*work(sum-1,a[i]-1)%mod;
sum-=a[i];
}
cout<<ans;
}codeforces 553 A Kyoya and Colored Balls
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原文地址：http://blog.csdn.net/stl112514/article/details/46632839
