CodeForces #309 Div.2 C. Kyoya and Colored Balls

C. Kyoya and Colored Balls

题目链接:http://codeforces.com/problemset/problem/554/C

题意:题目给出k种颜色的球,每个求各有Ci个,且要求第i种球其后紧跟第i+1种球.

求一共有多少种排法.

题解:

//由于题目要求第i种球其后紧跟第i+1种球.

//意味着必须确定每种颜色最后一个球的方法

//所以对于最后一种颜色,假设这个颜色有x个,总球数为y.

//那么最后一种颜色的球,必须有一个球,要放在最后一个位置,那么剩下的球就有C(x-1,y-1)种方法.

//依次从后往前类推,将每种情况相乘即可.

#include<stdio.h>
#include<string.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define ll __int64
#define maxn 1000000007
int a[1600];
ll c[1050][1060];
ll sum;

int main() {
int n,m,sum1;
for(int i = 1; i <= 1000; i++)
c[i][0] = 1;
for(int i = 1; i <= 1000; i++) {
for(int j = 1; j <= i; j++) {
if(i == j)
c[i][j] = 1;
else if(i > j)
c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % maxn;
}
}
while(~scanf("%d",&n)) {
for(int i = 1; i <= n; i++) {
scanf("%d",&a[i]);
}
sum1 = a[1];
sum = 1;
for(int i = 2; i <= n; i++) {
sum1 += a[i];
sum = (sum * c[sum1 - 1][a[i] - 1]) % maxn;
}
printf("%I64d\n",sum);
}
return 0;
}