Codeforces Round #309 (Div. 2) C. Kyoya and colored Balls

Kyoya Ootori have a bag withNColored balls that is colored withkdifferent colors. The colors is labeled from1Tok. Balls of the same color is indistinguishable. He draws balls from the bag one by one until the bag is empty. He noticed that he drew the last ball of colorIBefore drawing the last ball of color i? +?1For allIFrom1To k?-? 1. Now he wonders how many different ways this can happen.

Input

The first line of input would have one integer k (1?≤? K≤?1000) The number of colors.

Then,kLines would follow. TheI-th line would contain Ci , the number of balls of theI-th Color (1?≤? C i? ≤?1000).

The total number of balls doesn ' t exceed 1000.

Output

A single Integer, the number of ways this Kyoya can draw the balls from the bag as described in the statement, modulo 1?000?000?007.

Sample Test (s) input

3221

Output

3

Input

41234

Output

1680

Note

In the first sample, we had 2 balls of color 1, 2 balls of color 2, and 1 ball of color 3. The three ways for Kyoya is:

1 2 1 2 31 1 2 2 3
2 1 1) 2 3
This problem let me learn the calculation of combinatorial number, because the direct use of the combination of number formula will lead to inaccurate results, such as C (100,50), if you multiply a number in a number method, it may result in an error can not be divided evenly.
Idea: If the total number of methods of the first I color is f (i), then the total number of methods of i+1 color is f (i+1) =f (i) *c (sum (i+1) -1,a[i+1]-1), where sum (i+1) is the sum of the number of previous i+1 colors.
#include <stdio.h> #include <string.h> #include <iostream> #include <string> #include <map > #include <algorithm>using namespace std; #define LL __int64#define maxn 1000000007int a[1600];ll c[1050][1060] ; ll Sum;int main () {int n,m,i,j,sum1;for (i=1;i<=1000;i++) c[i][0]=1;for (i=1;i<=1000;i++) {for (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)!=eof) {for (i=1;i<=n;i++) {scanf ("%d", &a[i]);} Sum1=a[1];sum=1;for (i=2;i<=n;i++) {sum1+=a[i];//printf ("%d%d\n", a[i]-1,sum1-1); sum= (Sum*c[sum1-1][a[i]-1])% Maxn;//sum= (Sum*f (a[i]-1,sum1-1))%maxn;//printf ("%lld\n", sum);} printf ("%i64d\n", sum);} return 0;}

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Codeforces Round #309 (Div. 2) C. Kyoya and colored Balls