标签：codeforces
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 ibefore 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 32 1 1 2 3
这道题让我学会了组合数的计算，因为直接用组合数公式会导致结果不准确，如C(100,50)这样，如果用乘一个数除一个数的方法，那么可能会导致不能整除而会发生误差。
思路：若前i种颜色的方法总数是f(i)，那么第i+1种颜色的方法总数是f(i+1)=f(i)*C(sum(i+1)-1,a[i+1]-1),其中sum(i+1)是前i+1种颜色的个数总和。
#include<stdio.h> #include<string.h> #include<iostream> #include<string> #include<map> #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,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
标签：codeforces
原文地址：http://blog.csdn.net/kirito_acmer/article/details/46830331
