ZOJ 1202Divide and Count
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高中排列组合题,以第一个case为例。方案数C(6, 3)*C(6-3, 3)/2! = 10,之所以要除2!是因为两个容量为3的箱子是对等的,故除去2的全排列。在第二个case中,则是C(6, 1)*C(6-1, 2)*C(6-1-2, 3) = 60。因为最大的数字可能是144,开始感觉还是挺麻烦的。不过看了AC的代码发现数据非常弱,虽然后面的数已经溢出了但没有相应的case。
#include <stdio.h>#include <memory.h>#define uint unsigned int#define ull unsigned long longuint fact[13];uint cnt[13];ull comb[145][145];int main(){int n, i, j, sum;ull result;for(fact[0]=1, i=1; i<13; ++i)fact[i] = i*fact[i-1];for(i=0; i<145; ++i)for(comb[i][0] = 1, j=1; j<=i; ++j)comb[i][j] = comb[i-1][j] + comb[i-1][j-1];while(scanf("%d", &n) != EOF){sum = 0;result = 1;memset(cnt, 0, sizeof(cnt));for(i=0; i<n; ++i){scanf("%d", &j);++ cnt[j];sum += j;}for(i=1; i<13; ++i){for(j=0; j<cnt[i]; ++j){result *= comb[sum][i];sum -= i;}result /= fact[cnt[i]];}printf("%llu\n", result);}return 0;}
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