codeforces#324(div2)B Kolya and Tanya(数论水)
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题意:有3n 个小人坐在一个圆环上,每个小人的硬币数为 a[i],(1 <= a[i] <= 3),问满足 a[i]+a[i+n]+a[i+2n] != 6 的硬币分法有多少种
一开始想着顺着推,一共n组人,其中某一组和不为六有20种,为六有7种。选一组和不为六为20*7^n-1,选两组20^2*7^n-2,选三组20^3*7^bn-3……
总和由二项式定理可得为(20+7)^n-7^n。
要求对1e9+7取摸,水同余
注意:27^n%(1e9+7)-7^n%(1e9+7)有可能会得出负数,所以要加上一个1e9+7再对1e9+7取摸,一开始在这上面wa了好几次……
#include<stdio.h>__int64 mm=1000000007;__int64 kuaisu(__int64 a,__int64 n) {__int64 ans=1; a=a%mm; while(n>0) {if(n%2==1)ans=ans%mm*a%mm; a=a%mm*a%mm%mm; n=n/2; } return ans; }int main() { __int64 n,i,j,k,m; __int64 ans; scanf("%I64d",&n); ans=(kuaisu(27,n)%mm-kuaisu(7,n)%mm+mm)%mm; printf("%I64d\n",ans); return 0; }
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