标签:names 递推 name pac http efi base iostream namespace
http://acm.xidian.edu.cn/problem.php?id=1111
只知道规律也做不做不出来,看来题解,用递推f(n)=f(n-1)+f(n-2)+n-2来推f(n)=3*f(n-1)-2*f(n-2)-f(n-3)+f(n-4),然后矩阵快速幂。
#include<iostream> #include<cstring> #include<cstdio> #include<string> #include<algorithm> #define MOD 1000000007 using namespace std; int n,anss[6] = {0,0,0,1,3,7}; struct matrix { long long m[4][4]; }; matrix ans = { 1,0,0,0, 0,1,0,0, 0,0,1,0, 0,0,0,1,}; matrix base = { 3,1,0,0, MOD-2,0,1,0, MOD-1,0,0,1, 1,0,0,0,}; matrix mul(matrix a, matrix b) { matrix tmp; for(int i = 0; i < 4;i++) { for(int j = 0; j < 4;j++) { tmp.m[i][j] = 0; for(int k = 0; k < 4;k++) tmp.m[i][j] = (tmp.m[i][j]+a.m[i][k]*b.m[k][j])%MOD; } } return tmp; } long long fast_mod(int n) { matrix x = ans,y = base; while(n) { if(n & 1) { x = mul(x,y); } y = mul(y,y); n >>= 1; } return (anss[5]*x.m[0][0]+anss[4]*x.m[1][0]+anss[3]*x.m[2][0]+anss[2]*x.m[3][0])%MOD; } int main() { while(~scanf("%d",&n)) { if(n <= 5) printf("%d\n",anss[n]); else printf("%lld\n",fast_mod(n-5)); } }
标签:names 递推 name pac http efi base iostream namespace
原文地址:http://www.cnblogs.com/zhurb/p/5998390.html