标签：script mod cst hit can scanf lld nts ott
蒜头君的兔子
蒜头君的小伙伴在 第一年 送给他一对 一岁 的兔子，并告诉他：这种兔子 刚生下来时算 00 岁，到了 22 岁时就可以繁殖了，它在 2-102?10 岁时，每年会生下来一对兔子，这些兔子到了 22 岁也可以繁殖，但这些兔子在 1010 岁那年 生完仔后 不久就会死亡，蒜头君想知道，第 nn 年兔子 产仔之后（第 nn 年 1010 岁的兔子此时已经死亡），他会有多少对兔子。结果对 10000000071000000007 取模。
共一行，一个正整数 nn，表示蒜头君想知道第 nn 年的兔子总对数。
输出一个整数，表示第 nn 年兔子总对数对 10000000071000000007 取模的值。
对于 3030% 的数据，满足 1 \le n \le 10^31≤n≤10?3??；
对于 6060% 的数据，满足 1 \le n \le 10^51≤n≤10?5??；
对于 100100% 的数据，满足 1 \le n \le 10^91≤n≤10?9??。
10
88
88
352138150
10086
405567313
裸的矩阵快速幂，复杂度O(11^3logn);
#include<iostream> #include<cstdio> #include<cstring> #include<cstdlib> #include<cmath> #include<algorithm> #define mod (1000000007) using namespace std; typedef long long lol; lol n,a[11][11]={{1,0,0,0,0,0,0,0,0,0,0}, {0,0,0,0,0,0,0,0,0,0,0}, {0,0,0,0,0,0,0,0,0,0,0}, {0,0,0,0,0,0,0,0,0,0,0}, {0,0,0,0,0,0,0,0,0,0,0}, {0,0,0,0,0,0,0,0,0,0,0}, {0,0,0,0,0,0,0,0,0,0,0}, {0,0,0,0,0,0,0,0,0,0,0}, {0,0,0,0,0,0,0,0,0,0,0}, {0,0,0,0,0,0,0,0,0,0,0}, {0,0,0,0,0,0,0,0,0,0,0}}; lol b[11][11]={{0,1,0,0,0,0,0,0,0,0,0}, {1,0,1,0,0,0,0,0,0,0,0}, {1,0,0,1,0,0,0,0,0,0,0}, {1,0,0,0,1,0,0,0,0,0,0}, {1,0,0,0,0,1,0,0,0,0,0}, {1,0,0,0,0,0,1,0,0,0,0}, {1,0,0,0,0,0,0,1,0,0,0}, {1,0,0,0,0,0,0,0,1,0,0}, {1,0,0,0,0,0,0,0,0,1,0}, {1,0,0,0,0,0,0,0,0,0,0}, {1,0,0,0,0,0,0,0,0,0,0}}; lol ans; struct matrix { lol a[11][11]; matrix(){for(lol i=0;i<=10;i++)for(lol j=0;j<=10;j++)a[i][j]=0;} matrix(lol b[11][11]){for(lol i=0;i<=10;i++)for(lol j=0;j<=10;j++)a[i][j]=b[i][j];} matrix operator*(matrix b) { matrix ans; for(lol i=0;i<=10;i++) for(lol j=0;j<=10;j++) for(lol k=0;k<=10;k++) ans.a[i][j]=(ans.a[i][j]+(a[i][k]*b.a[k][j])%mod)%mod; return ans; } }S,T; int main() { S=matrix(a);T=matrix(b); scanf("%lld",&n); while(n) { if(n&1)S=S*T; T=T*T; n>>=1; } for(lol i=0;i<=10;i++) ans=(ans+S.a[0][i])%mod; printf("%lld\n",ans); return 0; }
标签：script mod cst hit can scanf lld nts ott
原文地址：http://www.cnblogs.com/huangdalaofighting/p/7259893.html
