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蒜頭君的兔子

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蒜頭君的小伙伴在?第一年?送給他一對?一歲?的兔子，并告訴他：這種兔子?剛生下來時算?00?歲，到了?22?歲時就可以繁殖了，它在?2-102?10?歲時，每年會生下來一對兔子，這些兔子到了?22?歲也可以繁殖，但這些兔子在?1010?歲那年?生完仔后?不久就會死亡，蒜頭君想知道，第?nn?年兔子?產仔之后（第?nn?年?1010?歲的兔子此時已經死亡），他會有多少對兔子。結果對?10000000071000000007?取模。

輸入格式

共一行，一個正整數?nn，表示蒜頭君想知道第?nn?年的兔子總對數。

輸出格式

輸出一個整數，表示第?nn?年兔子總對數對?10000000071000000007?取模的值。

數據規模

對于?3030% 的數據，滿足?1 \le n \le 10^31≤n≤10?3??；

對于?6060% 的數據，滿足?1 \le n \le 10^51≤n≤10?5??；

對于?100100% 的數據，滿足?1 \le n \le 10^91≤n≤10?9??。

樣例輸入1

10

樣例輸出1

88

樣例輸入2

88

樣例輸出2

352138150

樣例輸入3

10086

樣例輸出3

405567313

題解

裸的矩陣快速冪，復雜度O(11^3logn);

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#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; }

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轉載于:https://www.cnblogs.com/huangdalaofighting/p/7259893.html

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