斐波那契数列

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http://blog.csdn.net/lvshubao1314/article/details/38013897

一：递推

二：矩阵

//斐波那契数列第N项后MOD位，矩阵快速幂求法； #include<iostream> #include<cstdio>#include<string.h>#define LL long longusing namespace std;const int MOD=10000;struct Matrix{LL m[3][3];};Matrix A={0,0,0,  0,1,1,  0,1,0,};Matrix E={0,0,0,  0,1,0,  0,0,1,};Matrix multi(Matrix a,Matrix b){Matrix ans;for(int i=1;i<=2;i++){for(int j=1;j<=2;j++){ans.m[i][j]=0;for(int k=1;k<=2;k++)ans.m[i][j]+=a.m[i][k]*b.m[k][j]%MOD;ans.m[i][j]%=MOD;}}return ans;}Matrix power(Matrix a,int p){Matrix ans=E,B=a;while(p){if(p%2==1)ans=multi(ans,B);p=p/2;B=multi(B,B);} return ans;}int main(){int n;while(scanf("%d",&n)!=EOF){Matrix ans=power(A,n-1);if(n==0)printf("0\n");elseprintf("%d\n",ans.m[1][1]);}return 0;}

三：利用数学对数计算求前几位数字

#include<iostream> #include<cstdio>#include<cmath>using namespace std;int main(){int f[25];f[0]=0,f[1]=1;for(int i=2;i<=20;i++){f[i]=f[i-1]+f[i-2];}int n;while(scanf("%d",&n)!=EOF){if(n<=20)printf("%d\n",f[n]);else{double f=(1.0+sqrt(5.0))/2.0;double p=-0.5*log10(5.0)+n*1.0*log(f)/log(10.0);p=p-floor(p);double ans=pow(10.0,p);while(ans<1000)ans=ans*10;printf("%d\n",(int)ans);}}return 0;}

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