poj3070 Fibonacci
Fibonacci
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 11510 | Accepted: 8179 |
Description
In the Fibonacci integer sequence, F0 = 0, F1 = 1, and Fn = Fn − 1 + Fn − 2 for n ≥ 2. For example, the first ten terms of the Fibonacci sequence are:
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, …
An alternative formula for the Fibonacci sequence is
.
Given an integer n, your goal is to compute the last 4 digits of Fn.
Input
The input test file will contain multiple test cases. Each test case consists of a single line containing n (where 0 ≤ n ≤ 1,000,000,000). The end-of-file is denoted by a single line containing the number −1.
Output
For each test case, print the last four digits of Fn. If the last four digits of Fn are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., print Fn mod 10000).
Sample Input
0 9 999999999 1000000000 -1
Sample Output
0 34 626
6875
这题可以用矩阵快速幂，算是入门题。
方法一：直接计算.
#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
typedef long long ll;
#define inf 0x7fffffff
#define pi acos(-1.0)
#define MOD 10000
ll fast_mod(ll n)
{
ll i,j,k;
ll ans[2][2]={1,0,0,1}; //初始化为单位矩阵，也是最终的答案
ll temp[2][2]; //做为矩阵乘法中的中间变量
ll a[2][2]={1,1,1,0};
while(n)
{
if(n&1){ //实现 ans*=t,其中要先把ans赋值给tmp然后用ans=tmp*t
for(i=0;i<2;i++){
for(j=0;j<2;j++){
temp[i][j]=ans[i][j];
}
}
ans[0][0]=ans[0][1]=ans[1][0]=ans[1][1]=0;
for(i=0;i<2;i++){
for(j=0;j<2;j++){
for(k=0;k<2;k++){
ans[i][j]=(ans[i][j]+(temp[i][k]*a[k][j]+MOD)%MOD+MOD )%MOD;
}
}
}
}
for(i=0;i<2;i++){
for(j=0;j<2;j++){
temp[i][j]=a[i][j];
}
}
a[0][0]=a[0][1]=a[1][0]=a[1][1]=0;
for(i=0;i<2;i++){
for(j=0;j<2;j++){
for(k=0;k<2;k++){
a[i][j]=(a[i][j]+(temp[i][k]*temp[k][j]+MOD)%MOD+MOD )%MOD; //这里先加MOD然后再取模是为了模完后不为负数
}
}
}
n>>=1;
}
return (ans[0][1]+MOD)%MOD;
}
int main()
{
ll n,m,i,j;
while(scanf("%lld",&n)!=EOF && n!=-1){
printf("%lld\n",fast_mod(n));
}
return 0;
} 方法二：考虑1×2的矩阵【f[n-2],f[n-1]】。我们可以通过乘以一个2×2的矩阵A，得到矩阵：【f[n-1],f[n]】。
即：【f[n-2],f[n-1]】*A = 【f[n-1],f[n]】=【f[n-1],f[n-1]+f[n-2]】
可以构造出这个2×2矩阵A，即：
０ １
１ １
所以，有【f[1],f[2]】×A＝【f[2],f[3]】
又因为矩阵乘法满足结合律，故有：
【f[1],f[2]】×A ^(n-1) =【f[n],f[n+1]】 <pre name="code" class="cpp">#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
typedef long long ll;
#define inf 0x7fffffff
#define pi acos(-1.0)
#define MOD 10000
ll ans[2][2];
void fast_mod(ll n)
{
ll i,j,k;
ll temp[2][2]; //做为矩阵乘法中的中间变量
ll a[2][2]={0,1,1,1};
while(n)
{
if(n&1){ //实现 ans*=t,其中要先把ans赋值给tmp然后用ans=tmp*t
for(i=0;i<2;i++){
for(j=0;j<2;j++){
temp[i][j]=ans[i][j];
}
}
ans[0][0]=ans[0][1]=ans[1][0]=ans[1][1]=0;
for(i=0;i<2;i++){
for(j=0;j<2;j++){
for(k=0;k<2;k++){
ans[i][j]=(ans[i][j]+(temp[i][k]*a[k][j]+MOD)%MOD+MOD )%MOD;
}
}
}
}
for(i=0;i<2;i++){
for(j=0;j<2;j++){
temp[i][j]=a[i][j];
}
}
a[0][0]=a[0][1]=a[1][0]=a[1][1]=0;
for(i=0;i<2;i++){
for(j=0;j<2;j++){
for(k=0;k<2;k++){
a[i][j]=(a[i][j]+(temp[i][k]*temp[k][j]+MOD)%MOD+MOD )%MOD; //这里先加MOD然后再取模是为了模完后不为负数
}
}
}
n>>=1;
}
}
int main()
{
ll n,m,i,j;
while(scanf("%lld",&n)!=EOF && n!=-1){
if(n==0){
printf("0\n");continue;
}
if(n==1){
printf("1\n");continue;
}
if(n==2){
printf("1\n");continue;
}
ans[0][0]=1;
ans[0][1]=0;
ans[1][0]=0;
ans[1][1]=1;
fast_mod(n-1);
printf("%lld\n",ans[1][1]%MOD );
}
return 0;
}
快速幂模板：
#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
typedef long long ll;
#define inf 0x7fffffff
#define pi acos(-1.0)
#define MOD 10000
struct matrix{
ll n,m,i;
ll data[99][99];
void init_danwei(){
for(i=0;i<n;i++){
data[i][i]=1;
}
}
}a,b,c,d,t;
matrix multi(matrix &a,matrix &b){
ll i,j,k;
matrix temp;
temp.n=a.n;
temp.m=b.m;
for(i=0;i<temp.n;i++){
for(j=0;j<temp.m;j++){
temp.data[i][j]=0;
}
}
for(i=0;i<a.n;i++){
for(k=0;k<a.m;k++){
if(a.data[i][k]>0){
for(j=0;j<b.m;j++){
temp.data[i][j]=(temp.data[i][j]+(a.data[i][k]*b.data[k][j])%MOD )%MOD;
}
}
}
}
return temp;
}
matrix fast_mod(matrix a,ll n){
matrix ans;
ans.n=a.n;
ans.m=a.m;
memset(ans.data,0,sizeof(ans.data));
ans.init_danwei();
while(n>0){
if(n&1)ans=multi(ans,a);
a=multi(a,a);
n>>=1;
}
return ans;
}
int main()
{
ll n,m,i,j;
while(scanf("%lld",&n)!=EOF && n!=-1)
{
a.data[0][0]=a.data[0][1]=a.data[1][0]=1;
a.data[1][1]=0;
a.n=a.m=2;
matrix ant=fast_mod(a,n);
printf("%lld\n",ant.data[0][1]%MOD);
}
return 0;
}
