2016弱校联盟十一专场10.5 F Fibonacci of Fibonacci(矩阵快速幂+找循环节)

题目分析

这道题第一步肯定是找循环节，因为要求的是Fn,但是这里的n = Fn，因此需要求循环节，其实就是用map处理一下而已，会发现循环正好是从（0,1）开始的为26880696，这题求余给的数很巧！！然后就可以先对下标进行矩阵快速幂，得到一个数之后再对上面进行矩阵快速幂即可。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
const LL mod = 20160519;
const LL MOD = 26880696;

struct Matirx{
LL mat[2][2];
};

Matirx Mul1(Matirx a, Matirx b){
Matirx ret;
for(int i = 0; i < 2; i++){
for(int j = 0; j < 2; j++){
ret.mat[i][j] = 0;
for(int k = 0; k < 2; k++)
ret.mat[i][j] += a.mat[i][k]*b.mat[k][j];
ret.mat[i][j] %= MOD;
}
}
return ret;
}

Matirx Mul2(Matirx a, Matirx b){
Matirx ret;
for(int i = 0; i < 2; i++){
for(int j = 0; j < 2; j++){
ret.mat[i][j] = 0;
for(int k = 0; k < 2; k++)
ret.mat[i][j] += a.mat[i][k]*b.mat[k][j];
ret.mat[i][j] %= mod;
}
}
return ret;
}

Matirx Mat_pow1(Matirx a, int n){
Matirx ret;
memset(ret.mat, 0, sizeof(ret.mat));
for(int i = 0; i < 2; i++) ret.mat[i][i] = 1;
while(n){
if(n&1) ret = Mul1(ret, a);
a = Mul1(a, a);
n >>= 1;
}
return ret;
}

Matirx Mat_pow2(Matirx a, int n){
Matirx ret;
memset(ret.mat, 0, sizeof(ret.mat));
for(int i = 0; i < 2; i++) ret.mat[i][i] = 1LL;
while(n){
if(n&1) ret = Mul2(ret, a);
a = Mul2(a, a);
n >>= 1;
}
return ret;
}

void init(Matirx &temp){
temp.mat[0][0] = 1;
temp.mat[0][1] = 1;
temp.mat[1][0] = 1;
temp.mat[1][1] = 0;
}

int main(){
int T, n;
scanf("%d", &T);
while(T--){
scanf("%d", &n);
Matirx temp;
init(temp);
Matirx ret = Mat_pow1(temp, n-1);
init(temp);
Matirx ans = Mat_pow2(temp, ret.mat[0][0]-1);
printf("%lld\n", ans.mat[0][0]);
}
return 0;
}

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define LL long long
const int maxn = 25;
LL bit[maxn], dp[maxn][15][5];

LL dfs(int pos, int pre, int x, bool zero, bool limit){
if(pos < 1) return 1LL;
if(!limit && dp[pos][pre][x] != -1) return dp[pos][pre][x];
int len = limit?bit[pos]:9;
LL ret = 0;
for(int i = 0; i <= len; i++){
if(zero) ret += dfs(pos-1, i, x, zero&&i==0, limit&&i==len);
else{
int temp = x;
if(i < pre){
if(temp == 0) continue;
temp = 1;
}
else if(i > pre){
if(temp == 1) continue;
temp = 0;
}
ret += dfs(pos-1, i, temp, false, limit&&i==len);
}
}
if(!limit) dp[pos][pre][x] = ret;
return ret;
}

LL solve(LL n){
int len = 0;
while(n){
bit[++len] = n%10;
n /= 10;
}
return dfs(len, 0, 2, true, true);
}

int main(){
int T;
LL l, r;
scanf("%d", &T);
memset(dp, -1, sizeof(dp));
while(T--){
scanf("%lld%lld", &l, &r);
printf("%lld\n", solve(r) - solve(l-1));
}
return 0;
}