Jzzhu and Sequences 【矩陣快速冪】
阿新 • • 發佈:2019-01-24
Jzzhu and Sequences
Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u
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Status
Description
Jzzhu has invented a kind of sequences, they meet the following property:
You are given x and y, please calculate fn modulo 1000000007(109 + 7).
Input
The first line contains two integers x and y(|x|, |y| ≤ 109). The second line contains a single integer n(1 ≤ n ≤ 2·109).
Output
Output a single integer representing fn modulo 1000000007(109 + 7).
Sample Input
Input
2 3
3
Output
1
Input
0 -1
2
Output
1000000006
Hint
In the first sample, f2 = f1 + f3, 3 = 2 + f3, f3 = 1.
In the second sample, f2 = - 1; - 1 modulo (109 + 7) equals (109 + 6).
構造矩陣之後套模板即可;
#include <stdio.h>
#include <iostream>
#include <string.h>
#include <algorithm>
#include <math.h>
#include <ctype.h>
#include <time.h>
#include <queue>
using namespace std;
const long long MOD = 1000000007;
struct node
{
long long m[2][2];
}ans,base;
long long a,b;
int n;
node multi(node a,node b)
{
node tmp;
for (int i=0;i<2;i++)
for(int j=0;j<2;j++)
{
tmp.m[i][j] = 0;
for(int k=0;k<2;k++)
{
tmp.m[i][j] = ((tmp.m[i][j] + a.m[i][k] * b.m[k][j]) % MOD + MOD)%MOD;
}
}
return tmp;
}
void fast_mod(int n)// 求矩陣 base 的 n 次冪
{
base.m[0][0] = 0;base.m[0][1] = 1;
base.m[1][0] = -1;base.m[1][1] = 1;
ans.m[0][0] = ans.m[1][1] = 1;// ans 初始化為單位矩陣
ans.m[0][1] = ans.m[1][0] = 0;
while (n)
{
if (n&1) //實現 ans *= t; 其中要先把 ans賦值給 tmp,然後用 ans = tmp * t
ans = multi(ans,base);
base = multi(base,base);
n>>=1;
}
}
int main()
{
while (scanf("%lld %lld %d",&a,&b,&n)!=EOF)
{
if (n==1)
{
printf("%lld\n",(a%MOD + MOD)%MOD);
continue;
}
if (n==2)
{
printf("%lld\n",(b%MOD + MOD)%MOD);
continue;
}
fast_mod(n-2);
long long anss = (((ans.m[1][0]*a+ans.m[1][1]*b) % MOD) + MOD) % MOD;
printf("%lld\n",anss);
}
return 0;
}