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Jzzhu and Sequences 【矩陣快速冪】

Jzzhu and Sequences
Time Limit:1000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u
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Description
Jzzhu has invented a kind of sequences, they meet the following property:

這裡寫圖片描述
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).

構造矩陣這裡寫圖片描述之後套模板即可;


#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; }