UVA_10229

    将斐波那契数用矩阵形式表示,然后再做快速幂取模即可。

#include<stdio.h>
#include<string.h>
#define MAXD 50
long long int N, M, D, a[MAXD][4], b[4];
void pow_mod(long long int n, int e)
{
if(n == 1)
{
a[e][0] = a[e][1] = a[e][2] = 1;
a[e][3] = 0;
return ;
}
pow_mod(n / 2, e + 1);
a[e][0] = (a[e + 1][0] * a[e + 1][0] + a[e + 1][1] * a[e + 1][2]) % D;
a[e][1] = (a[e + 1][0] * a[e + 1][1] + a[e + 1][1] * a[e + 1][3]) % D;
a[e][2] = (a[e + 1][2] * a[e + 1][0] + a[e + 1][3] * a[e + 1][2]) % D;
a[e][3] = (a[e + 1][2] * a[e + 1][1] + a[e + 1][3] * a[e + 1][3]) % D;
if(n % 2)
{
b[0] = (a[e][0] * a[0][0] + a[e][1] * a[0][2]) % D;
b[1] = (a[e][0] * a[0][1] + a[e][1] * a[0][3]) % D;
b[2] = (a[e][2] * a[0][0] + a[e][3] * a[0][2]) % D;
b[3] = (a[e][2] * a[0][1] + a[e][3] * a[0][3]) % D;
a[e][0] = b[0], a[e][1] = b[1], a[e][2] = b[2], a[e][3] = b[3];
}
}
void solve()
{
int i;
D = 1;
for(i = 0; i < M; i ++)
D <<= 1;
pow_mod(N - 1, 1);
printf("%lld\n", a[1][0]);
}
int main()
{
a[0][0] = a[0][1] = a[0][2] = 1;
a[0][3] = 0;
while(scanf("%lld%lld", &N, &M) == 2)
{
if(N == 0 || M == 0)
printf("0\n");
else if(N == 1)
printf("1\n");
else
solve();
}
return 0;
}


posted on 2011-12-15 14:35  Staginner  阅读(440)  评论(0编辑  收藏  举报