UVA 11582 Colossal Fibonacci Numbers!（模运算）

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题意：对A的B次方项斐波那契序列求余。

解题思路：根据递推关系找到余数的循环周期，再利用快速求余确定位置。

注意n = 0，和 1的情况

https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=26

#include<algorithm>#include<cstdio>#include<cstdlib>#include<cstring>#include<cmath>#include<cctype>#include<list>#include<iostream>#include<map>#include<queue>#include<set>#include<stack>#include<vector>using namespace std;#define FOR(i, s, t) for(int i = (s) ; i <= (t) ; ++i)#define REP(i, n) for(int i = 0 ; i < (n) ; ++i)int buf[10];inline long long read(){    long long x=0,f=1;    char ch=getchar();    while(ch<'0'||ch>'9')    {        if(ch=='-')f=-1;        ch=getchar();    }    while(ch>='0'&&ch<='9')    {        x=x*10+ch-'0';        ch=getchar();    }    return x*f;}inline void writenum(int i){    int p = 0;    if(i == 0) p++;    else while(i)        {            buf[p++] = i % 10;            i /= 10;        }    for(int j = p - 1 ; j >= 0 ; --j) putchar('0' + buf[j]);}/**************************************************************/#define MAX_N 1010 * 1010const int INF = 0x3f3f3f3f;int ans[MAX_N];int init(int n){    ans[0] = 0;    ans[1] = 1;    for(int i = 2 ; i <= n * n ; i++)    {        ans[i] = (ans[i - 1] % n + ans[i - 2] % n) % n;        if(ans[i] == 1 && ans[i - 1] == 0)        {            return i - 1;        }    }}int pow_mod(unsigned long long a, unsigned long long n, int m){    a %= m;    for(int i = 1 ; ; a = a * a % m)    {        if(n & 1) i = i * a % m;        if(!(n >>= 1)) return i;    }}int main(){    int t = read();    unsigned long long a, b;    while(t--)    {        scanf("%llu%llu", &a, &b);        int n = read();        if(n == 1 || n == 0)        {            printf("0\n");            continue;        }        int tmp = init(n);        int num = pow_mod(a, b, tmp);        printf("%d\n", ans[num]);    }    return 0;}

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