Lightoj 1070 Algebraic Problem（矩阵快速幂）

题意：给你a+b和a*b的值，求a^n+b^n

思路：

因为是膜2^64，所以直接开unsigned long long 就行

代码：

#include<iostream>#include<cstdio>#include<cstring>using namespace std;typedef unsigned long long ll;int n, m;struct node{    ll s[3][3];    node() {}    node(ll s1, ll s2, ll s3, ll s4)    {        s[1][1] = s1;        s[1][2] = s2;        s[2][1] = s3;        s[2][2] = s4;    }};node mul(node a, node b){    node t;    memset(t.s, 0, sizeof(t.s));    for(int i = 1; i <= 2; i++)        for(int j = 1; j <= 2; j++)            for(int k = 1; k <= 2; k++)                t.s[i][j] = (t.s[i][j]+a.s[i][k]*b.s[k][j]);    return t;}node mt_pow(node p, int k){    node q;    memset(q.s, 0, sizeof(q.s));    for(int i = 1; i <= 2; i++)        q.s[i][i] = 1;    while(k)    {        if(k%2) q = mul(p, q);        p = mul(p, p);        k /= 2;    }    return q;}int main(void){    int t, ca = 1;    cin >> t;    while(t--)    {        ll p, q;        int n;        scanf("%llu%llu%d", &p, &q, &n);        node base = node(p, 1, -q, 0);        printf("Case %d: ", ca++);        ll f1 = p;        ll f2 = p*p-2*q;        if(n == 0) puts("2");        else if(n == 1) printf("%llu\n", f1);        else if(n == 2) printf("%llu\n", f2);        else        {            node ans = mt_pow(base, n-2);            printf("%llu\n", f2*ans.s[1][1]+f1*ans.s[2][1]);        }    }    return 0;}