UVA10518——水矩阵+细节

传送门：https://vjudge.net/problem/UVA-10518

题意：题目里面说的b进制很神奇似的，其实就是对原数模b。。。

  所以就是矩阵快速幂了，递推公式f[n] = f[n-1] + f[n-2] +1，矩阵也很简单了

代码：

#include<iostream>#include<sstream>#include<cstdio>#include<cstring>#include<string>#include<algorithm>#include<cmath>#include<ctime>#include<list>#include<vector>#include<set>#include<map>#include<stack>#include<queue>#pragma GCC optimize("02")#define cl(a,b) memset(a,b,sizeof(a))#define in freopen("F://1.txt","r",stdin)#define out freopen("F://2.txt","w",stdout)using namespace std;typedef unsigned long long llu;typedef long long ll;const ll inf=(ll)1<<60;const int maxn=1e5+7;const int MAXN = 3;ll n, mod;struct Matrix{    long long mat[MAXN][MAXN];    void Init(){        cl(mat, 0);        //memset(mat, 0, sizeof(mat));    }    void Unit(){        //memset(mat, 0, sizeof(mat));        cl(mat, 0);        for (int i = 0; i < MAXN; i++)            mat[i][i] = 1;    }    void output(){        for (int i = 0; i < MAXN; i++){            for (int j = 0; j < MAXN; j++){                printf("%d ", mat[i][j]);            }            printf("\n");        }    }};Matrix operator*(Matrix &a, Matrix &b){    Matrix tmp;    tmp.Init();///初始化    for (int k = 0; k < MAXN; k++){        for (int i = 0; i < MAXN; i++){            if (!a.mat[i][k])                continue;            for (int j = 0; j < MAXN; j++){                tmp.mat[i][j] += a.mat[i][k] * b.mat[k][j] % mod;                if ( tmp.mat[i][j] >= mod)                    tmp.mat[i][j] -= mod;                if ( tmp.mat[i][j] < 0)                    tmp.mat[i][j] += mod;            }        }    }    return tmp;}Matrix operator ^(Matrix a, ll k){    Matrix tmp;///单位矩阵    tmp.Unit();    if(k<=1)        return a;    for (; k; k >>= 1){        if (k & 1)            tmp = tmp * a;        a = a * a;    }    return tmp;}int main(){    int i,j;    Matrix a, b;    a.Init();b.Init();    a.mat[0][0]=1,a.mat[0][1]=1,a.mat[0][2]=1;    a.mat[1][0]=1;    a.mat[2][2]=1;    b.mat[0][0]=1;b.mat[1][0]=1;b.mat[2][0]=1;    int cas = 1;    while(scanf("%lld %lld",&n, &mod)){        if(!n&&!mod)            return 0;        else if(n==1 || n==0){            printf("Case %d: %lld %lld %d\n", cas++,n,mod,1%mod);            continue;        }        else{            Matrix ans;            ans = a ^ (n-1);            ans = ans * b;            printf("Case %d: %lld %lld %lld\n", cas++,n,mod,ans.mat[0][0]);            continue;        }    }    return 0;}