LightOJ 1065 Number Sequence（矩阵快速幂）

题意：

就是用矩阵快速幂求斐波那契数列并取模。顺便学了一下矩阵快速幂。至于矩阵快速幂是什么就去看看讲解的博客吧。

代码：

////  Created by  CQU_CST_WuErli//  Copyright (c) 2016 CQU_CST_WuErli. All rights reserved.//#include <iostream>#include <cstring>#include <cstdio>#include <cstdlib>#include <cctype>#include <cmath>#include <string>#include <vector>#include <list>#include <map>#include <queue>#include <stack>#include <set>#include <algorithm>#include <sstream>#define CLR(x) memset(x,0,sizeof(x))#define OFF(x) memset(x,-1,sizeof(x))#define MEM(x,a) memset((x),(a),sizeof(x))#define BUG cout << "I am here" << endl#define lookln(x) cout << #x << "=" << x << endl#define SI(a) scanf("%d",&a)#define SII(a,b) scanf("%d%d",&a,&b)#define SIII(a,b,c) scanf("%d%d%d",&a,&b,&c)#define rep(flag,start,end) for(int flag=start;flag<=end;flag++)#define Rep(flag,start,end) for(int flag=start;flag>=end;flag--)#define Lson l,mid,rt<<1#define Rson mid+1,r,rt<<1|1#define Root 1,n,1#define BigInteger bignconst int MAX_L=2005;// For BigIntegerconst int INF_INT=0x3f3f3f3f;const long long INF_LL=0x7fffffff;const int MOD=1e9+7;const double eps=1e-9;const double pi=acos(-1);typedef long long  ll;using namespace std;int mod;int a,b,n,m;struct Matrix {    int mat[2][2];}initM,ansM;Matrix multi(Matrix& a,Matrix& b) {    Matrix ret;    rep(i,0,1) rep(j,0,1) {        ret.mat[i][j]=0;        rep(k,0,1) ret.mat[i][j]+=(a.mat[i][k]*b.mat[k][j])%mod;        ret.mat[i][j]=ret.mat[i][j]%mod;    }    return ret;}void calc(int n) {    ansM.mat[0][0]=1;    ansM.mat[0][1]=0;    ansM.mat[1][0]=0;    ansM.mat[1][1]=1;    initM.mat[0][0]=0;    initM.mat[0][1]=1;    initM.mat[1][0]=1;    initM.mat[1][1]=1;    while (n) {        if (n&1) ansM=multi(initM,ansM);        initM=multi(initM,initM);        n>>=1;    }}int main(int argc, char const *argv[]) {#ifdef LOCAL    freopen("C:\\Users\\john\\Desktop\\in.txt","r",stdin);    // freopen("C:\\Users\\john\\Desktop\\out.txt","w",stdout);#endif    for (int T_T,kase=SI(T_T);kase<=T_T;kase++) {        SII(a,b);        SII(n,m);        mod=1;        rep(i,1,m) mod*=10;        cout << "Case " << kase << ": ";        if (n==0) cout << a%mod;        else if (n==1) cout << b%mod;        else if (n==2) cout << (a+b)%mod;        else {            calc(n-1);            cout << ((ansM.mat[1][0]*a)%mod+(ansM.mat[1][1]*b)%mod)%mod << endl;        }    }    return 0;}