【矩阵快速幂】 HDOJ 5015 233 Matrix

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构造矩阵，进行矩阵快速幂即可。。。
#include <iostream>  #include <queue>  #include <stack>  #include <map>  #include <set>  #include <bitset>  #include <cstdio>  #include <algorithm>  #include <cstring>  #include <climits>#include <cstdlib>#include <cmath>#include <time.h>#define maxn 55#define maxm 200005#define eps 1e-10#define mod 10000007#define INF 1e9#define lowbit(x) (x&(-x))#define mp make_pair#define ls o<<1#define rs o<<1 | 1#define lson o<<1, L, mid  #define rson o<<1 | 1, mid+1, R  typedef long long LL;//typedef int LL;using namespace std;LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}void scanf(int &__x){__x=0;char __ch=getchar();while(__ch==' '||__ch=='\n')__ch=getchar();while(__ch>='0'&&__ch<='9')__x=__x*10+__ch-'0',__ch = getchar();}LL gcd(LL _a, LL _b){if(!_b) return _a;else return gcd(_b, _a%_b);}// headint num[maxn], ans[maxn];int mat[maxn][maxn], mid[maxn][maxn], res[maxn][maxn];int n, m, t, tt;void init(void){memset(mat, 0, sizeof mat);memset(res, 0, sizeof res);memset(ans, 0, sizeof ans);}void read(void){t = n + 2;num[1] = 1, num[2] = 233;for(int i = 3; i <= t; i++) scanf("%d", &num[i]);}void calculate(void){while(m) {if(m % 2) {for(int i = 1; i <= t; i++)for(int j = 1; j <= t; j++) {LL tmp = 0;for(int k = 1; k <= t; k++)tmp = (tmp + (LL)res[i][k] * mat[k][j]) % mod;mid[i][j] = tmp;}for(int i = 1; i <= t; i++)for(int j = 1; j <= t; j++)res[i][j] = mid[i][j];}for(int i = 1; i <= t; i++)for(int j = 1; j <= t; j++) {LL tmp = 0;for(int k = 1; k <= t; k++)tmp = (tmp + (LL)mat[i][k] * mat[k][j]) % mod;mid[i][j] = tmp;}for(int i = 1; i <= t; i++)for(int j = 1; j <= t; j++)mat[i][j] = mid[i][j];m /= 2;}}void work(void){mat[1][1] = 1, mat[1][2] = 3, mat[2][2] = 10;for(int i = 3; i <= t; i++)for(int j = 2; j <= i; j++)mat[j][i] = 1;for(int i = 1; i <= t; i++) res[i][i] = 1;calculate();for(int i = 1; i <= t; i++)for(int j = 1; j <= t; j++)ans[i] = ((LL)ans[i] + (LL)num[j] * res[j][i]) % mod;printf("%d\n", ans[t]);}int main(void){while(scanf("%d%d", &n, &m)!=EOF) {init();read();work();}return 0;}
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