FZU Problem 1692 Key problem(循环矩阵)
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循环矩阵,这里有解说:http://wenku.baidu.com/link?
url=zcJ-sxrj0QDqzz8xCnHTnB7gxjoNRyOZzS4_4ZA22c8Bs9inYn6vVkqTVr_w-riLa8oRnYA9SRcCZ9f4UciCUNGeNAG4dCGclYRPS18YLGa
推出第一层以下依据性质就能够得到。
Problem 1692 Key problem
Accept: 144 Submit: 663
Time Limit: 1000 mSec Memory Limit : 32768 KB
Problem Description
Input
Output
Each case separated by a space. See sample.
Sample Input
1 3 2 3 4 10000 1 2 3
Sample Output
120 133 131
Source
FOJ月赛-2009年3月--- Coral#include <set> #include <map> #include <queue> #include <math.h> #include <vector> #include <string> #include <stdio.h> #include <string.h> #include <stdlib.h> #include <iostream> #include <algorithm> #define eps 1e-8 #define pi acos(-1.0) #define LL __int64 using namespace std; const int maxn = 110; LL a[maxn], b[maxn], f[maxn]; LL mod, n; void mul(LL a[], LL b[]) { LL c[maxn]; memset(c, 0, sizeof(c)); for(int i = 0; i < n; i++) for(int j = 0; j < n; j++) c[i] = (a[j]*b[(i-j+n)%n]+c[i])%mod; memcpy(a, c, sizeof(c)); } void pow_mod(LL a[], LL b) { LL c[maxn]; memset(c, 0, sizeof(c)); c[0] = 1LL; while(b) { if(b&1) mul(c, a); mul(a, a); b >>= 1; } memcpy(a, c, sizeof(c)); } int main() { int T; cin>>T; LL m, l, r; while(T--) { cin>>n>>m>>l>>r>>mod; for(int i = 0;i < n;i++) cin>>a[i]; memset(f, 0, sizeof(f)); f[0] = 1; f[1] = r; f[n-1]=l; pow_mod(f, m); LL ans[maxn]; for(int i = 0; i < n; i++) { ans[i] = 0; for(int j = 0;j < n;j++) ans[i] = (ans[i]+a[j]*f[(i-j+n)%n])%mod; } cout<<ans[0]; for(int i = 1; i < n; i++) cout<<" "<<ans[i]; cout<<endl; } return 0; }
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