3231: [Sdoi2008]遞歸數列
Time Limit: 1 Sec Memory Limit: 256 MBSubmit: 502 Solved: 217
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Description
Input
Output
Sample Input
1 1
1 1
2 10 1000003
Sample Output
HINT
對於100%的測試數據:
1<= k<=15
1 <= m <= n <= 1018
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <vector>
#include <queue>
#include <set>
#include <algorithm>
#define LL long long
using namespace std;
const long long MAXN = 20;
long long n, m, p, k;
long long read()
{
long long x = 0, f = 1; char ch = getchar();
while(ch < '0' || ch > '9'){if(ch == '-') f *= -1; ch = getchar();}
while(ch >= '0' && ch <= '9'){x = x * 10 + ch - '0'; ch = getchar();}
return x * f;
}
struct Matrix
{
long long mat[MAXN][MAXN];
Matrix operator * (const Matrix &b)
{
Matrix rs;memset(rs.mat, 0, sizeof(rs.mat));
for(long long i=0;i<=k;i++)
{
for(long long j=0;j<=k;j++)
{
for(long long x=0;x<=k;x++)
{
(rs.mat[i][j] += (mat[i][x] * b.mat[x][j])%p) %= p;
}
}
}
return rs;
}
};
Matrix pow_mod(Matrix a, long long b)
{
Matrix rs;
for(long long i=0;i<=k;i++) rs.mat[i][i] = 1;
while(b)
{
if(b & 1) rs = rs * a;
a = a * a;
b >>= 1;
}
return rs;
}
long long b[20], c[20];
int main()
{
k = read();
for(long long i=1;i<=k;i++) b[i] = read();
for(long long i=1;i<=k;i++) c[i] = read();
m = read(), n = read(), p = read();
if(n <= k)
{
long long l = 0, r = 0;
for(long long i=1;i<m;i++) (l += b[i]) %= p;
for(long long i=m;i<=n;i++) (r += b[i]) %= p;
long long ans = (r - l + p) % p;
while(ans < 0) ans = (ans + p) % p;
printf("%lld\n", ans % p);
}
else
{
Matrix a; memset(a.mat, 0, sizeof(a.mat));
a.mat[0][0] = 1; long long sk = 0;
for(long long i=1;i<=k;i++) (sk += b[i]) %= p;
for(long long i=1;i<=k;i++) a.mat[0][i] = a.mat[1][i] = c[i];
for(long long i=2;i<=k;i++) a.mat[i][i-1] = 1;
Matrix ma = pow_mod(a, n - k);
long long r = 0;
for(long long i=0;i<=k;i++)
{
if(i == 0) (r += ma.mat[0][0] * sk) %= p;
else (r += (ma.mat[0][i] * (b[k-i+1])) % p) %= p;
}
long long l = 0;
if(m - 1<= k)
{
for(long long i=1;i<m;i++) (l += b[i]) %= p;
}
else
{
Matrix mb = pow_mod(a, m - 1 - k);
for(long long i=0;i<=k;i++)
{
if(i == 0) (l += mb.mat[0][0] * sk) %= p;
else (l += (mb.mat[0][i] * b[k-i+1]) % p) %= p;
}
}
long long ans = (r - l + p) % p;
while(ans < 0) ans = (ans + p) % p;
printf("%lld\n", ans % p);
}
return 0;
}