Nice Prefixes
时间： 1ms 内存：128M
描述：
Consider strings formed from characters from an alphabet of size K. For example, if K = 4, our alphabet might be {a,b,c,d}, and an example string is bbcac.
For a string S, define count(S, k) to be the number of occurrences of the symbol k in S. For example, count(bbcac, b) = 2 and count(bbcac, a) = 1.
A prefix of a string S is any string obtained from S by deleting some (possibly none) of the trailing characters of S. For example, the prefixes of acb are the empty string, a, ac, and acb.
A string S has "nice prefixes" if for every prefix P of S and for every two characters k1 and k2 in the alphabet, |count(P, k1) - count(P, k2)| <= 2. For example, bbcac has nice prefixes, but abbbc does not because count(abbb, b) = 3 and count(abbb, c) = 0.
Count the number of strings of length L on an alphabet of size K that have nice prefixes. This number can be large, so print its remainder when divided by 1000000007.
输入：
The input is a single line containing the two integers L and K, separated by spaces, with 1 <= L <= 1018 and 1 <= K <= 50.
输出：
Output a single line containing the number of strings of length L on an alphabet of size K that have nice prefixes, modulo 1000000007.
示例输入：
4 2
示例输出：
12
提示：
参考答案（内存最优[1516]）：
#include <cstring>
#include <algorithm>
#include <cassert>
#include <cstdio>
#include <cstdlib>
#include <deque>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <utility>
#include <vector>
using namespace std;
#define FR(i,a,b) for(int i=(a);i<(b);++i)
#define FOR(i,n) FR(i,0,n)
#define FORALL(i,v) for(typeof((v).end())i=(v).begin();i!=(v).end();++i)
#define CLR(x,a) memset(x,a,sizeof(x))
#define BEND(v) (v).begin(),(v).end()
#define MP make_pair
#define PB push_back
#define A first
#define B second
#define dump(x) cerr << #x << " = " << (x) << endl
typedef long long ll; typedef long double ld;
const ll MOD = 1000000007;
ll L;
int K;
ll dp[2][101];
int dodp(int l) {
FOR(i,l) {
int i1 = i%2, i2 = (i+1)%2;
CLR(dp[i2],0);
FOR(k2,K+1) {
int k1 = ((i-2*k2)%K+K)%K;
int k0 = K-k1-k2;
if (k0+k1+k2 > K) continue;
if (k0 < 1) continue;
int k0p = k0-1, k1p = k1+1, k2p = k2;
if (k0p == 0) k1p = k2p, k2p = 0;
dp[i2][k2p] = (dp[i2][k2p] + k0*dp[i1][k2])%MOD;
if (k1 > 0) {
k1p = k1-1, k2p = k2+1;
dp[i2][k2p] = (dp[i2][k2p] + k1*dp[i1][k2])%MOD;
}
}
}
return l%2;
}
struct Matrix {
ll dat[101][101];
Matrix() {
CLR(dat,0);
FOR(k,K+1) dat[k][k] = 1;
}
};
const Matrix operator*(const Matrix &A, const Matrix &B) {
Matrix C;
FOR(i,K+1) FOR(j,K+1) {
C.dat[i][j] = 0;
FOR(k,K+1) C.dat[i][j] = (C.dat[i][j] + A.dat[i][k]*B.dat[k][j]) % MOD;
}
return C;
}
const Matrix expmod(Matrix X, ll n) {
Matrix ret;
while (n) {
if (n%2) ret = ret*X;
X = X*X;
n /= 2;
}
return ret;
}
int main() {
scanf("%lld%d",&L,&K);
assert(1 <= L && L <= 1000000000000000000LL);
assert(1 <= K && K <= 100);
ll matpart = L/K;
Matrix X;
FOR(k,K+1) {
CLR(dp,0);
dp[0][k] = 1;
int idx = dodp(K);
FOR(kp,K+1) {
X.dat[kp][k] = dp[idx][kp];
}
}
Matrix Y = expmod(X, matpart);
CLR(dp,0);
FOR(k,K+1) dp[0][k] = Y.dat[k][0];
ll extpart = L-K*matpart;
assert(extpart >= 0);
int finidx = dodp(extpart);
ll ans = 0;
FOR(k2,K+1) ans = (ans + dp[finidx][k2]) % MOD;
printf("%lld\n",ans);
}
参考答案（时间最优[36]）：
#include <cstring>
#include <algorithm>
#include <cassert>
#include <cstdio>
#include <cstdlib>
#include <deque>
#include <iostream>
#include <map>
#include <queue>
#include <set>
#include <utility>
#include <vector>
using namespace std;
#define FR(i,a,b) for(int i=(a);i<(b);++i)
#define FOR(i,n) FR(i,0,n)
#define FORALL(i,v) for(typeof((v).end())i=(v).begin();i!=(v).end();++i)
#define CLR(x,a) memset(x,a,sizeof(x))
#define BEND(v) (v).begin(),(v).end()
#define MP make_pair
#define PB push_back
#define A first
#define B second
#define dump(x) cerr << #x << " = " << (x) << endl
typedef long long ll; typedef long double ld;
const ll MOD = 1000000007;
ll L;
int K;
ll dp[2][101];
int dodp(int l) {
FOR(i,l) {
int i1 = i%2, i2 = (i+1)%2;
CLR(dp[i2],0);
FOR(k2,K+1) {
int k1 = ((i-2*k2)%K+K)%K;
int k0 = K-k1-k2;
if (k0+k1+k2 > K) continue;
if (k0 < 1) continue;
int k0p = k0-1, k1p = k1+1, k2p = k2;
if (k0p == 0) k1p = k2p, k2p = 0;
dp[i2][k2p] = (dp[i2][k2p] + k0*dp[i1][k2])%MOD;
if (k1 > 0) {
k1p = k1-1, k2p = k2+1;
dp[i2][k2p] = (dp[i2][k2p] + k1*dp[i1][k2])%MOD;
}
}
}
return l%2;
}
struct Matrix {
ll dat[101][101];
Matrix() {
CLR(dat,0);
FOR(k,K+1) dat[k][k] = 1;
}
};
const Matrix operator*(const Matrix &A, const Matrix &B) {
Matrix C;
FOR(i,K+1) FOR(j,K+1) {
C.dat[i][j] = 0;
FOR(k,K+1) C.dat[i][j] = (C.dat[i][j] + A.dat[i][k]*B.dat[k][j]) % MOD;
}
return C;
}
const Matrix expmod(Matrix X, ll n) {
Matrix ret;
while (n) {
if (n%2) ret = ret*X;
X = X*X;
n /= 2;
}
return ret;
}
int main() {
scanf("%lld%d",&L,&K);
assert(1 <= L && L <= 1000000000000000000LL);
assert(1 <= K && K <= 100);
ll matpart = L/K;
Matrix X;
FOR(k,K+1) {
CLR(dp,0);
dp[0][k] = 1;
int idx = dodp(K);
FOR(kp,K+1) {
X.dat[kp][k] = dp[idx][kp];
}
}
Matrix Y = expmod(X, matpart);
CLR(dp,0);
FOR(k,K+1) dp[0][k] = Y.dat[k][0];
ll extpart = L-K*matpart;
assert(extpart >= 0);
int finidx = dodp(extpart);
ll ans = 0;
FOR(k2,K+1) ans = (ans + dp[finidx][k2]) % MOD;
printf("%lld\n",ans);
}
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