【矩阵快速幂】 Codeforces Round #291 (Div. 2) E. Darth Vader and Tree

附上CF的官方题解。。。

It's easy to realize that

, where
dp[i] is number of vertices, which are situated on a distance
i from the root, and
cnt[j] is number of children, which are situated on a distance
j. Answer

.

Let the dynamics condition



Let's build a transformation matrix of 101 × 101 size



Now, to move to the next condition, we need to multiply A by B. So, if matrix
C = A·Bx - 100, then the answer will be situated in the very right cell of this matrix. For
x < 100 we'll find the answer using dynamics explained in the beginning.

In order to find Bk let's use binary power.

#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 105
#define maxm 200005
#define eps 1e-7
#define mod 1000000007
#define INF 0x3f3f3f3f
#define PI (acos(-1.0))
#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
#define pii pair<int, int>
#pragma comment(linker, "/STACK:16777216")
typedef long long LL;
typedef unsigned long long ULL;
//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;}
//head

LL mat[maxn][maxn];
LL res[maxn][maxn];
LL mid[maxn][maxn];
LL cnt[maxn];
LL ans[maxn];
LL dp[maxn];
int n, x;

void read()
{
int xx;
scanf("%d%d", &n, &x);
for(int i = 1; i <= n; i++) {
scanf("%d", &xx);
cnt[xx]++;
}
}

void calc(int r)
{
int o = 101;
while(r) {
if(r % 2) {
for(int i = 1; i <= o; i++)
for(int j = 1; j <= o; j++) {
LL t = 0;
for(int k = 1; k <= o; k++)
t = (t + res[i][k] * mat[k][j]) % mod;
mid[i][j] = t;
}
memcpy(res, mid, sizeof mid);
}
for(int i = 1; i <= o; i++)
for(int j = 1; j <= o; j++) {
LL t = 0;
for(int k = 1; k <= o; k++)
t = (t + mat[i][k] * mat[k][j]) % mod;
mid[i][j] = t;
}
memcpy(mat, mid, sizeof mid);
r /= 2;
}
}

void work()
{
dp[0] = 1;
for(int i = 1; i <= 100; i++)
for(int j = 1; j <= i; j++)
dp[i] = (dp[i] + cnt[j] * dp[i - j]) % mod;
for(int i = 0; i <= 100; i++) dp[101] = (dp[101] + dp[i]) % mod;
for(int i = 1; i <= 101; i++) res[i][i] = 1;
for(int i = 2; i <= 100; i++) mat[i][i-1] = 1;
for(int i = 1; i <= 100; i++) mat[i][100] = mat[i][101] = cnt[101-i];
mat[101][101] = 1;
if(x <= 100) {
LL t = 0;
for(int i = 0; i <= x; i++) t = (t + dp[i]) % mod;
printf("%I64d\n", t);
return;
}
calc(x - 100);
LL t = 0;
for(int i = 1; i <= 101; i++) {
t = (t + dp[i] * res[i][101]) % mod;
//printf("%d KKK\n", res[i][101]);
}
printf("%I64d\n", t);
}

int main()
{
read();
work();

return 0;
}