感觉是个很裸的矩阵快速幂, 搞个100 × 100 的矩阵, 直接转移就好啦。
#include<bits/stdc++.h> #define LL long long #define fi first #define se second #define mk make_pair #define PLL pair<LL, LL> #define PLI pair<LL, int> #define PII pair<int, int> #define SZ(x) ((int)x.size()) #define ull unsigned long long using namespace std; const int N = 1e5 + 7; const int inf = 0x3f3f3f3f; const LL INF = 0x3f3f3f3f3f3f3f3f; const int mod = 1e9 + 7; const double eps = 1e-8; const double PI = acos(-1); int n, x, c[101]; struct Matrix { int a[101][101]; Matrix() { memset(a, 0, sizeof(a)); } void init() { for(int i = 0; i < 101; i++) a[i][i] = 1; } Matrix operator * (const Matrix &B) const { Matrix C; for(int i = 0; i < 101; i++) for(int j = 0; j < 101; j++) for(int k = 0; k < 101; k++) C.a[i][j] = (C.a[i][j] + 1ll * a[i][k] * B.a[k][j]) % mod; return C; } Matrix operator ^ (int b) { Matrix C; C.init(); Matrix A = (*this); while(b) { if(b & 1) C = C * A; A = A * A; b >>= 1; } return C; } } M; int main() { scanf("%d%d", &n, &x); for(int i = 1; i <= n; i++) { int v; scanf("%d", &v); c[v]++; } for(int j = 0; j < 100; j++) M.a[0][j] = c[j + 1]; M.a[0][100] = 1; for(int i = 1; i < 100; i++) M.a[i][i - 1] = 1; M.a[100][100] = 1; Matrix mat = M ^ (x); printf("%d ", (mat.a[0][0] + mat.a[0][100]) % mod); return 0; } /* */