用的dls的板子，因为看不懂调了好久...果然用别人的板子就是这么蛋疼- -||
num数组0~k+1储存了k+2个值，且这k+2个值是自然数i的k次方而不是次方和，dls的板子自己帮你算和的...搞得我弄了好久
1 #include <iostream> 2 #include <string.h> 3 #include <cstdio> 4 #include <vector> 5 #include <queue> 6 #include <assert.h> 7 #include <math.h> 8 #include <string> 9 #include <algorithm> 10 #include <functional> 11 12 #define SIGMA_SIZE 26 13 #define lson rt<<1 14 #define rson rt<<1|1 15 #define lowbit(x) (x&-x) 16 #define foe(i, a, b) for(int i=a; i<=b; i++) 17 #define fo(i, a, b) for(int i=a; i<b; i++) 18 #define pii pair<int,int> 19 #pragma warning ( disable : 4996 ) 20 21 using namespace std; 22 typedef long long LL; 23 inline LL LMax(LL a, LL b) { return a>b ? a : b; } 24 inline LL LMin(LL a, LL b) { return a>b ? b : a; } 25 inline LL lgcd(LL a, LL b) { return b == 0 ? a : lgcd(b, a%b); } 26 inline LL llcm(LL a, LL b) { return a / lgcd(a, b)*b; } //a*b = gcd*lcm 27 inline int Max(int a, int b) { return a>b ? a : b; } 28 inline int Min(int a, int b) { return a>b ? b : a; } 29 inline int gcd(int a, int b) { return b == 0 ? a : gcd(b, a%b); } 30 inline int lcm(int a, int b) { return a / gcd(a, b)*b; } //a*b = gcd*lcm 31 const LL INF = 0x3f3f3f3f3f3f3f3f; 32 const LL mod = 1e9+7; 33 const double eps = 1e-8; 34 const int inf = 0x3f3f3f3f; 35 const int maxk = 3e6 + 5; 36 const int maxn = 1e6+10; 37 38 /// 注意mod，使用前须调用一次 polysum::init(int M); 39 namespace polysum { 40 #define rep(i,a,n) for (int i=a;i<n;i++) 41 #define per(i,a,n) for (int i=n-1;i>=a;i--) 42 typedef long long ll; 43 const ll mod = 1e9 + 7; /// 取模值 44 ll powmod(ll a, ll b) { ll res = 1; a %= mod; assert(b >= 0); for (; b; b >>= 1) { if (b & 1)res = res*a%mod; a = a*a%mod; }return res; } 45 46 const int D = 1010000; /// 最高次限制 47 ll a[D], f[D], g[D], p[D], p1[D], p2[D], b[D], h[D][2], C[D]; 48 ll calcn(int d, ll *a, ll n) { 49 if (n <= d) return a[n]; 50 p1[0] = p2[0] = 1; 51 rep(i, 0, d + 1) { 52 ll t = (n - i + mod) % mod; 53 p1[i + 1] = p1[i] * t%mod; 54 } 55 rep(i, 0, d + 1) { 56 ll t = (n - d + i + mod) % mod; 57 p2[i + 1] = p2[i] * t%mod; 58 } 59 ll ans = 0; 60 rep(i, 0, d + 1) { 61 ll t = g[i] * g[d - i] % mod*p1[i] % mod*p2[d - i] % mod*a[i] % mod; 62 if ((d - i) & 1) ans = (ans - t + mod) % mod; 63 else ans = (ans + t) % mod; 64 } 65 return ans; 66 } 67 void init(int M) { /// M：最高次 68 f[0] = f[1] = g[0] = g[1] = 1; 69 rep(i, 2, M + 5) f[i] = f[i - 1] * i%mod; 70 g[M + 4] = powmod(f[M + 4], mod - 2); 71 per(i, 1, M + 4) g[i] = g[i + 1] * (i + 1) % mod; 72 } 73 ll polysum(ll n, ll *arr, ll m) { /// a[0].. a[m] \sum_{i=0}^{n-1} a[i] 74 for (int i = 0; i <= m; i++) 75 a[i] = arr[i]; 76 a[m + 1] = calcn(m, a, m + 1); 77 rep(i, 1, m + 2) a[i] = (a[i - 1] + a[i]) % mod; 78 return calcn(m + 1, a, n - 1); 79 } 80 ll qpolysum(ll R, ll n, ll *a, ll m) { /// a[0].. a[m] \sum_{i=0}^{n-1} a[i]*R^i 81 if (R == 1) return polysum(n, a, m); 82 a[m + 1] = calcn(m, a, m + 1); 83 ll r = powmod(R, mod - 2), p3 = 0, p4 = 0, c, ans; 84 h[0][0] = 0; h[0][1] = 1; 85 rep(i, 1, m + 2) { 86 h[i][0] = (h[i - 1][0] + a[i - 1])*r%mod; 87 h[i][1] = h[i - 1][1] * r%mod; 88 } 89 rep(i, 0, m + 2) { 90 ll t = g[i] * g[m + 1 - i] % mod; 91 if (i & 1) p3 = ((p3 - h[i][0] * t) % mod + mod) % mod, p4 = ((p4 - h[i][1] * t) % mod + mod) % mod; 92 else p3 = (p3 + h[i][0] * t) % mod, p4 = (p4 + h[i][1] * t) % mod; 93 } 94 c = powmod(p4, mod - 2)*(mod - p3) % mod; 95 rep(i, 0, m + 2) h[i][0] = (h[i][0] + h[i][1] * c) % mod; 96 rep(i, 0, m + 2) C[i] = h[i][0]; 97 ans = (calcn(m, C, n)*powmod(R, n) - c) % mod; 98 if (ans<0) ans += mod; 99 return ans; 100 } 101 } 102 103 LL num[maxn]; 104 LL n, k; 105 106 void init() 107 { 108 for( int i = 0; i <= k+1; i++ ) 109 num[i] = polysum::powmod((LL)i+1, k); 110 } 111 112 int main(){ 113 cin >> n >> k; 114 115 polysum::init(k); 116 init(); 117 118 LL ans = polysum::polysum(n, num, k+1)%mod; 119 printf("%lld\n", ans); 120 return 0; 121 }