51nod 1161 Partial Sums,1172 Partial Sums V2

 1 n,k=map(int,raw_input().split())
 2 c=[1]*n
 3 p=10**9+7
 4 a=[int(input()) for i in xrange(n)]
 5 inv=[1]*5005
 6 for i in range(2,5005):
 7     inv[i] =(p-p//i) * inv[p%i] % p
 8 for i in xrange(1,n):
 9     c[i]=c[i-1]*(k-1+i)*inv[i]%p
10 c.reverse()
11 for i in xrange(1,n+1):
12     s=0
13     for j in xrange(i):
14         s=s+a[j]*c[-(i-j)]
15     print(s%p)

  1 #include <iostream>
  2 using namespace std;
  3 
  4 #define N 340030
  5 #define M 340020
  6 #define LL long long
  7 #define mod 1000000007
  8 #define K 3
  9 
 10 const int m[K] = {1004535809, 998244353, 104857601};
 11 const int G = 3;
 12 
 13 LL qpow(LL x, LL k, LL p) {
 14     int ret = 1;
 15     while(k) {
 16         if(k & 1) ret = 1LL * ret%p * x % p;
 17         k >>= 1;
 18         x = 1LL * x%p * x % p;
 19     }
 20     return ret;
 21 }
 22 struct _NTT {
 23     LL wn[25], p;
 24 
 25     void init(LL _p) {
 26         p = _p;
 27         for(int i = 1; i <= 21; ++i) {
 28             int t = 1 << i;
 29             wn[i] = qpow(G, (p - 1) / t, p);
 30         }
 31     }
 32     void change(LL *y, int len) {
 33         for(int i = 1, j = len / 2; i < len - 1; ++i) {
 34             if(i < j) swap(y[i], y[j]);
 35             int k = len / 2;
 36             while(j >= k) j -= k, k /= 2;
 37             j += k;
 38         }
 39     }
 40     void NTT(LL y[], int len, int on) {
 41         change(y, len);
 42         int id = 0;
 43         for(int h = 2; h <= len; h <<= 1) {
 44             ++id;
 45             for(int j = 0; j < len; j += h) {
 46                 LL w = 1;
 47                 for(int k = j; k < j + h / 2; ++k) {
 48                     int u = y[k];
 49                     int t = y[k+h/2] * w % p;
 50                     y[k] = u + t;
 51                     if(y[k] >= p) y[k] -= p;
 52                     y[k+h/2] = u + p - t;
 53                     if(y[k+h/2] >= p) y[k+h/2] -= p;
 54                     w = w * wn[id] % p;
 55                 }
 56             }
 57         }
 58         if(on == -1) {
 59             for(int i = 1; i < len / 2; ++i) swap(y[i], y[len-i]);
 60             int inv = qpow(len, p - 2, p);
 61             for(int i = 0; i < len; ++i)
 62                 y[i] = 1LL * y[i] * inv % p;
 63         }
 64     }
 65     void mul(LL A[], LL B[], LL len) {
 66         NTT(A, len, 1);
 67         NTT(B, len, 1);
 68         for(int i = 0; i < len; ++i) A[i] = 1LL * A[i] * B[i] % p;
 69         NTT(A, len, -1);
 70     }
 71 }ntt[K];
 72 
 73 LL tmp[N][K], t1[N], t2[N];
 74 LL x1[N], x2[N];
 75 LL r[K][K];
 76 
 77 LL CRT(LL a[]) {
 78     LL x[K];
 79     for(int i = 0; i < K; ++i) {
 80         x[i] = a[i];
 81         for(int j = 0; j < i; ++j) {
 82             int t = (x[i] - x[j]) % m[i];
 83             if(t < 0) t += m[i];
 84             x[i] = 1LL * t * r[j][i] % m[i];
 85         }
 86     }
 87     LL mul = 1, ret = x[0] % mod;
 88     for(int i = 1; i < K; ++i) {
 89         mul = 1LL * mul * m[i-1] % mod;
 90         ret += 1LL * x[i] * mul % mod;
 91         if(ret >= mod) ret -= mod;
 92     }
 93     return ret;
 94 }
 95 
 96 void mul(LL A[], LL B[], LL len) {
 97     for(int id = 0; id < K; ++id) {
 98         for(int i = 0; i < len; ++i) {
 99             t1[i] = A[i], t2[i] = B[i];
100         }
101         ntt[id].mul(t1, t2, len);
102         for(int i = 0; i < len; ++i) {
103             tmp[i][id] = t1[i];
104         }
105     }
106     for(int i = 0; i < len; ++i) A[i] = CRT(tmp[i]);
107 }
108 
109 LL a[N],b[N];
110 LL inv[50005];
111 
112 void init() {
113     for(int i = 0; i < K; ++i) {
114         for(int j = 0; j < i; ++j)
115             r[j][i] = qpow(m[j], m[i] - 2, m[i]);
116     }
117     for(int i = 0; i < K; ++i) ntt[i].init(m[i]);//ntt的初始化到这里结束
118     inv[1]=1;
119     for(int x=2;x<50003;x++)inv[x]=(mod-mod/x)*inv[mod%x]%mod;
120 }
121 
122 
123 int main()
124 {
125     init();
126     int n,k,i;
127     cin>>n>>k;
128     b[0]=1;
129     for(i=1;i<n;i++)
130     {
131         b[i]=b[i-1]*(k-1+i)%mod*inv[i]%mod;//求矩阵第一列系数
132         b[i]=(b[i]+mod)%mod;
133     }
134     for(i=0;i<n;i++)
135     {
136         cin>>a[i];
137     }
138     mul(a,b,1<<17);//做卷积
139     for(i=0;i<n;i++)
140     {
141         cout<<(a[i]+mod)%mod<<endl;
142     }
143     return 0;
144 }