zoj 1864
题目链接：
题意：
求自然数幂和。
限制：
0 <= n <= 10^50; 1 <= k <= 100
思路：
k不大，而且答案不取模，直接搞
(n+1)^(k+1) - n^(k+1) = C(k+1,1)*n^k + C(k+1,2)*n^(k-1) + ... + C(k+1,k)*n + 1;
(n+1)^(k+1) - 1 = ( (n+1)^(k+1) - n^(k+1) ) + ( n^(k+1) - (n-1)^(k+1) ) + ... + ( 2^(k+1) - 1^(k+1) )
=C(k+1,1)*(1^k+2^k+...+n^k) + C(k+1,2)*(1^(k-1)+2^(k-1)+...+n^(k-1)) + C(k+1,k)*(1+2+...+n) + n;
令S(k,n)=1^k+2^k+...+n^k 得：
(n+1)^(k+1) - 1 = C(k+1,1)*S(k,n) + C(k+1,2)*S(k-1,n) + ... + C(k+1,k)*S(1,n) + n;
移项得：
C(k+1,1)*S(k,n) = ...
S(k,n) = 1/C(k+1,1) * ...
能O(k^2)得出结果
其实也可以用伯努利数来做。
Java Code
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/*zoj 1864 题意： 求自然数幂。 限制： 0 <= n <= 10^50; 1 <= k <= 100 思路： k不大，而且答案不取模，直接搞 (n+1)^(k+1) - n^(k+1) = C(k+1,1)*n^k + C(k+1,2)*n^(k-1) + ... + C(k+1,k)*n + 1; (n+1)^(k+1) - 1 = ( (n+1)^(k+1) - n^(k+1) ) + ( n^(k+1) - (n-1)^(k+1) ) + ... + ( 2^(k+1) - 1^(k+1) ) =C(k+1,1)*(1^k+2^k+...+n^k) + C(k+1,2)*(1^(k-1)+2^(k-1)+...+n^(k-1)) + C(k+1,k)*(1+2+...+n) + n; 令S(k,n)=1^k+2^k+...+n^k 得： (n+1)^(k+1) - 1 = C(k+1,1)*S(k,n) + C(k+1,2)*S(k-1,n) + ... + C(k+1,k)*S(1,n) + n; 移项得： C(k+1,1)*S(k,n) = ... S(k,n) = 1/C(k+1,1) * ... 能O(k^2)得出结果 其实也可以用伯努利数来做。 */ import java.math.BigInteger; import java.util.Scanner; public class Main { static final int N = 105; static final BigInteger _one = BigInteger.valueOf(-1); static BigInteger[][] C = new BigInteger[N][N]; static BigInteger[] ans = new BigInteger[N]; static BigInteger[] n1k1 = new BigInteger[N]; static void predo() { for(int i = 0; i < N; ++i) C[i][0] = C[i][i] = BigInteger.ONE; for(int i = 2; i < N; ++i) for(int j = 1; j < i; ++j) C[i][j] = C[i - 1][j].add(C[i - 1][j - 1]); } static void init(BigInteger n, int k) { for(int i = 0; i <= k; ++i) ans[i] = _one; n1k1[0] = n.add(BigInteger.ONE); for(int i = 1; i <= k; ++i) n1k1[i] = n1k1[i - 1].multiply(n.add(BigInteger.ONE)); } static BigInteger gao(BigInteger n, int k) { if(!ans[k].equals(_one)) return ans[k]; if(k == 1) ans[k] = n.add(BigInteger.ONE).multiply(n).divide(BigInteger.valueOf(2)); else { ans[k] = n1k1[k].subtract(n.add(BigInteger.ONE)); for(int i = 2; i <= k; ++i) ans[k] = ans[k].subtract(C[k + 1][i].multiply(gao(n, k - i + 1))); ans[k] = ans[k].divide(BigInteger.valueOf(k + 1)); } return ans[k]; } public static void main(String[] args) { predo(); Scanner in = new Scanner(System.in); BigInteger n; int k; while (in.hasNext()) { n = in.nextBigInteger(); k = in.nextInt(); init(n, k); System.out.println(gao(n, k)); } } } |