題目地址：srm#397_div1_500

題目描述：

Problem Statement
?	NOTE: This problem statement contains superscripts that may not display properly if viewed outside of the applet. You are given ints n and k. Return the value of the sum 1k + 2k + 3k + ... + nk modulo 1000000007.
Definition
?	Class: SumOfPowers Method: value Parameters: int, int Returns: int Method signature: int value(int n, int k) (be sure your method is public)
Limits
?	Time limit (s): 2.000 Memory limit (MB): 64
Constraints
-	n will be between 1 and 109, inclusive.
-	k will be between 1 and 50, inclusive.
Examples
0)	?
?	5 1 Returns: 15 Here, we have arithmethic progression: 1 + 2 + 3 + 4 + 5 = 15.
1)	?
?	4 2 Returns: 30 Just a little bit more complicated example here: 12 + 22 + 32 + 42 = 1 + 4 + 9 + 16 = 30.
2)	?
?	13 5 Returns: 1002001 This one would be harder to check by hand.
3)	?
?	123456789 1 Returns: 383478132

This problem statement is the exclusive and proprietary property of TopCoder, Inc. Any unauthorized use or reproduction of this information without the prior written consent of TopCoder, Inc. is strictly prohibited. (c)2003, TopCoder, Inc. All rights reserved. ? ??

構造一個k+2階矩陣；

這里n比較大 顯然我們接受不了O（n）的復雜度

最好的方法就是像求Fibonacci數列第n項一樣用矩陣乘法+快速冪做

要實現求和長度在k左右的遞推式

首選（n+1）^k-n^k=sigma(c[k][i]*n^i);

?

直接看代碼里面的矩陣構造吧

wa了幾次

1 組合數是要求到c[52][i]的

2 因為mod 10^9+7 所以存在的數都是可能接近int上限的 ?要進行+運算 所以要用long long 存儲

代碼：

#include<iostream>typedef long long inta ;using namespace std;struct Matrix {inta m[60][60];};inta n; // 用來表示維度inta c[60][60]; //組合數const inta mod=1000000007;void init() {for(inta i=0;i<=55;i++)c[i][0]=1;for(inta i=0;i<55;i++)for(inta j=0;j<=i;j++)c[i+1][j+1]=c[i][j+1]+c[i][j];}Matrix multi(Matrix a,Matrix b) {Matrix ans;for(inta i=0;i<n;i++)for(inta j=0;j<n;j++){inta c=0;for(inta k=0;k<n;k++)c=(c+a.m[i][k]*b.m[k][j])%mod;ans.m[i][j]=c;}return ans;}Matrix quick_mod(Matrix a,inta b) {Matrix ans;Matrix p=a;for(inta i=0;i<n;i++)for(inta j=0;j<n;j++)ans.m[i][j]=(i==j?1:0);while(b){if(b&1){ans=multi(ans, p);b--;}b>>=1;p=multi(p, p);}return ans;}class SumOfPowers {public :inta value(inta nn,inta k){init();n=k+2;Matrix A;for(inta i=0;i<n;i++) // so importantfor(inta j=0;j<n;j++)A.m[i][j]=0;for(inta i=0;i<k+1;i++)for(inta j=0;j<=i;j++)A.m[i][j]=c[i][j]%mod;for(inta i=0;i<k+1;i++)A.m[k+1][i]=c[k][i]%mod;A.m[k+1][k+1]=1;A=quick_mod(A, nn-1);inta ans=0;for(inta i=0;i<k+2;i++)ans=(ans+A.m[k+1][i])%mod;return ans;} };int main() {inta nn,k;cin>>nn>>k;SumOfPowers obj;cout<<obj.value(nn, k)<<endl;}
tc上提交沒有main()

轉載于:https://www.cnblogs.com/jingqi814/p/3644368.html

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