Sum of powers UVA

有一定难度的一道题目，具体的需要找出等式之间的关系，然后递推求解即可，推荐博客：点击打开链接，这里作者将需要用到的等式关系列举的很清楚了，然后递推求解就行了，具体实现见如下代码：

#include<iostream>#include<vector>#include<string>#include<set>#include<stack>#include<queue>#include<map>#include<algorithm>#include<cmath>#include<iomanip>#include<cstring>#include<sstream>#include<cstdio>#include<deque>#include<functional>using namespace std;const int N = 25;typedef long long LL;int C[N][N];LL res[N][N];LL m[N];LL getAbs(LL a){if (a < 0) return -a;return a;}int getSym(LL a){if (a < 0) return -1;else if(a>0) return 1;return 0;}LL gcd(LL a,LL b){if (b == 0) return a;return gcd(b, a%b);}void Reduct(LL& data,LL* arr,int n){LL d = gcd(data, getAbs(arr[0]));for (int i = 1; i <= n; i++){if (arr[i] == 0) continue;d = gcd(d,getAbs(arr[i]));}for (int i = 0; i <= n; i++)arr[i] = getAbs(arr[i])/d*getSym(arr[i]);data /= d;}void Cal(LL* aim,LL* source,LL mul,LL& basea,LL bases,int n){for (int i = 0; i <= n; i++){aim[i] = aim[i] * bases - source[i] * mul*basea;}basea *= bases;Reduct(basea, aim, n);}void Init(){C[0][0] = 1;for (int i = 1; i < N; i++){C[0][i] = C[i][i] = 1;for (int j = 1; j < i; j++)C[j][i] = C[j - 1][i - 1] + C[j][i - 1];}memset(m, 0, sizeof(m));memset(res, 0, sizeof(res));m[0] = 1;res[0][1] = 1;for (int i = 1; i <=20; i++){m[i] = 1;int n = i + 1;for (int j = 1; j <= n; j++){res[i][j] += C[j][n];}for (int j = 0; j < i; j++)Cal(res[i], res[j], C[j][n], m[i], m[j], n);m[i] *= C[i][n];Reduct(m[i], res[i], n);}}int main(){Init();int T;cin >> T;while (T--){int ind;cin >> ind;cout << m[ind];for (int i = ind+1; i >=0 ; i--)cout << " " << res[ind][i];cout << endl;if (T) cout << endl;}return 0;}