UVA766 Sum of powers(1到n的自然数幂和 伯努利数)

自然数幂和：

 

(1)

 

伯努利数的递推式：

B0 = 1

 

(要满足(1)式,求出Bn后将B1改为1 /2)

参考：https://en.wikipedia.org/wiki/Bernoulli_number

http://blog.csdn.net/acdreamers/article/details/38929067

 

使用分数类，代入求解

 

 

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#include<cstdio> #include<iostream> #include<cstdlib> #include<cstring> #include<string> #include<algorithm> #include<map> #include<queue> #include<vector> #include<cmath> #include<utility> using namespace std; typedef long long LL; const int N = 25, INF = 0x3F3F3F3F;   LL gcd(LL a, LL b){     while(b){         LL t = a % b;         a = b;         b = t;     }     return a; }   LL lcm(LL a, LL b){     return a / gcd(a, b) * b; }   struct frac{     LL x, y;     frac(){         x = 0;         y = 1;     }     frac(LL x1, LL y1){         x = x1;         y = y1;     }     frac operator*(const frac &tp)const{         LL a = x * tp.x;         LL b = y * tp.y;         LL d = gcd(a, b);         a /= d;         b /= d;         if(a >= 0 && b < 0){             a = -a;             b = -b;         }         return frac(a, b);     }       frac operator+(const frac &tp)const{         LL a = x * tp.y + tp.x * y;         LL b = y * tp.y;         LL d = gcd(a, b);         a /= d;         b /= d;         if(a >= 0 && b < 0){             a = -a;             b = -b;         }           return frac(a, b);     }   }ans[N][N], bo[N];   LL cm[N][N]; void init(){     memset(cm, 0, sizeof(cm));     cm[0][0] = 1;     for(int i = 1; i < N; i++){         cm[i][0] = 1;         for(int j = 1; j <= i; j++){             cm[i][j] = cm[i - 1][j - 1] + cm[i - 1][j];         }     }       bo[0].x = 1, bo[0].y = 1;     for(int i = 1; i < N; i++){         bo[i].x = 0;         bo[i].y = 1;         for(int j = 0; j < i; j++){             bo[i] = bo[i] + frac(cm[i + 1][j], 1) * bo[j];         }         bo[i] = bo[i] * frac(-1, i + 1);     }     bo[1].x = 1; bo[1].y = 2;     for(int m = 0; m < N; m++){         for(int k = 0; k <= m; k++){             ans[m][m + 1 - k] = frac(cm[m + 1][k], 1) * bo[k] * frac(1, m + 1);         }         LL lc = ans[m][0].y;         for(int k = 1; k <= m; k++){             lc = lcm(ans[m][k].y, lc);         }         for(int k = 0; k <= m + 1; k++){             LL d = lc / ans[m][k].y;             ans[m][k].x *= d;             ans[m][k].y *= d;         }     }   }   int main(){     init();     int t;     cin >> t;     while(t--){         int n;         cin >>n;         printf("%lld ", ans[n][0].y);         for(int i = n + 1; i >= 0; i--){             if(i == 0){                 printf("%lld\n", ans[n][i].x);             }else{                 printf("%lld ", ans[n][i].x);             }         }         if(t){             printf("\n");         }     }       return 0; }