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关于Hilbert矩阵的几道编程题
最后一个敲了好久，写篇日志纪念一下，有需要的自取吧。
2.以浮点数的形式打印 n 阶 Hilbert 矩阵
//editor: Jan Tang//problem: 2#include <iostream>#include <cstdio>using namespace std;#define set0(a) memset(a,0,sizeof(a));#define CIN(a,n) for(int i=1;i<=n;i++) cin>>a[i];typedef long long ll;typedef unsigned long long ull;const int Mod = 1e9+7;const int maxn = 100005;const int inf = 0x3f3f3f3f;int m,n;/*==============================head==========================*/int main(){ while(scanf("%d",&n)!=EOF){ for(int i = 1; i <= n; i++){ putchar(‘[‘); for(int j = 1; j <= n; j++){ printf("%.3lf", 1.0/(i+j-1)); if(j!=n) printf("\t "); } putchar(‘]‘); printf("\n"); if(i==n) continue; putchar(‘[‘); for(int j = 1; j <= n-1; j++){ putchar(‘\t‘); } printf(" "); putchar(‘]‘); printf("\n"); } } return 0;}
4.以有理数的形式打印 n 阶 Hilbert 矩阵
输出格式挺难控制的，用-左对齐
//editor: Jan Tang//problem: 4#include <iostream>#include <cstdio>using namespace std;#define set0(a) memset(a,0,sizeof(a));#define CIN(a,n) for(int i=1;i<=n;i++) cin>>a[i];typedef long long ll;typedef unsigned long long ull;const int Mod = 1e9+7;const int maxn = 100005;const int inf = 0x3f3f3f3f;int m,n;/*==============================head==========================*/int main(){ while(scanf("%d",&n)!=EOF){ for(int i = 1; i <= n; i++){ putchar(‘[‘); for(int j = 1; j <= n; j++){ printf("1/%-2d", j+i-1); if(j!=n) printf("\t "); } putchar(‘]‘); printf("\n"); if(i==n) continue; putchar(‘[‘); for(int j = 1; j <= n-1; j++){ putchar(‘\t‘); } printf(" "); putchar(‘]‘); printf("\n"); } } return 0;}
8&9.求n阶 Hilbert 矩阵的逆矩阵，并用有理数的形式打印
要用高斯消元，结合分数类来做，输出格式不想管了，放弃治疗了。
//editor: Jan Tang//problem: 8 & 9#include <iostream>#include <cstdio>#include <string>#include <cstring>#include <algorithm>#include <cmath>#include <map>#include <stack>#include <queue>#include <vector>#include <set>#include <cstdlib>using namespace std;#define set0(a) memset(a,0,sizeof(a));#define CIN(a,n) for(int i=1;i<=n;i++) cin>>a[i];typedef long long ll;typedef unsigned long long ull;const int Mod = 1e9+7;const int maxn = 1005;const int inf = 0x3f3f3f3f;int m,n;ll gcd(ll s, ll t){ return t==0 ? s : gcd(t, s%t);}struct fraction{ //分数类 long long num, den; //分子 numerator, 分母 denominator fraction(long long num = 0, long long den = 1){ //构造函数，默认值为0/1 if(den < 0){ //保持分母为正 num = -num; den = -den; }// assert(den != 0); //断言：如果不满足，程序直接停止 long long g =gcd(abs(num), den); this->num = num / g; this->den = den / g; } fraction operator +(const fraction &o) const{ return fraction(num * o.den + den * o.num, den * o.den); } fraction operator -(const fraction &o) const{ return fraction(num * o.den - den * o.num, den * o.den); } fraction operator *(const fraction &o) const{ return fraction(num * o.num, den * o.den); } fraction operator /(const fraction &o) const{ return fraction(num * o.den, den * o.num); } bool operator <(const fraction &o) const{ return num * o.den < den * o.num; } bool operator ==(const fraction &o) const{ return num * o.den == den * o.num; }}; //不能漏了分号vector<fraction> aa[maxn], bb[maxn];inline vector<fraction> operator * (vector<fraction> a, fraction b){ int n = a.size(); vector<fraction> res; for(int i = 0; i < n ;i++){ res.push_back(b*a[i]); } return res;}inline vector<fraction> operator - (vector<fraction> a, vector<fraction> b){ int n = a.size(); vector<fraction> res; for(int i = 0; i < n ;i++){ res.push_back(a[i]-b[i]); } return res;}inline void inverse(vector<fraction> A[], vector<fraction> C[], int n){ for(int i = 0; i < n; i++){ for(int j = 0; j < n; j++){ if(j == i) C[i].push_back(fraction(1,1)); else C[i].push_back(fraction(0,1)); } } for(int i = 0; i < n; i++){ for(int j = i; j < n; j++){ if(A[j][i].num != 0){ swap(A[i], A[j]); swap(C[i], C[j]); break; } } C[i] = C[i] * (fraction(1,1) / A[i][i]); A[i] = A[i] * (fraction(1,1) / A[i][i]); for(int j = 0; j < n; j++) if(j != i && A[j][i].num != 0){ C[j] = C[j] - C[i] * A[j][i]; A[j] = A[j] - A[i] * A[j][i]; } }}/*==============================head==========================*/int main(){ while(scanf("%d",&n)!=EOF){ for(int i = 0; i < n; i++){ aa[i].clear(); bb[i].clear(); } for(int i = 0; i < n; i++){ for(int j = 0; j < n; j++){ aa[i].push_back(fraction(1, i+j+1));// printf("%lld/%lld\t", aa[i][j].num, aa[i][j].den); }// cout<<endl; } // fraction tmp1 = fraction(2,2), tmp2 = fraction(2,4); // swap(tmp1, tmp2); // printf("%lld/%lld\t", tmp1.num, tmp1.den); inverse(aa, bb, n); //这里不用加方括号 for(int i = 0; i < n; i++){ putchar(‘[‘); for(int j = 0; j < n; j++){ printf("%lld/%lld\t", bb[i][j].num, bb[i][j].den); } putchar(‘]‘); printf("\n"); if(i==n-1) continue; putchar(‘[‘); for(int j = 1; j <= n-1; j++){ putchar(‘\t‘); } printf(" "); putchar(‘]‘); printf("\n"); } } return 0;}
关于Hilbert矩阵的几道编程题
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