c语言编程求多项式的和（两个多项式相加运算（用c语言））

：暂无数据 2026-04-25 14:40:01 ：0

老铁们，关于c语言编程求多项式的和，你可能听过不少说法。今天，咱们就坐下来好好聊聊两个多项式相加运算（用c语言），保证让你豁然开朗。

本文目录

两个多项式相加运算（用c语言）
C语言求多项式的和y=a1x1+a2x2+a3x3+a4x4
多项式相加C语言
求大佬c语言编程求多项式和，题目在图片里
C语言实验--多项式求和
C语言写多项式相加怎么写
用C语言编写多项式相加的程序，下面图中前两个是要相加的多项式第三个是相加后的结果，求好心人拜托了
用c语言实现多项式相加
C语言求多项式的和y=a1*x1+a2*x2+a3*x3+a4*x4
题目:计算如样图所示的多项式求和,输入一个x就可以得到求和要求x<****.编写一个C语言程序，谢谢

两个多项式相加运算（用c语言）

#include《*****》
#include《*****》
#define Null 0
typedef struct Node
{
int coeff;
int expo;
Node *next;
}listNode,*list;
list createList()
{
list head;
head = (list)malloc(sizeof(listNode));
head = NULL;
printf("want to create a new node?y/n\n");
char ch;
fflush(stdin);
scanf("%c",&ch);
while(ch==’Y’ || ch== ’y’)
{
list p;
p = (list)malloc(sizeof(listNode));
printf("input coeff\n");
int coeff;
scanf("%d",&coeff);
p-》coeff = coeff;
printf("input expo\n");
int expo;
scanf("%d",&expo);
p-》expo = expo;
p-》next = NULL;
//链表为空的时候，即添加首个元素
if(head == NULL)
{
head=p;//添加代码
}
else
{
list prev,curr;
curr = head;
prev = NULL;
//找到添加位置
while(curr!=NULL && curr-》expo》p-》expo)
{
prev=curr;
curr=curr-》next;//添加代码
}
if(curr!=NULL && curr-》expo == p-》expo)
{
curr-》coeff = curr-》coeff + p-》coeff;
printf("want to create a new node?y/n\n");
fflush(stdin);
scanf("%c",&ch);
if(ch==’Y’ || ch== ’y’)
continue;
else
return head;
}
//插入结点，结点非首
if(prev != NULL)
{
p-》next=curr;
prev-》next=p;
//添加代码
}
//插入结点，结点为首
else
{
p-》next=curr;
head=p;
//添加代码
}
}
printf("want to create a new node?y/n\n");
fflush(stdin);
scanf("%c",&ch);
}
return head;
}
list add(list head1,list head2)
{
list head,newNode,ptr1,ptr2,ptr3;
ptr1 = head1;
ptr2 = head2;
head = NULL;
while(ptr1 != NULL && ptr2 != NULL)
{
newNode = (list)malloc(sizeof(listNode));
if(ptr1-》expo 》 ptr2-》expo)
{
newNode-》coeff = ptr1-》coeff;
newNode-》expo = ptr1-》expo;
newNode-》next = NULL;
ptr1 = ptr1-》next;
}
else if(ptr1-》expo 《 ptr2-》expo)
{
newNode-》coeff = ptr2-》coeff;
newNode-》expo = ptr2-》expo;
newNode-》next = NULL;
ptr2 = ptr2-》next;//添加代码
}
else
{
newNode-》coeff = ptr1-》coeff + ptr2-》coeff;
newNode-》expo = ptr1-》expo;
newNode-》next = NULL;
ptr1 = ptr1-》next;
ptr2 = ptr2-》next;
}
if(head==NULL)
{
head = newNode;
}
else
{
ptr3 = head;
//添加代码
while(ptr3-》next != NULL)
ptr3 = ptr3-》next;
ptr3-》next = newNode;
}
}
while(ptr1 != NULL)
{
newNode = (list)malloc(sizeof(listNode));
newNode-》coeff = ptr1-》coeff;
newNode-》expo = ptr1-》expo;
newNode-》next = NULL;
ptr3 = head;
if(ptr3 == NULL)
head = ptr3 = newNode;
else
{
while(ptr3-》next != NULL)
ptr3 = ptr3-》next;
ptr3-》next = newNode;//添加代码
}
ptr1 = ptr1-》next;
}
while(ptr2 != NULL)
{
newNode = (list)malloc(sizeof(listNode));
newNode-》coeff = ptr2-》coeff;
newNode-》expo = ptr2-》expo;
ptr3 = head;
if(ptr3 == NULL)
head = ptr3 = newNode;
else
{
while(ptr3-》next != NULL)
ptr3 = ptr3-》next;
ptr3-》next = newNode;
}
ptr2 = ptr2-》next;
}
return head;
}
void display(list head)
{
list ptr = head;
while(ptr != NULL)
{
if(ptr != head )
printf("+");
printf("%d",ptr-》coeff);
printf("x^");
printf("%d",ptr-》expo);
ptr = ptr-》next;
}
printf("\n");
}
int main(int argc, char* argv)
{
list head,head1,head2;
printf("input the first list\n");
head1 = createList();
display(head1);
printf("input the second list\n");
head2 = createList();
display(head2);
head = add(head1,head2);
display(head);
return 0;
}

C语言求多项式的和y=a1x1+a2x2+a3x3+a4x4

#include《*****》
void main()
{
float a1,a2,a3,a4,y;
printf("enter a1,a2,a3,a4\n");
scanf("%f%f%f%f",&a1,&a2,&a3,&a4);
y=a1*1+a2*2+a3*3+a4*4;
printf("%f",y);
return 0;
}

多项式相加C语言

#include 《*****》
#include 《*****》
struct SLL
{
// ax^r
int a;
int r;
struct SLL* next;
};
struct SLL *add(struct SLL *head, int a, int r)
{
struct SLL *t;
struct SLL *tt;
for(t=head, tt=NULL; t; tt = t, t = t-》next) if(t-》r 《= r) break;
if(!t)
{
struct SLL *node = (struct SLL*)malloc(sizeof(struct SLL));
node-》a = a;
node-》r = r;
node-》next = NULL;
if(!tt) return node;
tt-》next = node;
return head;
}
if(t-》r == r)
{
t-》a += a;
if(t-》a == 0)
{
if(!tt)
{
tt = head;
head = head -》 next;
****(tt);
return head;
}
tt-》next = t-》next;
****(t);
return head;
}
}
if(t-》r 《 r)
{
struct SLL *node = (struct SLL*)malloc(sizeof(struct SLL));
node-》a = a;
node-》r = r;
node-》next = t;
if(!tt) return node;
tt-》next = node;
}
return head;
}
void print(struct SLL *head)
{
struct SLL *t = head;
for(; t; t=t-》next)
{
printf("%d %d\n", t-》a, t-》r);
}
}
int main()
{
struct SLL *head = NULL;
int m, n, a, r, i;
scanf("%d%d", &m, &n);
for(i=0; i《m+n; i++)
{
scanf("%d %d", &a, &r);
head = add(head, a, r);
}
print(head);
return 0;
}

求大佬c语言编程求多项式和，题目在图片里

//以下这个函数可以计算前Num项的和，Num必须大于等于4
double CalByNum(int Num)
{
double nUp,nDown,theResult;
double nUpLast,nDownLast;
nUp = 2;
nDown = 3;
nUpLast = 1;
nDownLast = 2;
theResult = 1 + nUpLast / nDownLast + nUp / nDown;
for(int n = 4 ; n 《= Num; n++)
{
double theUp = nUp + nUpLast;
double theDown = nDown + nDownLast;
theResult += theUp / theDown;
nUpLast = nUp;
nDownLast = nDown;
nUp = theUp;
nDown = theDown;
}
return theResult;
}

C语言实验--多项式求和

main()
{
int i,j,n,m;
float s=***;
printf("input ceshishiligeshu(1《=m《100):");
scanf("%d",&m);
printf("input a int num(1《=n《1000):");
scanf("%d",&n);
for(j=1;j《m;j++)
{for(i=1;i《=n;i++)
{if(i%2!=0)
s+=s1/n;
else
s-=1/n;
}
printf("this is %d xiang shili:",j);
printf(the sum is%f\n",s);
}

C语言写多项式相加怎么写

作者：pfgmylove
***隐藏网址***
/*
   2007-3-22
   一元多项式的加法
*/

# include 《*****》
# include 《*****》
# include 《*****》
typedef struct  PolyNode
{
  int  coef;
  int  exp;
  struct  PolyNode *next;
}node;

node *CreatePoly(void)
{
  node *h,*tail,*s;
  int  coef, exp;
  h = (node *)malloc(sizeof(node));
  if (!h)
  {
    exit(-1);
  }
  h-》next = NULL;
  tail = h;
  printf("请输入每一项额系数和指数(中间以逗号隔开):/n");
  printf("coef,exp: ");
  scanf("%d,%d",&coef,&exp);
  while (coef)
  {
    s = (node *)malloc (sizeof(node));
    if (!s)
    {
      exit(-1);
    }
    s-》coef = coef;
    s-》exp = exp;

    s-》next = tail-》next;
    tail-》next = s;
    tail = s;

    printf("ceof,exp: ");
    scanf("%d,%d",&coef,&exp);
  }
  return h;
}

void PolyAdd(node *polya, node *polyb)
{
  node *p, *q, *pre, *temp;
  int  sum = 0;;
  p = polya-》next;
  q = polyb-》next;
  pre = polya;
  while (p && q)
  {
    if (p-》exp 《 q-》exp)
    {
      pre-》next = p;
      pre = p;
      p = p-》next;
    }
    else if (p-》exp 》 q-》exp)
    {
      pre-》next = q;
      pre = q;
      q = q-》next;
    }
    else
    {
      sum = p-》coef + q-》coef;
      if (sum)
      {
        p-》coef = sum;
        pre-》next = p;
        pre = p;
        p = p-》next;

        temp = q-》next;
        ****(q);
        q = temp;
      }
      else
      {
        temp = p-》next;
        ****(p);
        p = temp;
        temp = q-》next;
        ****(q);
        q = temp;
      }
    }
  }
  pre-》next = p?p:q;
}
int Prin(node *h)
{
  node *p = h-》next;
  while (p)
  {
    printf("%d*x^%d  ",p-》coef, p-》exp);
p = p-》next;
  }
  printf("/n");
  return 1;
}

int main(void)
{
  node *polya, *polyb;
  printf("请输入第一个一元多项式的系数和指数（假定以输入系数为0来结束）:/n");
  polya = CreatePoly();
  printf("请输入的第一个一元多项式为:/n");
  Prin(polya);
  printf("请输入第二个一元多项式的系数和指数（假定以输入系数为0来结束）:/n");
  polyb = CreatePoly();
  printf("请输入的第二个一元多项式为:/n");
  Prin(polyb);
  printf("这两个一元多项式相加后的结果为:/n");
  PolyAdd(polya, polyb);
  Prin(polya);
  return 1;
}

用C语言编写多项式相加的程序，下面图中前两个是要相加的多项式第三个是相加后的结果，求好心人拜托了

#include 《*****》
#include 《*****》
#include 《*****》
#define EPS 1E-6
typedef struct item {
double coefficient;
int power;
struct item *next;
} *POLYNOMIAL,*pItem;
POLYNOMIAL Create() { // 创建多项式
pItem head,p;
double coe;
int pwr;
head = p = (pItem)malloc(sizeof(item));
while(1) {
printf("系数幂次(0 0结束) : ");
scanf("%lf%d",&coe,&pwr);
if(fabs(coe) 《= EPS && !pwr) break;
p-》next = (pItem)malloc(sizeof(item));
p-》next-》coefficient = coe;
p-》next-》power = pwr;
p = p-》next;
}
p-》next = NULL;
return head;
}
void Sort(POLYNOMIAL head) { // 按幂次降排序
pItem pt,q,p = head;
while(p-》next) {
q = p-》next;
while(q-》next) {
if(p-》next-》power 《 q-》next-》power) {
pt = p-》next;
p-》next = q-》next;
q-》next = p-》next-》next;
p-》next-》next = pt;
}
else q = q-》next;
}
p = p-》next;
}
}
void Show(POLYNOMIAL head) { // 显示多项式
POLYNOMIAL p = head-》next;
int flag = 1;
if(p == NULL) return;
while(p) {
if(flag) {
if(fabs(p-》coefficient) 》= EPS) {
if(p-》power == 0) printf("%.2lf ",p-》coefficient);
else if(p-》power == 1) {
if(p-》coefficient == ***) printf("x ");
else if(p-》coefficient == -1.0) printf("-x ");
else printf("%.2lfx ",p-》coefficient);
}
else if(p-》coefficient == ***) printf("x^%d ",p-》power);
else if(p-》coefficient == -1.0) printf("-x^%d ",p-》power);
else printf("%.2lfx^%d ",p-》coefficient,p-》power);
flag = 0;
}
}
else if(p-》coefficient 》 *** && fabs(p-》coefficient) 》= EPS) {
if(p-》power == 0) printf("+ %.2lf ",p-》coefficient);
else if(p-》power == 1) {
if(p-》coefficient == ***) printf("+ x ");
else printf("+ %.2lfx ",p-》coefficient);
}
else if(p-》coefficient == ***) printf("+ x^%d ",p-》power);
else printf("+ %.2lfx^%d ",p-》coefficient,p-》power);
}
else if(p-》coefficient 《 *** && fabs(p-》coefficient) 》= EPS) {
if(p-》power == 0) printf("- %.2lf ",-p-》coefficient);
else if(p-》power == 1) {
if(p-》coefficient == -1.0) printf("- x ");
else printf("- %.2lfx ",-p-》coefficient);
}
else if(p-》coefficient == -1.0) printf("- x^%d ",p-》power);
else printf("- %.2lfx^%d ",-p-》coefficient,p-》power);
}
p = p-》next;
}
printf("\n");
}
double Power(double x,int n) {
double value = ***;
int i;
for(i = 0; i 《 n; ++i) value *= x;
return value;
}
double Value(POLYNOMIAL head,double x) { // 多项式求值
POLYNOMIAL p;
double value = ***;
for(p = head-》next; p; p = p-》next)
value += p-》coefficient * Power(x,p-》power);
return value;
}
POLYNOMIAL Copy(POLYNOMIAL A) {
POLYNOMIAL head,t,p;
head = t = (pItem)malloc(sizeof(item));
for(p = A-》next; p; p = p-》next) {
t-》next = (pItem)malloc(sizeof(item));
t-》next-》coefficient = p-》coefficient;
t-》next-》power = p-》power;
t = t-》next;
}
t-》next = NULL;
return head;
}
POLYNOMIAL Additive(POLYNOMIAL A, POLYNOMIAL B) { // 多项式相加
POLYNOMIAL head,p,q,t;
head = Copy(A);
for(p = B; p-》next; p = p-》next) {
q = head;
while(q-》next) {
if(p-》next-》power == q-》next-》power) {
q-》next-》coefficient += p-》next-》coefficient;
if(fabs(q-》next-》coefficient) 《= EPS) {
t = q-》next;
q-》next = t-》next;
****(t);
}
break;
}
q = q-》next;
}
if(q-》next == NULL) {
q-》next = (pItem)malloc(sizeof(item));
q-》next-》coefficient = p-》next-》coefficient;
q-》next-》power = p-》next-》power;
q-》next-》next = NULL;
}
}
Sort(head);
return head;
}
POLYNOMIAL Subtract(POLYNOMIAL A, POLYNOMIAL B) { // 多项式相减
POLYNOMIAL head,p,q,t;
head = Copy(A);
for(p = B; p-》next; p = p-》next) {
q = head;
while(q-》next) {
if(p-》next-》power == q-》next-》power) {
q-》next-》coefficient -= p-》next-》coefficient;
if(fabs(q-》next-》coefficient) 《= EPS) {
t = q-》next;
q-》next = t-》next;
****(t);
}
break;
}
q = q-》next;
}
if(q-》next == NULL) {
q-》next = (pItem)malloc(sizeof(item));
q-》next-》coefficient = -p-》next-》coefficient;
q-》next-》power = p-》next-》power;
q-》next-》next = NULL;
}
}
Sort(head);
return head;
}
POLYNOMIAL Multiplication(POLYNOMIAL A, POLYNOMIAL B) { // 多项式相乘
POLYNOMIAL head,t,p,q;
head = t = (pItem)malloc(sizeof(item));
for(p = A-》next; p; p = p-》next) { // 完成相乘过程
for(q = B-》next; q; q = q-》next) {
t-》next = (pItem)malloc(sizeof(item));
t-》next-》coefficient = p-》coefficient * q-》coefficient;
t-》next-》power = p-》power + q-》power;
t = t-》next;
}
}
t-》next = NULL;
Sort(head); // 排序
p = head;
while(p-》next) { // 合并同类项
q = p-》next;
while(q-》next) {
if(p-》next-》power == q-》next-》power) {
p-》next-》coefficient += q-》next-》coefficient;
t = q-》next;
q-》next = t-》next;
****(t);
}
else q = q-》next;
}
p = p-》next;
}
return head;
}
void FreeMemory(POLYNOMIAL head) {
POLYNOMIAL q,p = head;
while(p) {
q = p;
p = q-》next;
****(q);
}
}
int main() {
printf("创建多项式A:\n");
POLYNOMIAL A = Create();
Sort(A);
printf("A(x) = ");Show(A);
printf("创建多项式B:\n");
POLYNOMIAL B = Create();
Sort(B);
printf("B(x) = ");Show(B);
POLYNOMIAL C = Additive(A,B);
printf("C(x) = ");Show(C);
POLYNOMIAL D = Subtract(A,B);
printf("D(x) = ");Show(D);
POLYNOMIAL E = Multiplication(A,B);
printf("E(x) = ");Show(E);
printf("A(%.2lf) = %.4lf\n",***,Value(A,***));
printf("B(%.2lf) = %.4lf\n",***,Value(B,***));
printf("C(%.2lf) = %.4lf\n",***,Value(C,***));
printf("D(%.2lf) = %.4lf\n",***,Value(D,***));
printf("E(%.2lf) = %.4lf\n",***,Value(E,***));
FreeMemory(A);
FreeMemory(B);
FreeMemory(C);
FreeMemory(D);
FreeMemory(E);
return 0;
}

用c语言实现多项式相加

#include《*****》
#include《*****》
typedef struct node
{
float xi;
int n;
struct node *next;
}term;
void sort(term *head)
{
term *p,*q,*s;
p=(term *)malloc(sizeof(term));
q=(term *)malloc(sizeof(term));
s=(term *)malloc(sizeof(term));
p=head;

while(p-》next!=NULL)
{
s=p-》next;
while(s!=NULL)
{
if(p-》n》s-》n)
{
q-》n=p-》n;
p-》n=s-》n;
s-》n=q-》n;
q-》xi=p-》xi;
p-》xi=s-》xi;
s-》xi=q-》xi;
}
s=s-》next;
}
p=p-》next;
}
}
term *polynadd(term *head1,term *head2)
{
term *p,*q,*s,*r,*head3;
float x;
p=head1-》next;
q=head2-》next;
head3=(term *)malloc(sizeof(term));
r=head3;
while(p!=NULL&&q!=NULL)
{
if(p-》n==q-》n)
{
x=p-》xi+q-》xi;
if(x!=0)
{
s=(term *)malloc(sizeof(term));
s-》xi=x;
s-》n=p-》n;
r-》next=s;
r=s;
}
p=p-》next;
q=q-》next;
}
else if(p-》n》q-》n)
{
s=(term *)malloc(sizeof(term));
s-》n=q-》n;
s-》xi=q-》xi;
r-》next=s;
r=s;
q=q-》next;
}
else
{
s=(term *)malloc(sizeof(term));
s-》n=p-》n;
s-》xi=p-》xi;
r-》next=s;
r=s;
p=p-》next;
}
}
while(p!=NULL)
{
s=(term *)malloc(sizeof(term));
s-》n=p-》n;
s-》xi=p-》xi;
r-》next=s;
r=s;
p=p-》next;
}
while(q!=NULL)
{
s=(term *)malloc(sizeof(term));
s-》n=q-》n;
s-》xi=q-》xi;
r-》next=s;
r=s;
q=q-》next;
}
r-》next=NULL;
return head3;

}
term *createpolyn(int m)
{
term *p,*head,*q;
int i;
head=(term *)malloc(sizeof(term));
q=head;
for(i=0;i《m;i++)
{
p=(term *)malloc(sizeof(term));
printf("请输入第%d项数的系数和指数\n",i+1);
scanf("%f%d",&p-》xi,&p-》n);
q-》next=p;
q=p;
}
p-》next=NULL;
return head;
}
dayin(term *head)
{
term *p,*q;
q=head-》next;
if(q-》xi==0)
printf(" ");
if(q-》n==0&&q-》xi》0&&q-》xi!=1)
printf("%****",q-》xi);
if(q-》n==0&&q-》xi==1)
printf("1");
if(q-》n==0&&q-》xi《0)
printf("%****",q-》xi);
if(q-》n==1&&q-》xi==1)
printf("x");
if(q-》n==1&&q-》xi!=1&&q-》xi》0)
printf("%*****",q-》xi);
if(q-》n==1&&q-》xi!=1&&q-》xi《0)
printf("%*****",q-》xi);
if(q-》n!=1&&q-》n!=0&&q-》xi》0&&q-》xi!=1)
printf("%*****^%d",q-》xi,q-》n);
if(q-》n!=1&&q-》n!=0&&q-》xi《0)
printf("%*****^%d",q-》xi,q-》n);
if(q-》n!=0&&q-》n!=1&&q-》xi==1)
printf("x^%d",q-》n);
p=q-》next;
while(p!=NULL)
{
if(p-》xi==0)
printf(" ");
if(p-》n==0&&p-》xi》0&&p-》xi!=1)
printf("+%****",p-》xi);
if(p-》n==0&&p-》xi==1)
printf("+1");
if(p-》n==0&&p-》xi《0)
printf("%****",p-》xi);
if(p-》n==1&&p-》xi==1)
printf("+x");
if(p-》n==1&&p-》xi!=1&&p-》xi》0)
printf("+%*****",p-》xi);
if(p-》n==1&&p-》xi!=1&&p-》xi《0)
printf("%*****",p-》xi);
if(p-》n!=1&&p-》n!=0&&p-》xi》0&&p-》xi!=1)
printf("+%*****^%d",p-》xi,p-》n);
if(p-》n!=1&&p-》n!=0&&p-》xi《0)
printf("%*****^%d",p-》xi,p-》n);
if(p-》n!=0&&p-》n!=1&&p-》xi==1)
printf("+x^%d",p-》n);
p=p-》next;
}
printf("\n");
}
main()
{
int i,j;
term *head1,*head2,*head3;
printf("请输入第一个多项式的项数：\n");
scanf("%d",&i);
head1=createpolyn(i);
sort(head1);
printf("\n");
dayin(head1);
printf("\n");
printf("请输入第二个多项式的项数：\n");
scanf("%d",&j);
head2=createpolyn(j);
sort(head2);
printf("\n");
dayin(head2);
printf("\n\n");
head3=polynadd(head1,head2);
sort(head3);
printf("\n");
dayin(head3);
printf("\n");
}

C语言求多项式的和y=a1x1+a2x2+a3x3+a4x4

#include 《*****》
double f(double x, double a,int n) {
int i;
double sum = 0,factor = x;
for(i = 0; i 《 n; i++) {
sum += a * factor;
factor *= x;
}
return sum;
}
int main() {
double a,x;
printf("a b c d x:\n");
while(scanf("%lf%lf%lf%lf%lf",&a,&x) == 5) {
printf("%lf\n",f(x,a,4));
printf("a b c d x(q to quit):\n");
}
return 0;
}

题目:计算如样图所示的多项式求和,输入一个x就可以得到求和要求x<****.编写一个C语言程序，谢谢

AS3核心程序，
//2.计算公式：Sn=1+Σ(i=1→n)*x^i
定义变量x，键盘输入数值，并赋值给变量x，
var ds:Vector.《Number》=new《Number》=1
var i:uint=1,sum:Number=ds;//定义循环变量i和前n项和变量sum，
fanc();
function fanc(){
for(;i《3||*****(ds)》=*****;i++){
*****(ds*(0.5-i+1)*x/i);
sum+=ds;}
trace("i="+(i-1),"sum="+sum,abs(ds)》=)}}//输出多项式的项数、前n项和、最后两项之差，
这是一个交错级数，收敛极其缓慢，下面是我用大数据计算的结果，计算到 i 等于40000，才精确到小数点后530位：
末两相差的绝对值：78(78组小数，每组7位),1(正数),530,1302196（末 10 位小数，前面都是 0）
第40000项： 78(78组小数，每组7位),-1(负数),261,236164（末 10 位小数，前面都是 0）
i=40000 sum=1.( 1 是整数部分，以下是小数部分，每 7 位一组，)
4035668,8476181,9963051,9461386,4070457,0856344,2943674,1345992,
7122615,9136061,7375075,6372266,2090550,5615077,5827676,7835481,
3814115,0754727,9633933,4537566,8850233,5583102,4307483,8331096,
4027367,9081933,3346101,4856449,5261156,4548201,7489263,5929258,
2237721,5594412,7937400,2501248,4837222,9164999,8503667,1480225,
4040625,1352937,6476018,1897496,2680550,9503637,1608375,5589647,
1124717,8373119,8092354,4666333,3267575,5702612,2112539,3440904,
8470967,7586541,4907781,7580708,6709928,8182895,3952037,0473265,
2007532,1141026,6553492,8939546,1635330,1408768,7526179,9595999,
2396644,0262953,4053159,0634852,5383427,6549369
(546位小数部分，但末10位无效)
程序与上面的程序类似，不过就是引用了大数据的计算，也就是构造了一个类，