希望高手给我画出流程分析图，谢谢谢谢谢谢了

多项式相加（应该是C语言的程序）
#include<stdio.h>
#include<stdlib.h>
#include<math.h>

typedef struct polynode
{
float coef;
int exp;
struct polynode *next;
}NODE;
 void printpoly(NODE*);
NODE *createploy(int);
NODE *polyadd(NODE*,NODE*);

int main(void)
{
NODE *poly1,*poly2,*poly3;
int keyin=0;
printf(“\n===多项式相加===\n”)
printf(“\n 1.第一个多项式 poly1(x) 共有多少项:”);
scanf(“%d”,&keyin);
printf(“\n”);
poly1=creatpoly(keyin);
prinf(“\n”);
printf(“\n 2.第二个多项式 poly2(x) 共有多少项:”);
scanf(“%d”,&keyin);
poly2=createpoly(keyin);
poly3=polyadd(poly1,poly2);
prinf(“\n 1. poly1(x)=”);
printpoly(poly1);
prinf(“\n 2 . poly2(x)=”);
printpoly(poly2);
orintf(“\n 3. poly(x)+poly2(x)”);
printf(“\n poly3(x)”);
printf(“\n’);
return o;


void printpoly(NODE *prt)
{
char opr;
while (ptr!=NULL)
{
if(ptr!=NULL)
{
opr=’+’;
}
else
{
opr=’-‘;
}
printf(“%c%4.1fx^%d,opr,fabs(pyr->coef),ptr->exp);
ptr=ptr->next;
}
}


NODE *createpoly(int n)
{
float coef=0;
int exp=0;
int i=0;
NODE *head,*tail,*ptr;
head=tail=NULL;
do{
i++;
ptr=malloc(sizeof(NODE));
printf(“   输入第（%d）项的系数与指数（以空白隔开）：”,i);
scanf(“%f  %d”,&coef,&exp);
ptr->coef=coef;
ptr->exp=exp;
pyr->next=NULL;
if(head==NULL)
head=ptr;
else
tail->next=ptr;
tail=ptr;
}while(n>i);
return head;
}



NODE *polyadd(NODE  *poly1,NODE  *poly2)
{
NODE *tail,*head,*ptrl,*ptr2,*ptr3;
ptr1=poly1;
ptr2=poly2;
head=tail=NULL;
while(ptr1!=NULL && ptr2!=NULL)
{
ptr3=malloc(sizeof(NODE));
if(ptr1->exp>ptr2->coef;)
{
ptr3->coef=ptr1->coef;
ptr3->exp=ptr1->exp;
ptr1=ptr1->next;
}
else if(ptr1->exp<ptr2->exp)
{
ptr3->coef=ptr2->coef;
ptr3->exp=ptr2->exp;
ptr2=ptr2->next;
}
else
{
ptr3->coef=ptr1->coef+ptr2->coef;
ptr3->exp=ptr1->exp;
ptr1=ptr1->next;
ptr2=ptr2->next;
}
ptr3->next=NULL;
if(head==NULL)
head==ptr3;
else
tail->next=ptr3;
tail=ptr3;
}
if(ptr1==NULL)
{
while(ptr2!=NULL)
{
ptr3=malloc(sizeof(NODE));
ptr3->coef=ptr2->coef;
ptr3->exp=ptr2->exp;
ptr1=ptr2->next;
tail->next=ptr3;
tail=ptr3;
}
}
else
{
while(ptr1!=NULL)
{
ptr3=malloc(sizeof(NODE));
ptr3->coef=ptr1->coef;
ptr3->exp=ptr1->exp;
ptr1=ptr1->next;
tail->next=ptr3;
tail=ptr3;
}
}
return head;
}