标签：col otto def inner end eset one ble while
This time, you are supposed to find A+B where A and B are two polynomials.
Each input file contains one test case. Each case occupies 2 lines, and each line contains the information of a polynomial:
K N?1?? a?N?1???? N?2?? a?N?2???? ... N?K?? a?N?K????
where K is the number of nonzero terms in the polynomial, N?i?? and a?N?i???? (i=1,2,?,K) are the exponents and coefficients, respectively. It is given that 1≤K≤10，0≤N?K??<?<N?2??<N?1??≤1000.
For each test case you should output the sum of A and B in one line, with the same format as the input. Notice that there must be NO extra space at the end of each line. Please be accurate to 1 decimal place.
2 1 2.4 0 3.2
2 2 1.5 1 0.5
3 2 1.5 1 2.9 0 3.2
题目很简单，说一下可能的wa点：
1. 末尾不要有空格。
2. 两项相加可能为零，此时要删除此项。
代码如下：
#include<stdio.h> #include<string.h> #include<stdlib.h> typedef struct node *ptonext; typedef ptonext ploy, list; struct node{ float n; int e; ptonext next; ptonext pro; }; list A, B; list res; ptonext A_head, A_tail; ptonext B_head, B_tail; ptonext r_head, r_tail; int cnt; list init() { ptonext T; T = (list)malloc(sizeof(list)); T->n = 0; T->e = 0; T->pro = NULL; T->next = NULL; return T; } void caculate() { ptonext a, b, r; ptonext T; a = A_head; b = B_head; r = r_head; a = a->next; b = b->next; while(a != A_tail && b != B_tail) { if(a->e == b->e){ T = (ptonext)malloc(sizeof(ptonext)); T->e = a->e; T->n = a->n + b->n; if(T->n == 0){ a = a->next; b = b->next; continue; } T->pro = r; r->next = T; r = r->next; a = a->next; b = b->next; cnt++; } else if(a->e > b->e){ T = (ptonext)malloc(sizeof(ptonext)); T->e = a->e; T->n = a->n; r->next = T; T->pro = r; r = r->next; a = a->next; cnt++; } else{ T = (ptonext)malloc(sizeof(ptonext)); T->e = b->e; T->n = b->n; r->next = T; T->pro = r; r = r->next; b = b->next; cnt++; } } while(a != A_tail){ T = (ptonext)malloc(sizeof(ptonext)); T->e = a->e; T->n = a->n; r->next = T; T->pro = r; r = r->next; a = a->next; cnt++; } while(b != B_tail){ T = (ptonext)malloc(sizeof(ptonext)); T->e = b->e; T->n = b->n; r->next = T; T->pro = r; r = r->next; b = b->next; cnt++; } r->next = r_tail; r_tail->pro = r; } int main() { int k; float n; int e; A = init(); B = init(); res = init(); A_tail = init(); B_tail = init(); r_tail = init(); A_head = A; B_head = B; r_head = res; cnt = 0; scanf("%d", &k); while(k--){ scanf("%d %f", &e, &n); ptonext T; T = (ptonext)malloc(sizeof(ptonext)); T->n = n; T->e = e; A->next = T; T->pro = A; A = A->next; } A->next = A_tail; A_tail->pro = A; scanf("%d", &k); while(k--){ scanf("%d %f", &e, &n); ptonext T; T = (ptonext)malloc(sizeof(ptonext)); T->n = n; T->e = e; B->next = T; T->pro = B; B = B->next; } B->next = B_tail; B_tail->pro = B; caculate(); printf("%d", cnt); while(res->next != r_tail){ printf(" %d %.1f", res->next->e, res->next->n); res = res->next; } printf("\n"); // system("pause"); return 0; }
1002 A+B for Polynomials （PAT (Advanced Level) Practice）
标签：col otto def inner end eset one ble while
原文地址：https://www.cnblogs.com/Ido-911/p/10732348.html
