02-线性结构2：一元多项式的乘法与加法运算
Description:
设计函数分别求两个一元多项式的乘积与和。
Input:
输入分2行，每行分别先给出多项式非零项的个数，再以指数递降方式输入一个多项式非零项系数和指数（绝对值均为不超过1000的整数）。数字间以空格分隔。
Output:
输出分2行，分别以指数递降方式输出乘积多项式以及和多项式非零项的系数和指数。数字间以空格分隔，但结尾不能有多余空格。零多项式应输出0 0。
SampleInput:
4 3 4 -5 2 6 1 -2 0
3 5 20 -7 4 3 1
SampleOutput:
15 24 -25 22 30 21 -10 20 -21 8 35 6 -33 5 14 4 -15 3 18 2 -6 1
5 20 -4 4 -5 2 9 1 -2 0
Codes:
//#define LOCAL
#include <cstdio>
#define M 2010
int A[M], B[M], C[M], D[M];
void add() {
int i, j, f = 0;
for(i=0; i<M; ++i)
C[i] = A[i]+B[i];
for(i=M; i>=0; --i) {
if(C[i] != 0) {
if(!f) f = 1;
else printf(" ");
printf("%d %d", C[i], i);
}
}
if(f == 0) printf("0 0");
printf("
");
}
void mul() {
int i, j, f = 0;
for(i=0; i<M; ++i)
for(j=0; j<M; ++j)
D[i+j] += A[i]*B[j];
for(i=M; i>=0; --i) {
if(D[i] != 0) {
if(!f) f = 1;
else printf(" ");
printf("%d %d", D[i], i);
}
}
if(f == 0) printf("0 0");
printf("
");
}
int main()
{
#ifdef LOCAL
freopen("E:\Temp\input.txt", "r", stdin);
freopen("E:\Temp\output.txt", "w", stdout);
#endif
int n1, n2, i, p, q;
scanf("%d", &n1);
for(i=0; i<n1; ++i) {
scanf("%d%d", &q, &p);
A[p] = q;
}
scanf("%d", &n2);
for(i=0; i<n2; ++i) {
scanf("%d%d", &q, &p);
B[p] = q;
}
mul(); add();
return 0;
}
PAT-1074:Reversing Linked List.
Description:
Given a constant K and a singly linked list L, you are supposed to reverse the links of every K elements on L. For example, given L being 1→2→3→4→5→6, if K = 3, then you must output 3→2→1→6→5→4; if K = 4, you must output 4→3→2→1→5→6.
Input:
Each input file contains one test case. For each case, the first line contains the address of the first node, a positive N (<= 105) which is the total number of nodes, and a positive K (<=N) which is the length of the sublist to be reversed. The address of a node is a 5-digit nonnegative integer, and NULL is represented by -1.
Then N lines follow, each describes a node in the format:
Address Data Next
where Address is the position of the node, Data is an integer, and Next is the position of the next node.
Output:
For each case, output the resulting ordered linked list. Each node occupies a line, and is printed in the same format as in the input.
SampleInput:
00100 6 4
00000 4 99999
00100 1 12309
68237 6 -1
33218 3 00000
99999 5 68237
12309 2 33218
SampleOutput:
00000 4 33218
33218 3 12309
12309 2 00100
00100 1 99999
99999 5 68237
68237 6 -1
Codes:
//#define LOCAL
#include <cstdio>
#include <algorithm>
#define M 100010
struct Node { int d, p; };
Node A[M]; int B[M];
int main()
{
#ifdef LOCAL
freopen("E:\Temp\input.txt", "r", stdin);
freopen("E:\Temp\output.txt", "w", stdout);
#endif
int a, b, c, f, n, i, k, j = 0;
scanf("%d%d%d", &f, &n, &k);
for(i=0; i<n; ++i) {
scanf("%d%d%d", &a, &b, &c);
A[a].d = b, A[a].p = c;
}
while(f != -1) { B[j++] = f; f = A[f].p; } i = 0;
while(i+k <= j) { std::reverse(&B[i], &B[i+k]); i+=k; }
for(i=0; i<j-1; ++i) printf("%05d %d %05d
", B[i], A[B[i]].d, B[i+1]);
printf("%05d %d -1
", B[i], A[B[i]].d);
return 0;
}
PAT-1051:Pop Sequence.
Description:
Given a stack which can keep M numbers at most. Push N numbers in the order of 1, 2, 3, ..., N and pop randomly. You are supposed to tell if a given sequence of numbers is a possible pop sequence of the stack. For example, if M is 5 and N is 7, we can obtain 1, 2, 3, 4, 5, 6, 7 from the stack, but not 3, 2, 1, 7, 5, 6, 4.
Input:
Each input file contains one test case. For each case, the first line contains 3 numbers (all no more than 1000): M (the maximum capacity of the stack), N (the length of push sequence), and K (the number of pop sequences to be checked). Then K lines follow, each contains a pop sequence of N numbers. All the numbers in a line are separated by a space.
Output:
For each pop sequence, print in one line "YES" if it is indeed a possible pop sequence of the stack, or "NO" if not.
SampleInput:
5 7 5
1 2 3 4 5 6 7
3 2 1 7 5 6 4
7 6 5 4 3 2 1
5 6 4 3 7 2 1
1 7 6 5 4 3 2
SampleOutput:
YES
NO
NO
YES
NO
Codes:
//#define LOCAL
#include <cstdio>
#define M 1010
int A[M], B[M];
int main()
{
#ifdef LOCAL
freopen("E:\Temp\input.txt", "r", stdin);
freopen("E:\Temp\output.txt", "w", stdout);
#endif
int m, n, k, i, j, t;
scanf("%d%d%d", &m, &n, &k);
while(k--) {
for(i=0; i<n; ++i) scanf("%d", &A[i]);
j = 1, i = t = 0;
while(i < n) {
if(A[i] == j) ++i, ++j;
else if(B[t] == A[i]) ++i, --t;
else if(t < m-1) B[++t] = j++;
else break;
}
if(i == n) printf("YES
");
else printf("NO
");
}
return 0;
}