本篇文章适用于考研和复试上机的同学，建议去ACWING买考研算法辅导课同步刷题，上岸的c++选手贼多，我都拿的是Java写的，有些拿c++写的，简单做法和最优做法基本都有，仅供参考！！！
import java.util.*;
class Main{
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int k = sc.nextInt();
int[][] arr = new int[n][2];
String[] s = new String[n];
for(int i = 0; i < n; i++){
s[i] = sc.next();
arr[i][0] = i;
arr[i][1]= sc.nextInt();
}
if(k == 1){
Arrays.sort(arr,(o1,o2)->{
return o1[1]-o2[1];
});
}else{
Arrays.sort(arr,(o1,o2)->{
return o2[1]-o1[1];
});
}
for(int[] x: arr){
System.out.print(s[x[0]]+" ");
System.out.println(x[1]);
}
}
}
import java.util.*;
class Main{
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int[][] arr = new int[n][2];
for(int i = 0; i < n; i++){
arr[i][0] = sc.nextInt();
arr[i][1] = sc.nextInt();
}
Arrays.sort(arr,(o1,o2)->{
if(o1[1] != o2[1]){
return o1[1] - o2[1];
}
return o1[0] - o2[0];
});
for(int[] x: arr){
System.out.print(x[0]+" ");
System.out.println(x[1]);
}
}
}
package Acwing;
import java.util.Deque;
import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner;
public class AC3373_2 {
static Deque<Integer> div(Queue<Integer> queue, int b){
Deque<Integer> tmp = new LinkedList<>();
int cnt = queue.size();
int r = 0;
while(cnt-- != 0){
r = r * 10 + queue.poll();
tmp.offer(r/b);
r %=b;
}
while(tmp.size() != 0 && tmp.peek()==0){
tmp.poll();
}
return tmp;
}
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
while(sc.hasNext()){
String x = sc.next();
Deque<Integer> queue = new LinkedList<>();
for(char c: x.toCharArray()){
queue.offer(c-'0');
}
StringBuilder res = new StringBuilder();
if("0".equals(x)){
res.append("0");
}else{
while(!queue.isEmpty()){
System.out.println(queue.peekLast());
res.append(queue.peekLast()%2);
queue = div(queue,2);
}
}
System.out.println(res.reverse());
}
}
}
import java.util.*;
import java.math.BigInteger;
class Main{
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
while(sc.hasNext()){
BigInteger b = sc.nextBigInteger();
System.out.println(b.toString(2));
}
}
}
import java.util.*;
import java.math.BigInteger;
class Main{
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
while(sc.hasNext()){
int m = sc.nextInt();
int n = sc.nextInt();
BigInteger b = sc.nextBigInteger(m);
System.out.println(b.toString(n));
}
}
}
import java.util.*;
class Main{
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
int a = sc.nextInt();
int b = sc.nextInt();
String s = sc.next();
Deque<Integer> queue = new LinkedList<>();
for(char c: s.toCharArray()){
int x = 0;
if(c>='A'){
x = c-'A'+10;
}else{
x = c-'0';
}
queue.offer(x);
}
StringBuilder sb = new StringBuilder();
if("0".equals(s)){
sb.append("0");
}else{
while(!queue.isEmpty()){
Deque<Integer> tmp = new LinkedList<>();
int cnt = queue.size();
int r = 0;
for(int i = 0; i < cnt; i++){
int t = queue.poll();
//计算t的十进制
t += r * a;
//记录余数
r= t%b;
//记录除数
t = t/b;
tmp.offer(t);
}
while(!tmp.isEmpty() && tmp.peek()== 0){
tmp.poll();
}
queue=tmp;
if(r <10){
sb.append(r);
}else{
sb.append((char)(r-10+'a'));
}
}
System.out.println(sb.reverse());
}
}
}
这个题有两种做法：
链表相加：
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
class Solution {
public ListNode findFirstCommonNode(ListNode headA, ListNode headB) {
ListNode h1 = headA;
ListNode h2 = headB;
while(h1!=h2){
h1 = h1!=null? h1.next:headB;
h2 = h2!=null? h2.next:headA;
}
return h1;
}
}
链表相减找后半段：
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
class Solution {
public ListNode findFirstCommonNode(ListNode headA, ListNode headB) {
ListNode h1 = headA;
ListNode h2 = headB;
int len1 = 0;
int len2 = 0;
while(h1 != null){
len1++;
h1 = h1.next;
}
while(h2 != null){
len2++;
h2 = h2.next;
}
if(len1 <= len2){
for(int i = 0; i < len2 - len1; i++){
headB = headB.next;
}
}else{
for(int i = 0; i < len1 - len2; i++){
headA = headA.next;
}
}
while(headA!=null&&headB!=null){
if(headA.val==headB.val){
return headA;
}
headA = headA.next;
headB = headB.next;
}
return null;
}
}
作者：福尔摩DONG
链接：https://www.acwing.com/activity/content/code/content/1414690/
来源：AcWing
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* filterList(ListNode* head) {
bool st[10001] = {};
ListNode *dummy = new ListNode(-1);
ListNode *cur = dummy, *p = new ListNode(-1);
while(head){
int val = abs(head->val);
if(!st[val]){
cur->next = head;
p = head;
cur = cur->next;
st[val] = true;
}
head = head->next;
}
p->next = NULL;
return dummy->next;
}
};
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode filterList(ListNode head) {
HashSet<Integer> set = new HashSet<>();
ListNode dummy = new ListNode(-1);
ListNode cur = dummy,h1 = new ListNode(-1);
while(head!=null){
int x = Math.abs(head.val);
if(!set.contains(x)){
cur.next = head;
h1 = head;
cur = cur.next;
set.add(x);
}
head = head.next;
}
h1.next = null;
return dummy.next;
}
}
ListNode在 = 赋值的时候，传递的是地址。前半段要记得终止，就要是的a->next = null，而在遍历的时候使用的是指针。地址是全局的。相当于链表的最后一个肯定要指向空
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public void rearrangedList(ListNode head) {
//计算链表长度
ListNode p = head;
int len = 0;
while(p!=null){
len++;
p = p.next;
}
int left = (len+1)/2;
ListNode a = head;
for(int i = 0; i < left-1; i++){
a = a.next;
}
ListNode b = a.next;
a.next = null;
ListNode pre = null;
while(b!=null){
ListNode c = b.next;
b.next = pre;
pre = b;
b = c;
}
while(pre!=null){
ListNode ta = head.next, tb = pre.next;
head.next = pre;
pre.next = ta;
head = ta;
pre = tb;
}
}
}
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
void rearrangedList(ListNode* head) {
ListNode *h = head;
int len = 0;
while(h){
len++;
h = h->next;
}
int left = (len+1)/2;
ListNode *l = head;
for(int i = 0; i < left-1; i++){
l = l->next;
}
ListNode *r = l->next;
l->next = NULL;
ListNode *pre = NULL;
while(r){
ListNode *ne = r ->next;
r->next = pre;
pre = r;
r = ne;
}
while(pre){
ListNode *l1 = head->next;
ListNode *r1 = pre->next;
head ->next =pre;
pre->next = l1;
head = l1;
pre = r1;
}
}
};
一年中的天数为
1,3,5,7,8,10,12为31天，2月的话闰年为29天，平年为28天。
闰年的话分为普通闰年和世纪闰年，4年一闰，百年不闰或400年一闰。
这个比较恶心的是yyyy-mm-dd，
可以用prinf试试，卧槽，printf可以补前导0，绝了，以前不知道，一直特判，人傻了。。。
import java.util.*;
class Main{
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
int[] arr1 = {31,28,31,30,31,30,31,31,30,31,30,31};
int[] arr2 = {31,29,31,30,31,30,31,31,30,31,30,31};
while(sc.hasNext()){
int x = sc.nextInt();
int y = sc.nextInt();
if((x%4==0&&x%100!=0)||(x%400==0)){
for(int i = 1; i <= 12; i++){
y = y - arr2[i-1];
if(y<=0){
System.out.printf("%04d-%02d-%02d\n",x,i,(y+arr2[i-1]));
break;
}
}
}else{
for(int i = 1; i <= 12; i++){
y = y - arr1[i-1];
if(y<=0){
System.out.printf("%04d-%02d-%02d\n",x,i,(y+arr1[i-1]));
break;
}
}
}
}
}
}
import java.util.*;
class Main{
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
int[] arr1 = {31,28,31,30,31,30,31,31,30,31,30,31};
int[] arr2 = {31,29,31,30,31,30,31,31,30,31,30,31};
while(sc.hasNext()){
int x = sc.nextInt();
int y = sc.nextInt();
String xx = String.valueOf(x);
if(xx.length()<4){
for(int i = 0; i < 4-xx.length();i++){
xx = "0"+xx;
}
}
if((x%4==0&&x%100!=0)||(x%400==0)){
for(int i = 1; i <= 12; i++){
y = y - arr2[i-1];
if(y<=0){
System.out.println(xx+"-"+(i<10?"0":"")+i+"-"+(arr2[i-1]+y<10?"0":"")+(arr2[i-1]+y));
break;
}
}
}else{
for(int i = 1; i <= 12; i++){
y = y - arr1[i-1];
if(y<=0){
System.out.println(xx+"-"+(i<10?"0":"")+i+"-"+(arr1[i-1]+y<10?"0":"")+(arr1[i-1]+y));
break;
}
}
}
}
}
}
import java.util.*;
class Main{
//日期问题三板斧
static int[] month = {0,31,28,31,30,31,30,31,31,30,31,30,31};
static int is_leap(int year){
if((year %4 == 0&& year %100 !=0)||(year % 400 == 0)){
return 1;
}
return 0;
}
static int get_month(int y ,int m){
if(m == 2){
//获得月份天数
return month[m]+is_leap(y);
}
return month[m];
}
public static void main(String[]args){
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
while(n-- != 0){
int y = sc.nextInt();
int m = sc.nextInt();
int d = sc.nextInt();
int a = sc.nextInt();
int sum = 0;
//算总数从一年的开始开始算
for(int i = 1; i < m; i++){
sum +=get_month(y,i);
}
sum+=d+a;
//然后按年算
while(sum > 365+is_leap(y)){
sum -= 365+is_leap(y);
y++;
}
m = 1;
//按月算
while(sum > get_month(y,m)){
sum -=get_month(y,m);
m++;
}
//最后剩下的就是天数
System.out.printf("%04d-%02d-%02d\n",y,m,sum);
}
}
}