「算法刷题宝典」必须知道的知识点和技巧

enum 枚举名{ 枚举值表 };

enum weekday{ sun,mou,tue,wed,thu,fri,sat };

#include <stdio.h>int main(void){  int i,t,sum=0;  for(i=1;;i++)  {    for(t=i;t!=0;t/=10)      if(t%10==1)        sum++;    if(sum>=2021)      goto end;  }  end:printf("%d",i);  return 0;}

#include <iostream>using namespace std;
int main(){    long int i=0;    long int sum = 2021,tmp=0;    long int b=0;    for(i=1; i<20210; i++) {        //记住需要一个数把i的值给记下来，因为后面的相除那一个数是发生变化         tmp=i;        while(tmp) {            b = tmp%10;            if(b==1 && sum>0) {                sum-=1;            }            tmp/=10;        }        if (sum==0) {            break;        }    }    cout<<i;    return 0;}

import java.util.Scanner;// 1:无需package// 2: 类名必须Main, 不可修改
public class Main {    public static void main(String[] args) {        int count=0;        for (int i = 1; i < 20210; i++) {            String str=i+"";            for (int j = 0; j < str.length(); j++) {                if(str.charAt(j)=='1')                {                    count++;                }            }            if(count==2021)            {                System.out.println(str);                break;            }        }                scan.close();    }}

num=0  #用num用来累计用过的数字“1”的次数for i in range(1,10000):       num+=str(i).count("1")  #出现过几次1就 给num加上多少       if num>2021:  #当num数量超过2021时卡牌数量不足              breakprint(i-1)

//分析：是否合法（年月日）-->不合法，跳过//                          -->合法--是否是闰年-->分解日期（装入数组）-->判断是否满足回文-->设置变量只打印下一个 

#include<stdio.h>int mon[12]={31,28,31,30,31,30,31,31,30,31,30,31};int arr[8]={0};int year,month,day,j,flag=0,temp; int legal()//判断是否合法{    if(year%400==0)      mon[1]=29;    else      mon[1]=28;    if(year<1000||year>9999||month<=0||month>12||day>mon[month])      return 0;    return 1;}int main(){  int time;  scanf("%d",&time);  for(int i=time+1;i<=99991231;i++)  {    year=i/10000;    month=i/100%100;    day=i%100;    if(!legal())      continue;    j=0;temp=i;    while(temp)    {        arr[j++]=temp%10;        temp/=10;    }    if(arr[0]==arr[7]&&arr[1]==arr[6]&&arr[2]==arr[5]&&arr[3]==arr[4])    {        if(flag==0)        {            for(int t=0;t<8;t++)              printf("%d",arr[t]);            printf("n");            flag=1;        }      if(arr[0]==arr[2]&&arr[1]==arr[3])        {          for(int t=0;t<8;t++)              printf("%d",arr[t]);          break;      }     }   }  return 0;}

#include <iostream>using namespace std;int main(){  // 请在此输入您的代码      int i,n,a,b,c,d,e,f,g,h;    int j=0;    scanf("%d",&n);    for(i=n+1;i<=89991231;i++)    {        a = i%10;        b = (i/10)%10;        c = (i/100)%10;        d = (i/1000)%10;        e = (i/10000)%10;        f = (i/100000)%10;        g = (i/1000000)%10;        h = (i/10000000)%10;        if((a==h)&&(b==g)&&(c==f)&&(d==e))        {            if(j==0)            {                printf("%dn",i);                j++;            }
        }        if((a==h)&&(b==g)&&(c==f)&&(d==e)&&(a==c)&&(b==d))        {            printf("%d",i);            break;        }    }  return 0;}

import java.util.Scanner;// 1:无需package// 2: 类名必须Main, 不可修改
public class Main {   public static void main(String[] args) {    Scanner sc = new Scanner(System.in);    int data = sc.nextInt();    huiwenriqi(data);    ABhuiwenriqi(data);    sc.close();}public static void huiwenriqi(int n) {    for (int i = n+1; i <= 89991231; i++) {        String str = i+"";        char[] ch = str.toCharArray();        char a = ch[0], b = ch[1], c = ch[2], d = ch[3],e = ch[4],f = ch[5],g = ch[6],h = ch[7];        if(a==h&&b==g&&c==f&&d==e) {            System.out.println(ch);            break;        }}}public static void ABhuiwenriqi(int n) {    for (int i = n+1; i <= 89991231; i++) {        String str = i+"";        char[] ch = str.toCharArray();        char a = ch[0], b = ch[1], c = ch[2], d = ch[3],e = ch[4],f = ch[5],g = ch[6],h = ch[7];        if(a==c&&c==f&&f==h&&b==d&&d==e&&e==g) {            System.out.println(ch);            break;        }}    }}

import datetime #导入日期库
idate=input()y=int(idate[:4])           #取出输入的年月日m=int(idate[4:6])d=int(idate[6:])dd=datetime.date(y,m,d)   #将输入的表示日期的字符串转换成日期flag=True                 #回文日期只输出一次
for n in range(9999999):  dd=dd+datetime.timedelta(days=1)     #日期不断增加1天  sd=str(dd).replace('-','')           #将日期中的-去掉    if sd[:]==sd[::-1]:                  #判断日期是否是回文    if flag:      print(int(sd))                   #输出回文日期      flag=False                       #下次不输出回文日期    if sd[0]==sd[2]==sd[5]==sd[7] and sd[1]==sd[3]==sd[4]==sd[6]: #判断是否是ABABBABA类型      print(int(sd))      break

#include <stdio.h>#include <stdlib.h>int a[102]={0},b[102]={0},N,i,max=0,j,sum=0,k,h,m,max1=0,l;void dat(){  for(i=0,sum=0,j=1;j<=max;)  {     if(a[i]<0)    {      i=0;    }     if(a[i]==j)    {      sum=sum+j;      j=1;            for(k=i;a[k+1]!=0;k++)      {        a[k]=a[k+1];      }      continue;    }    i++;j++;  }
  if(max1<=sum)    max1=sum;}int main(int argc, char *argv[]){
  scanf("%d",&N);  for(i=0;i<N;i++)  {    scanf("%d",&b[i]);    max=max>=b[i]?max:b[i];  }  b[i]=-1;  for(h=0;h<N;h++)  {    m=b[0];    for(i=0;i<N-1;i++)    {      b[i]=b[i+1];    }    b[N-1]=m;    for(l=0;l<N+1;l++)        a[l]=b[l];       dat();  }
  printf("%dn",max1);  return 0;}

#include <iostream>#include <algorithm> #include <string>#include <cstring>#include <queue>#include <utility>#include <cmath>#include <unordered_map>#include <unordered_set>#include <vector>#include <deque> 
#define io ios::sync_with_stdio(0), cin.tie(0), cout.tie(0)#define rep(i, a, b) for(int i = a; i <= b; i ++)#define x first#define y second
using namespace std;
const int N = 110,  M = 400010;typedef long long LL;typedef pair<LL, LL> PLL;typedef pair<int, int> PII;
int n, m;unordered_set<int> s;int a[N];
int main(){    io;    int maxn = 0, res = 0;    cin >> n;    rep(i, 0, n - 1)    cin >> a[i], res += a[i];    rep(i, 0, n - 1)    {        s.clear();        int cnt = 1;        int j = i;        int sum = 0;         int k = 0;        while(j != -1)        {            k ++;            if(k > 50000) break;            j = j % n;            //cout << a[j] << " " << cnt;            if(s.find(a[j]) != s.end())            {                //cout << " ?" <<endl;                 j ++;            //    if(cnt == n) break;                continue;            }            else            {                //cout << " !";                if(cnt == a[j])                {                    //cout << "?n";                    s.insert(cnt);          sum += cnt;                    cnt = 1;                                    }                else                 {                    cnt ++;                //    cout << "!n";                }                j ++;            }            if(cnt > n) break;            if(sum == res)            {                cout << sum << endl;                return 0;            }        }        maxn = max(maxn, sum);    }    cout << maxn;    return 0;    }

import java.util.Scanner;// 1:无需package// 2: 类名必须Main, 不可修改
public class Main {   static int num;    static int msize=0;    public static void main(String[] args) {        Scanner sc=new Scanner(System.in);        num=sc.nextInt();        int[] arr=new int[num];        for (int i = 0; i < arr.length; i++) {            arr[i]=sc.nextInt();        }        for (int i = 1; i <= num; i++) {            msize+=i;        }        for (int i = 0; i < arr.length; i++) {            barr=new boolean[num];            P(arr,i,1,0);            if(maxsize==msize)break;        }        System.out.println(maxsize);    }    static int maxsize=0;    static boolean[] barr;    public static void P(int[] arr,int k,int i,int size) {        if(i>num||size==msize) {            if(size>maxsize)maxsize=size;            return;        }        if(barr[k]) {            P(arr,k+1<arr.length?k+1:0,i,size);            return;        }        if(arr[k]==i) {            barr[k]=true;            P(arr,k+1<arr.length?k+1:0,1,size+arr[k]);        }else {            P(arr,k+1<arr.length?k+1:0,i+1,size);        }    }}

import osimport sys
# 请在此输入您的代码def f(n,nums):    ans = 0    for m in range(n):    #控制次数    n个数有n种顺序读数，每一个数可以打头阵        tmp = nums[m:] + nums[:m]    #重新拼接数组           j = 0   #数数，起点        while j < max(tmp):    #数的数比最大值还大，就跳出去，无法全部卡片收取            for i,v in enumerate(tmp,1+j):                if v == i:                    j = 0                    if len(tmp)==1:    #只有一个，直接return出来                        return cnt                       elif v == tmp[-1]:   #只剩一个，不好切片，所以用[:-1]                        tmp = tmp[:-1]                    else:                        r = tmp.index(v)    #i已经不能用，已经再转圈计数了，所以用拼接的数组的索引                        tmp = tmp[r+1:] + tmp[:r]                                          break            else:                j = i  #转圈的i索引记录下来        ans = max(ans, cnt - sum(tmp))    return ans
n = int(input())nums = [int(i) for i in input().split()]cnt = sum(nums)print(f(n,nums))

#include <stdio.h>#include <stdlib.h>
#define  N 500005long long arr[N];//原数组long long  diff[N];//差分数组
int main(int argc, char *argv[]){    // 请在此输入您的代码    int n, q,i=0;    scanf("%d %d",&n,&q);    while(++i<=n){        scanf("%lld",&arr[i]);        diff[i]=arr[i]-arr[i-1];///构造差分数组    }
    while(q--){        long long  l,r,x;        scanf("%lld %lld %lld",&l,&r,&x);        diff[l]+=x;///对差分数组的两个端点进行修改        if(r+1>=N)          continue;        diff[r+1]-=x;    }
    ///区间修改已完成，开始还原数组    for(i=1;i<=n;i++){        diff [i]=diff[i]+diff[i-1];//还原修改后的数组        if(diff [i]>=0)            printf("%lld ",diff [i]);        else            printf("0 ");    }    return 0;}

#include <iostream>const int maxn=500005;long long N[maxn]={0};long long b[maxn]={0};using namespace std;int main(){  int P,Q;  cin>>P>>Q;  for(int i=1;i<=P;i++){    cin>>N[i];    b[i]=N[i]-N[i-1];//定义差分和数组  }  while(Q>0){    long long l,r,x;    cin>>l>>r>>x;    b[l]+=x;    b[r+1]-=x;//对两端进行操作    Q--;  }  for(int i=1;i<=P;i++){    N[i]=b[i]+N[i-1];//最后回归原来数组     if(N[i]<0){      cout<<0<<" ";    }else    cout<<N[i]<<" ";  }  return 0;}

import java.io.BufferedReader;import java.io.IOException;import java.io.InputStreamReader;import java.io.StreamTokenizer;
public class Main {
    public static void main(String[] args) throws IOException {        // TODO Auto-generated method stub        StreamTokenizer sc=new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));        sc.nextToken();        int n=(int)sc.nval;        sc.nextToken();        int q=(int)sc.nval;        long a[]=new long[n+1];        long b[]=new long[n+1];        for(int i=1;i<=n;i++){            sc.nextToken();            a[i]=(long)sc.nval;            b[i]=a[i]-a[i-1];        }        for(int i=0;i<q;i++) {            sc.nextToken();            int l=(int)sc.nval;            sc.nextToken();            int r=(int)sc.nval;            sc.nextToken();            long c=(long)sc.nval;            b[l]+=c;            if(r<n)b[r+1]-=c;        }        for(int j=1;j<=n;j++)a[j]=a[j-1]+b[j];        for(int i=1;i<=n;i++){            if(a[i]<0)a[i]=0;            System.out.print(a[i]+" ");        }    }}

import osimport sys
N, Q = map(int, input().split())a = list(map(int, input().split()))op = [0 for _ in range(N+1)]for _ in range(Q):  l, r, x = map(int, input().split())  op[l-1] += x  op[r] -= x
for i in range(1, N):  op[i]+= op[i-1]print(' '.join(str(max(a[i] + op[i], 0)) for i in range(N)))