import java.util.*;
public class Main{static int N=100010;static int a[]=new int[N];public static void main(String []args){Scanner sc=new Scanner(System.in);int n=sc.nextInt();int q=sc.nextInt();for(int i=0;i<n;i++)a[i]=sc.nextInt();while(q-->0){int x=sc.nextInt();int l=0,r=n-1;while(l<r){int mid=l+r>>1;if(a[mid]>=x)r=mid;else l=mid+1;}if(x!=a[l])System.out.println("-1 -1");else {System.out.print(l+" ");l=0;r=n-1;while(l<r){int mid=l+r+1>>1;if(a[mid]<=x)l=mid;else r=mid-1;}System.out.println(l);}}}
}

谨记：无论是差分/前缀和，下标均从1开始，防止下标越界

4.前缀和

蓝桥杯省一你一定不能错过的模板大全（第二期）

一维

题目

写法1

import java.util.*;
public class Main{static int N=100010;static int a[]=new int[N];static int s[]=new int[N];public static void main(String []args){Scanner sc=new Scanner(System.in);int n=sc.nextInt();int m=sc.nextInt();for(int i=1;i<=n;i++)s[i]=sc.nextInt();for(int i=1;i<=n;i++){s[i]+=s[i-1];}while(m-->0){int l=sc.nextInt();int r=sc.nextInt();System.out.println(s[r]-s[l-1]);}}
}

写法2

import java.util.*;
public class Main{static int N=1000010;static int a[]=new int[N];static int s[]=new int[N];public static void main(String []args){Scanner sc=new Scanner(System.in);int n=sc.nextInt();int m=sc.nextInt();for(int i=1;i<=n;i++)a[i]=sc.nextInt();for(int i=1;i<=n;i++){s[i]=s[i-1]+a[i];}while(m-->0){int l=sc.nextInt();int r=sc.nextInt();System.out.println(s[r]-s[l-1]);}}
}

二维

步骤：
(1)先输入数据
(2)预处理前缀和

s[i][j]=s[i-1][j]+s[i][j-1]-s[i-1][j-1]+a[i][j];

(3)求出区域内的数的和

s[x2][y2]-s[x2][y1-1]-s[x1-1][y2]+s[x1-1][y1-1];

子矩阵的和

import java.util.*;
public class Main{static int N=1010;static int a[][]=new int[N][N];static int s[][]=new int[N][N];public static void main(String []args){Scanner sc=new Scanner(System.in);int n=sc.nextInt();int m=sc.nextInt();int q=sc.nextInt();for(int i=1;i<=n;i++){for(int j=1;j<=m;j++){a[i][j]=sc.nextInt();}}for(int i=1;i<=n;i++){for(int j=1;j<=m;j++){s[i][j]=s[i-1][j]+s[i][j-1]-s[i-1][j-1]+a[i][j];}}while(q-->0){int x1=sc.nextInt();int y1=sc.nextInt();int x2=sc.nextInt();int y2=sc.nextInt();System.out.println(s[x2][y2]-s[x2][y1-1]-s[x1-1][y2]+s[x1-1][y1-1]);}}
}

5.差分

作用：使得区间/区域内的数加上1/C
步骤：
(1)输入数据
(2)初始化，先自己插自己
(3)处理询问即差分
差分数组：b[],前缀和数组：a[]
(4)再求一遍前缀和
(5)输出前缀和数组的a[]即可

模板题

一维

import java.util.*;
public class Main{static int N=100010;static int a[]=new int[N];static int s[]=new int[N];static int b[]=new int[N];public static void main(String []args){Scanner sc=new Scanner(System.in);int n=sc.nextInt();int m=sc.nextInt();for(int i=1;i<=n;i++)a[i]=sc.nextInt();for(int i=1;i<=n;i++){insert(i,i,a[i]);}while(m-->0){int l=sc.nextInt();int r=sc.nextInt();int c=sc.nextInt();insert(l,r,c);}for(int i=1;i<=n;i++)a[i]=a[i-1]+b[i];for(int i=1;i<=n;i++){System.out.print(a[i]+" ");}}public static void insert(int l,int r,int c){b[l]+=c;b[r+1]-=c;}
}

二维

差分矩阵

import java.util.*;
public class Main{static int N=1010;static int a[][]=new int[N][N];static int b[][]=new int[N][N];public static void main(String []args){Scanner sc=new Scanner(System.in);int n=sc.nextInt();int m=sc.nextInt();int q=sc.nextInt();for(int i=1;i<=n;i++){for(int j=1;j<=m;j++){a[i][j]=sc.nextInt();insert(i,j,i,j,a[i][j]);}}while(q-->0){int x1=sc.nextInt();int y1=sc.nextInt();int x2=sc.nextInt();int y2=sc.nextInt();int c=sc.nextInt();insert(x1,y1,x2,y2,c);}for(int i=1;i<=n;i++){for(int j=1;j<=m;j++){a[i][j]=a[i-1][j]+a[i][j-1]-a[i-1][j-1]+b[i][j];}}for(int i=1;i<=n;i++){for(int j=1;j<=m;j++){System.out.print(a[i][j]+" ");}System.out.println();}}public static void insert(int x1,int y1,int x2,int y2,int c){b[x1][y1]+=c;b[x2+1][y1]-=c;b[x1][y2+1]-=c;b[x2+1][y2+1]+=c;
}
}

6.并查集

蓝桥杯上岸必背！！！(并查集补更)

合并集合

import java.util.*;
public class Main{static int N=100010;static int p[]=new int[N];public static int find(int x){if(x!=p[x])p[x]=find(p[x]);return p[x];}public static void main(String []args){Scanner sc=new Scanner(System.in);int n=sc.nextInt();int m=sc.nextInt();for(int i=1;i<=n;i++)p[i]=i;while(m-->0){String s=sc.next();if(s.equals("M")){int a=sc.nextInt();int b=sc.nextInt();if(find(a)!=find(b)){p[find(a)]=find(b);}}else{int a=sc.nextInt();int b=sc.nextInt();if(find(a)==find(b))System.out.println("Yes");else System.out.println("No");}}}
}

7.最大公约数

蓝桥杯上岸必背！！！(第八期简单数论)

最大公约数

返回的a即是最大公约数

import java.util.*;
public class Main{public static int gcd(int a,int b){while(b!=0){int c=a%b;a=b;b=c;}return a;}public static void main(String []args){Scanner sc=new Scanner(System.in);int n=sc.nextInt();while(n-->0){int a=sc.nextInt();int b=sc.nextInt();System.out.println(gcd(a,b));}
}
}

8.筛质数

筛质数

找出是i的倍数的数都筛掉，用st[]数组维护！

.import java.util.*;
public class Main{static int N=1000010;static int primes[]=new int[N];static boolean st[]=new boolean[N];static int cnt;public static int get_primes(int x){for(int i=2;i<=x;i++){if(!st[i]){primes[cnt++]=i;for(int j=i;j<=x;j+=i){st[j]=true;}}}return cnt;}public static void main(String []args){Scanner sc=new Scanner(System.in);int n=sc.nextInt();System.out.println(get_primes(n));}
}

背包问题

0-1背包

import java.util.*;
public class Main{static int N=1010;static int v[]=new int[N];static int w[]=new int[N];static int f[][]=new int[N][N];public static void main(String []args){Scanner sc=new Scanner(System.in);int n=sc.nextInt();int m=sc.nextInt();for(int i=1;i<=n;i++){v[i]=sc.nextInt();w[i]=sc.nextInt();}for(int i=1;i<=n;i++){for(int j=0;j<=m;j++){f[i][j]=f[i-1][j];if(j>=v[i])f[i][j]=Math.max(f[i][j],f[i-1][j-v[i]]+w[i]);}}System.out.println(f[n][m]);}
}

完全背包问题

import java.util.*;
public class Main{static int N=1010;static int f[][]=new int[N][N];static int w[]=new int[N];static int v[]=new int[N];public static void main(String []args){Scanner sc=new Scanner(System.in);int n=sc.nextInt();int m=sc.nextInt();for(int i=1;i<=n;i++){v[i]=sc.nextInt();w[i]=sc.nextInt();}for(int i=1;i<=n;i++){for(int j=1;j<=m;j++){f[i][j]=f[i-1][j];if(j>=v[i])f[i][j]=Math.max(f[i][j],f[i][j-v[i]]+w[i]);}}System.out.println(f[n][m]);}
}