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https://blog.csdn.net/JiangHxin/article/details/104032351

 

import java.util.Deque;
import java.util.HashMap;
import java.util.LinkedList;
import java.util.Map;
import java.util.Queue;
import java.util.Scanner;
import java.util.concurrent.LinkedBlockingDeque;

public class Main {
    static  int a[] =new int[2005];
    static  int mod;
    
    public static int gets(String s) {//把字符转化为十进制数 
        int ans=0;
        int n=s.length();
        for(int i=0;i<n;i++) {
            ans=ans*26+s.charAt(i)-'A';
            ans%=mod;//防止数据溢出 
        }
        return ans;
    }
    
    public static void main(String[] args) {
        Scanner cin=new Scanner(System.in);
        
        String s=cin.next();
        mod=cin.nextInt();
        int n=s.length();
       // System.out.println(n);
       
        int ans=gets(s);
        
        if(ans==0) {
            System.out.println("0 0");
            return ;
        }
        
        a[n-1]=1;
        
        for(int i=n-2;i>=0;i--) {//每一位的权值 
            a[i]=26*a[i+1]%mod;
            //System.out.println(a[i]);
        }
        for(int i=0;i<n;i++)
            for(int j=i+1;j<n;j++)
            {
                int res=(s.charAt(i)-s.charAt(j))*a[j]+(s.charAt(j)-s.charAt(i))*a[i];//变化值 
                
                if( (res+ans+mod)%mod==0) {
                    System.out.println(++i +" "+ ++j);
                    return ;
                }
            }
        System.out.println("-1 -1");
    }
}

 

#include<cstdio>
#include<cstring>
#include<string>
#include<cmath>
using namespace std;

char s[2005];
int a[2005];
int mod;

int ge(char s[])
{
    int ans=0;
    int len=strlen(s);
    for(int i=0;i<len;i++)
    {
        ans=ans*26+s[i]-'A';
        ans%=mod;
    }
    return ans;
}

int main()
{
    scanf("%s",s);
    scanf("%d",&mod);
    int ans=ge(s);
    
    if(ans==0)
    {
        printf("0 0\n");
        return 0;
    }

        int n=strlen(s);
        a[n-1]=1;
        for(int i=n-2;i>=0;i--)
        {
            a[i]=a[i+1]*26%mod;
        }
        for(int i=0;i<n;i++)
        {
            for(int j=i+1;j<n;j++)
            {
                int res=(s[i]-s[j])*a[j]+(s[j]-s[i])*a[i];
                
                if((res+ans+mod)%mod==0)
                {
                //    printf("!!!!!!!\n");
                    printf("%d %d\n",++i,++j);
                    return 0;
                }
            }
        }
        
    
    printf("-1 -1\n");
    return 0;
}