HDU-5685 Problem A (inverse elements, prefixes and preprocessing)

Link to the topic: Problem - 5685 (hdu.edu.cn)

topic:

Sample topic:

 

 Topic ideas:

Preprocess the hash values ​​from 1 to all lengths first, and store them in the multiplied prefix and array (the timeout will occur without preprocessing). When querying, take sum[b]/sum[a-1], which is the hash value from a to b. When dividing, inverse element processing is required.

AC：

#include<bits/stdc++.h>
using namespace std;
#define ll long long 
const int N=1e5+5;
int n,a,b,x,y;
int sum[N],ans[N];
string s;
int ex_gcd(int a,int b,int &x,int &y){
    if(b==0){
        x=1;
        y=0;
        return a;
    }
    int d=ex_gcd(b,a%b,x,y); 
    int tmp=x;
    x=y;
    y=tmp-(a/b)*y;
    return d;
}
int main(){
    while(cin>>n){
        sum[0]=1;
        ans[0]=1;
        cin>>s;
        int len=s.size();
        for(int i=0;i<len;i++){
            int nn=s[i];
            sum[i+1]=sum[i]*(nn-28)%9973;
        }
        while(n--){
            cin>>a>>b;
            ex_gcd(sum[a-1],9973,x,y);
            x=(x%9973+9973)%9973;
            cout<<(sum[b]*x)%9973<<endl;
        } 
    }
    
    
    return 0;
}