Code Lock

Code Lock

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)

Total Submission(s): 255 Accepted Submission(s): 114

Problem Description

A lock you use has a code system to be opened instead of a key. The lock contains a sequence of wheels. Each wheel has the 26 letters of the English alphabet 'a' through 'z', in order. If you move a wheel up, the letter it shows changes to the next letter in the English alphabet (if it was showing the last letter 'z', then it changes to 'a').
At each operation, you are only allowed to move some specific subsequence of contiguous wheels up. This has the same effect of moving each of the wheels up within the subsequence.
If a lock can change to another after a sequence of operations, we regard them as same lock. Find out how many different locks exist?

 

Input

There are several test cases in the input.

Each test case begin with two integers N (1<=N<=10000000) and M (0<=M<=1000) indicating the length of the code system and the number of legal operations.
Then M lines follows. Each line contains two integer L and R (1<=L<=R<=N), means an interval [L, R], each time you can choose one interval, move all of the wheels in this interval up.

The input terminates by end of file marker.

 

Output

For each test case, output the answer mod 1000000007

 

Sample Input

1 1
1 1
2 1
1 2

 

Sample Output

1
26

 

Author

hanshuai

 

Source

2010 ACM-ICPC Multi-University Training Contest（3）——Host by WHU

 

Recommend

zhouzeyong

/*
没有可操作区间的时候是26的n次方，然后多一个可操作区间，就会少26种
每一个不可操作的区间的种类数是26种，就是26^k种答案
*/
#include<bits/stdc++.h>
using namespace std;
const int mod=;
int n,m;
int l,r;
int x,y;
int bin[];
int findx(int x){
    int temp=x;
    while(x!=bin[x]){
        x=bin[x];
    }
    bin[temp]=x;
    return x;
}
int Union(int x,int y){
    int fx=findx(x);
    int fy=findx(y);
    if(fx!=fy){
        bin[fy]=fx;
        return ;
    }else return ;
}
/***********快速幂模板**************/
long long power(int n,int x){
    if(x==) return ;
    long long t=power(n,x/);
    t=t*t%mod;
    if(x%==)t=t*n%mod;
    return t;
}
/***********快速幂模板**************/
int main(){
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF){
        for(int i=;i<=n+;i++){
            bin[i]=i;
        }
        int ans=;
        for(int i=;i<m;i++){
            scanf("%d%d",&x,&y);
            ans+=Union(x,y+);
        }
        printf("%d\n",power(,n-ans));
    }
}