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Code Lock

2024-08-27 14:56:42   222人阅读

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 11 12 11 2
 
Sample Output
126
 
Author
hanshuai
 
Source
2010 ACM-ICPC Multi-University Training Contest（3）——Host by WHU
 
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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*/#include<bits/stdc++.h>using namespace std;const int mod=1000000007;int n,m;int l,r;int x,y;int bin[10000005];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 1;    }else return 0;}/***********快速幂模板**************/long long power(int n,int x){    if(x==0) return 1;    long long t=power(n,x/2);    t=t*t%mod;    if(x%2==1)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=0;i<=n+2;i++){            bin[i]=i;        }        int ans=0;        for(int i=0;i<m;i++){            scanf("%d%d",&x,&y);            ans+=Union(x,y+1);        }        printf("%d\n",power(26,n-ans));    }}

 

Code Lock


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