HDU 3461

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

 

大致题意：

　　一个密码锁上有编号为1到N的N个字母，每个字母可以取26个小写英文字母中的一个。

　　再给你M个区间[L,M]，表示该区间的字母可以一起同步“增加”（从'a'变为'b'为增1，'z'增1为'a'）。

　　假如一组密码按照给定的区间进行有限次的“增加”操作后可以变成另一组密码，那么我们认为这两组密码是相同的。

　　该题的目标就是在给定N、M和M个区间的前提下计算有多少种不同的密码。

解题思路：

　　根据题意，如果一个可调整的区间都没有的话，答案应该是26的N次方。

　　每当加入一个区间的时候，答案就减少为之前的26分之1（因为该区间的加入使得原本不同的26种情况变得等价了）。

　　因此当有x个“不同的”区间加入进来之后，答案应该为26^(N-x)。

　　最后在求26^(N-x)时需要用到两分快速幂；（某大神的代码一览）；

#include <cstdio>
#include <iostream>
using namespace std;
const int mod=1000000007;
int f[10000005];
int sf(int x){return x==f[x]? x: f[x]=sf(f[x]);}
int Union(int x,int y){
    x=sf(x),y=sf(y);
    if(x==y) return 0;
    return f[x]=y,1;
}
long long power(int x){
    if(x==0) return 1;
    long long t=power(x/2);
    t=t*t%mod;
    if(x%2==1)t=t*26%mod;
    return t;
}
int main(){
    int n,m;
    while(~scanf("%d%d",&n,&m)){
        for(int i=1;i<=n+2;i++) f[i]=i;
        int a,b,ans=0;
        for(int i=1;i<=m;i++){
            scanf("%d%d",&a,&b);
            ans+=Union(a,b+1);
        } printf("%lld
",power(n-ans));
    }
}