题目:
问题 : NEW RDSP MODE I
题目描述
Little A has became fascinated with the game Dota recently, but he is not a good player. In all the modes, the rdsp Mode is popular on online, in this mode, little A always loses games if he gets strange heroes, because, the heroes are distributed randomly.
Little A wants to win the game, so he cracks the code of the rdsp mode with his talent on programming. The following description is about the rdsp mode:
There are N heroes in the game, and they all have a unique number between 1 and N. At the beginning of game, all heroes will be sorted by the number in ascending order. So, all heroes form a sequence One.
These heroes will be operated by the following stages M times:
1.Get out the heroes in odd position of sequence One to form a new sequence Two;
2.Let the remaining heroes in even position to form a new sequence Three;
3.Add the sequence Two to the back of sequence Three to form a new sequence One.
After M times' operation, the X heroes in the front of new sequence One will be chosen to be Little A's heroes. The problem for you is to tell little A the numbers of his heroes.
输入
There are several test cases.
Each case contains three integers N (1<=N<1,000,000), M (1<=M<100,000,000), X(1<=X<=20).
Proceed to the end of file.
输出
For each test case, output X integers indicate the number of heroes. There is a space between two numbers. The output of one test case occupied exactly one line.
样例输入
5 1 2
5 2 2
样例输出
2 4
4 3
提示
In case two: N=5,M=2,X=2,the initial sequence One is 1,2,3,4,5.After the first operation, the sequence One is 2,4,1,3,5. After the second operation, the sequence One is 4,3,2,1,5.So,output 4 3.
题目描述:
可以发现,无论是奇数还是偶数,都是 X=2*X 之后对N取模,那么经过M次变换可以看成2的M次幂,这里写一个快速幂就行了。
还可以找出循环节t,然后t%m求余,在计算剩下的次数就可以了。
这里用的是快速幂写的。
代码:
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
long long m,n;
long long p(long long x,long long y,long long z)
{
long long t = 1;
while(y){
if(y&1) t = (t*x)%z;
x = (x*x)%z;
y>>=1;
}
return t;
}
int main()
{
int i,j,k,l;
while(scanf("%lld%lld%d",&m,&n,&k)!=EOF){
if(m%2==0) m++;
// n %=m;
long long an = p(2,n,m);
long long ans = an;
printf("%lld",ans);
for(i = 2;i <= k;i++){
ans+=an;
ans%=m;
printf(" %lld",ans);
}
printf("\n");
}
return 0;
}