OJ 1018 counting game

describe

 

n A circle of individuals （ The number is 1 - n）, From 1 Individual starts counting , To report for duty k The people of , The people in the back start from 1 Start counting . Ask the number of the last remaining person .

for example ：n = 3,k = 2.2 No. first out , And then there was 1 Number , The last thing left is 3 Number .

Input

 

Input is a single set of test data .

Input 2 Number n and k, Express n personal , Count to k List out .(2 <= n, k <= 200)

Output

 

Output an integer to represent the number of the last person left .

sample input 1 

10 3

sample output 1

4

The topic is a problem of yasuf ring , The title explains In the first n No. is about to delete a data , And then start again from the beginning , Cycle all the time , Until the end , That is to say, they are deleted step by step , If there is only one good analysis, the first one is the serial number 1, If it's two, it's in 1,2 Count off and know that a person has been kicked out , Such as n=4, So that is 1,2,1,2, It's counting 1,2,3,4 Last 2 Kicked out, so the serial number is 1 , Three is the same , Three numbers, that is 1,2,3,1//2,3,2,3 Last , With recursion, you will find a rule , Hypothetical use f It means to find the serial number , Yes k personal , newspaper n Then kick out , In the example n=4, that f(1)=0,f(2)=1,f(3)=2……-->f(2)=(f(1)+n)%2=0+1=1,f(3)=(f(2)+n)%3=1+1=2……f(k)=(f(k-1)+n)%k. That's the formula ,f(k)=(f(k-1)+n)%k, If k=1 Then it must be 0, So we can use 1 Special to build functions ,1 Pushable 2,2 Pushable 3, By analogy, we get the final answer .

 

#include <iostream>

using namespace std;
int n,k;
int f()
{
    int s=0;
    for(int i=2;i<=k;i++)
        s=(s+n)%i;
    return s+1;
}
int main()
{
    while(cin>>k>>n)
    {
        cout<<f()<<endl;

    }

    return 0;
}