Codeforces 459c pashmak and buses wit math

There are n people, K buses, M days, and N people are arranged to take the bus every day (there can be empty buses ), in order to make these N people not friends, as long as each person takes at least one day on the m-day bus, it will be OK.

I want you to determine whether there is such an arrangement. If so, I will output a specific arrangement scheme for everyone to take that car every day.

I did not think of any good questions at all .. Really, although it was so simple to scream after knowing it, at first I was really confused. Now it is clear that every student's vehicle in the M-day will form a series composed of M numbers (1-K stands for the bus number ), according to the question requirements, only the sequence of students is unique.

Then, we only need to convert the meaning of the question into a sequence composed of n Different M numbers (as long as there is a number that is different), then we will put this sequence, the K-base number of a m-bit is not enough, starting from 111111 every time ..

Start enumeration, always enumeration to kkkkkk... It's over... Ah, it's really witty.

Of course, at the beginning, you can directly take a ride to determine whether the M k-in-number is greater than or equal to N. If the M k-in-number is not enough, you can use a high-precision addition method to obtain the sequence and then output it.

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#define LL __int64using namespace std;LL n,k,d;int ans[1010][1010];int main(){    while (scanf("%I64d%I64d%I64d",&n,&k,&d)!=EOF)    {        LL tmp=k;        for (LL i=2;i<=d;i++){            tmp=tmp*k;            if (tmp>=n) break;        }        if (tmp<n){            puts("-1");            continue;        }        for (int i=d;i>=1;i--) ans[0][i]=1;        ans[0][d]=0;        for (int i=1;i<=n;i++){            int c=0;            ans[i][d]=ans[i-1][d]+1;            if (ans[i][d]>k){ans[i][d]=1;c++;}            for (int j=d-1;j>=1;j--){                ans[i][j]=ans[i-1][j]+c;                if (ans[i][j]>k){                    c=1;                    ans[i][j]=1;                }                else c=0;            }        }        for (int i=1;i<=d;i++){            printf("%d",ans[1][i]);            for (int j=2;j<=n;j++){                printf(" %d",ans[j][i]);            }            puts("");        }    }    return 0;}

 

Codeforces 459c pashmak and buses wit math