【贪心】【TOJ4107】【A simple problem】

Given three integers  n(1≤n≤1018) ,  m(1≤m≤105) ,  k(1≤k≤1018) . you should find a list of integer  A1,A2,…,Am  which satisfies three conditions:
1.  A1+A2+⋯+Am=n .
2.  1≤Ai≤k−1  for each  (1≤i≤m) .
3.  GCD(A1,A2,…,Am)=1 .GCD means the greatest common divisor
4. if  i<j  then  Ai≥Aj .
As the author is too lazy to write a special judge, if there's no answer ouput "I love ACM", And if there's more than one answer, output the one has the minimum  A1 , and if there still multiple answer make the A2  as small as possible, then  A3,A4…

m=1  直接 “I love ACM”

m=2

均摊 第二个给第一个不断给1

m=3 

均摊 最后一个给第一个1个1

很多边界数据注意下

3 3 1

1 1 1

1 1 4

等等..

代码如下：

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <ctime>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <string>
#define oo 0x13131313
using namespace std;
long long n,m,k;
long long gcd(long long a,long long b)
{
    long long r;
    while(b>0)
    {
         r=a%b;
         a=b;
         b=r;
    }
    return a;
}
long long A[100000+5];
void do1()
{
    if(n%2==1)
    {
        long long p=n/2+1;
        if(p<=k-1) printf("%lld %lld\n",p,p-1);
        else printf("I love ACM\n");
    }
    else
    {
        int ok=1;
        long long a=n/2,b=n/2;
        while(a+1<=k-1&&b-1>=1)
        {
            a++;
            b--;
            if(gcd(a,b)==1)
            {
                ok=0;
                printf("%lld %lld\n",a,b);
                break;
            }
        }
        if(ok)
        printf("I love ACM\n");
    }
}
void do2()
{
    memset(A,0,sizeof(A));
    for(int i=1;i<=m;i++)
    {
        A[i]=n/m;
    }
    for(int i=1;i<=n%m;i++)
    {
        A[i]++;
    }
    if(n%m==0&&n!=m)
    {
        A[1]++;A[m]--;
    }
    if(A[1]<=k-1&&A[m]>=1)
    {
     for(int i=1;i<=m;i++)
    {
        printf("%lld",A[i]);
        if(i!=m) printf(" ");
    }
    printf("\n");
    }
    else printf("I love ACM\n");
}
int main()
{
    while(cin>>n>>m>>k)
    {
        if(m==1&&n!=1||k==1) printf("I love ACM\n");
        else if(m==1&&n==1) printf("1\n");
        else
        {
                if(m==2) do1();
                else if(m>=3) do2();
        }
    }
}