正文
Friends and Cookies(思维)
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Abood's birthday has come, and his n friends are aligned in a single line from 1 to n , waiting for their cookies, Abood has x cookies to give to his friends.
Here is an example to understand how Abood gives away the cookies. Suppose Abood has 4 friends and x cookies, then Abood will do the following:
- Give a cookie to the 1 st friend.
- Give a cookie to the 2 nd friend.
- Give a cookie to the 3 rd friend.
- Give a cookie to the 4 th friend.
- Give a cookie to the 3 rd friend.
- Give a cookie to the 2 nd friend.
- Give a cookie to the 1 st friend.
- Give a cookie to the 2 nd friend.
- And so on until all the x cookies are given away.
Your task is to find how many cookies each friend will get. Can you?
Input
The first line contains an integer T (1 ≤ T ≤ 100) specifying the number of test cases.
Each test case consists of a single line containing two integers x and n (1 ≤ x ≤ 10 18 , 1 ≤ n ≤ 1000), in which x is the number of cookies Abood has, and n is the number of his friends.
Output
For each test case, print a single line containing n space-separated integers a 1 , ..., a n , in which a i represents how many cookies the i th friend got.
Example
Input
1
5 3
Output
2 2 1 解题思路：我将蛋糕的分配方式模拟一下
－ － － － － － －
－ － － － － －
－ － － － － －
－ － － － － －
－ － － － － －
－ － －
我们可以看到的是：第一行每一个人都有，然后开始的奇数个前N-1个有，偶数个后N-1有，最后一行再判断是第奇数个还是偶数个就可以得出结果。
注意的是：没跑完第一行的情况需要特判一下,跑完第一行但是m<n也需要特判一下。
#include<cstdio>
#include<cstring>
#include<algorithm>
#define LL long long int
using namespace std;
int main()
{
int t;
LL i,j,a[],m,n,x,y;
scanf("%d",&t);
while(t--)
{
memset(a,,sizeof(a));
scanf("%lld%lld",&m,&n);
if(n==)///特判一下当n为1的时候
{
printf("%lld\n",m);
}
else if(m<n)
{
for(i=;i<m;i++)
{
a[i]=;
}
for(i=;i<n;i++)///只有前m个人有蛋糕，后面的人没有了蛋糕
{
if(i==n-)
{
printf("%lld\n",a[i]);
}
else
{
printf("%lld ",a[i]);
}
} }
else
{
for(i=;i<n;i++)///第一行的分配
{
a[i]=;
}
m=m-n;///分配完第一行之后剩下的蛋糕数量
x=m/(n-);///x表示剩下的行数
y=m%(n-);///y表示最后一行余下的蛋糕份数
for(i=;i<n-;i++)
{
a[i]=a[i]+x;///除了第一个人和最后一个人其他的人再得到x份蛋糕
}
///下面是对第一个人和最后一个人以及最后一行的蛋糕分配
if(x%==)///行数为偶数
{
a[]=a[]+x/;
a[n-]=a[n-]+x/;
for(i=n-;i>=n--y+;i--)
{
a[i]++;
}
}
else///行数为奇数
{
a[]=a[]+x/+;
a[n-]=a[n-]+x/;
for(i=;i<=y;i++)
{
a[i]=a[i]+;
}
}
}
for(i=;i<n;i++)///输出
{
if(i==n-)
{
printf("%lld\n",a[i]);
}
else
{
printf("%lld ",a[i]);
}
}
}
return ;
}
