Robbers

Robbers

Time Limit: 2000/1000MS
(Java/Others) Memory
Limit: 128000/64000KB
(Java/Others)
Special Judge
Submit Status

Problem Description

N robbers
have robbed the bank. As the result of their crime they chanced to
get M golden coins. Before the robbery the band has made an
agreement that after the robbery i-th gangster would get
Xi/Y
of all money gained. However, it turned out that M may be not
divisible by Y.
The problem
which now should be solved by robbers is what to do with the coins.
They would like to share them fairly. Let us suppose that i-th
robber would get Ki coins.
In this case unfairness of this fact is |Xi/Y
- Ki/M|.
The total unfairness is the sum of all particular unfairnesses.
Your task as the leader of the gang is to spread money among
robbers in such a way that the total unfairness is minimized.

Input

The first
line of the input file contains numbers N, M and Y (1 ≤ N ≤ 1000, 1
≤ M, Y ≤ 10000). N integer numbers follow - Xi (1
≤ Xi ≤
10000, sum of all Xi is
Y).

Output

Output N
integer numbers - Ki (sum
of all Ki must
be M), so that the total unfairness is minimal.

Sample Input

3 10 4
1 1 2

Sample Output

2 3 5

强盗分钱，要求是不公平度最小，那么先按原来分配取整，对每个强盗最优解肯定是a[i]*m/y或是加一，然后再把他们加一后不公平度增加进行排序，选最小的分别加一，把剩下的硬币分配完。自己写的时候比较2比的先进行排序了，wa了好久。还是写的c语言，用sort会简单许多。

#include

#include

#include

struct robber{

int id;

int a;

double dis;

}dp[1010];

int ans[1010];

int main()

{

int m, n, i, j, y, sum1, sum2, leap, pre,
temp;

double temp1;

while (scanf("%d %d %d", &n, &m, &y)
!= EOF){

sum1 = 0;

for (i = 1; i <= n;
i++)

{

scanf("%d",
&dp[i].a);

dp[i].id =
i;

}



//没有必要排序

for (i = 1; i <= n;
i++)

{

ans[i] =
m*dp[i].a / y;

sum1 +=
ans[i];

}

for (i = 1; i <= n;
i++)

{

dp[i].dis =
-fabs(dp[i].a*1.0 / y - ans[dp[i].id] * 1.0 / m) + fabs(dp[i].a*1.0
/ y - (ans[dp[i].id] + 1) *1.0 / m);

}

for (i = 1; i <=
n;i++)

for (j = 1; j <= n -
i;j++)

if (dp[j].dis > dp[j +
1].dis)

{

temp =
dp[j].a; dp[j].a = dp[j + 1].a; dp[j + 1].a = temp;

temp =
dp[j].id; dp[j].id = dp[j + 1].id; dp[j + 1].id = temp;

temp1 =
dp[j].dis; dp[j].dis = dp[j + 1].dis; dp[j + 1].dis = temp1;

}

for (i = 1; i <= m - sum1;
i++)

ans[dp[i].id]++;

for (i = 1; i <= n;
i++)

printf("%d ",
ans[i]);

printf("\n");

}

return 0;

}