TC609 DIV1 （500）

Problem Statement

We have balls of K different colors. The colors are numbered 0 through K-1, and the number of balls of color i is X[i]. We want to divide the balls into as few packages as possible. Each package must contain between 1 and K balls, inclusive. Additionally, each package must be either a "normal set" (all balls in the package have the same color), or a "variety set" (no two balls have the same color).
You are given the int K. You are also given ints A, B, C, and D. Use these to generate the array X using the following pseudocode:

X[0] = A
for i = 1 to K-1:
    X[i] = (X[i-1] * B + C) % D + 1

Compute and return the smallest possible number of packages.

Definition

Class:	PackingBallsDiv1
Method:	minPacks
Parameters:	int, int, int, int, int
Returns:	int
Method signature:	int minPacks(int K, int A, int B, int C, int D)
(be sure your method is public)

Limits

Time limit (s):	2.000
Memory limit (MB):	256

Notes

-

You can assume that the answer will fit into a signed 32-bit integer.

Constraints

-

K will be between 1 and 100,000, inclusive.

-

B, C and D will be between 1 and 1,000,000,000, inclusive.

-

A will be between 1 and D, inclusive.

Examples

0)

3

4

2

5

6

Returns: 4

There are three colors of balls. Using the pseudocode in the problem statement, we can compute that X[0]=4, X[1]=2, and X[2]=4. As there are 10 balls and we can only put at most 3 into each package, we need at least 4 packages. One possible solution with 4 packages is {0,1,2}, {0,0}, {0,1}, and {2,2,2}. (That is, the first package contains one ball of each color, the second package contains two balls of color 0, and so on.)

1)

1

58

23

39

93

Returns: 58

All the balls have the same color, and each package can only contain one ball. Thus, the number of packages is the same as the number of balls.

2)

23

10988

5573

4384

100007

Returns: 47743

3)

100000

123456789

234567890

345678901

1000000000

Returns: 331988732

Watch out for integer overflow when generating X.

 

郁闷了 记得昨天就是这思路写的 硬是没过去样例 今天打了一遍直接A了

尽量的让每个背包装满 先把除得进的加上 再选一个最优的方案放余数  方案两种 要么全相同要么全不同 选一个最优的

#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<stdlib.h>
#include<cmath>
using namespace std;
#define LL long long
#define N 100010
#define INF 1e9
LL x[N];
int f[N];
class PackingBallsDiv1
{
    public :
    int minPacks(int K, int A, int B, int C, int D)
    {
        int i;
        x[0] = A;
        for(i = 1; i < K ; i++)
        x[i] = (x[i-1]*B+C)%D+1;
        LL s=0;
        for(i = 0; i < K ; i++)
        {
            s+=x[i]/K;
            x[i] = x[i]%K;f[x[i]]++;
        }
        LL ans = INF;
        sort(x,x+K);
        for(i = 0; i < K ; i++)
        {
            ans = min(ans,x[i]+s+K-i-1);
        }
        return ans;
    }
};