Hackrank Equal DP

Christy is interning at HackerRank. One day she has to distribute some chocolates to her colleagues. She is biased towards her friends and may have distributed the chocolates unequally. One of the program managers gets to know this and orders Christy to make sure everyone gets equal number of chocolates.

But to make things difficult for the intern, she is ordered to equalize the number of chocolates for every colleague in the following manner,

For every operation, she can choose one of her colleagues and can do one of the three things.

1. She can give one chocolate to every colleague other than chosen one.
2. She can give two chocolates to every colleague other than chosen one.
3. She can give five chocolates to every colleague other than chosen one.

Calculate minimum number of such operations needed to ensure that every colleague has the same number of chocolates. 

Input Format

First line contains an integer  denoting the number of testcases.  testcases follow. 
Each testcase has  lines. First line of each testcase contains an integer  denoting the number of colleagues. Second line contains N space separated integers denoting the current number of chocolates each colleague has.

Constraints

Number of initial chocolates each colleague has < 

Output Format

 lines, each containing the minimum number of operations needed to make sure all colleagues have the same number of chocolates.

Sample Input

1
4
2 2 3 7

Sample Output

2

Explanation

1st operation: Christy increases all elements by 1 except 3rd one 
2 2 3 7 -> 3 3 3 8
2nd operation: Christy increases all element by 5 except last one
3 3 3 8 -> 8 8 8 8

 

 

题意：

　　给你n个数

　　你每次操作可以选n-1个位置，每个位置+（1，2，5）

　　问你最少多少次操作使得n个数一致

题解：

　　选n-1个位置+上(1,2,5)

　　也就是选一个位置-(1,2,5)

　　按照这个找到最佳的一致的那个数就行了

#include<bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define ls i<<1
#define rs ls | 1
#define mid ((ll+rr)>>1)
#define pii pair<int,int>
#define MP make_pair
typedef long long LL;
const long long INF = 1e18+1LL;
const double Pi = acos(-1.0);
const int N = 6e3+10, M = 1e3+20, mod = 2017,inf = 2e9;

int dp[N][N],a[10005];
int main() {
    for(int i = 0;i <= 6000; ++i)
        for(int j = 0; j <= 6000; ++j) dp[i][j] = inf;

    for(int i = 0; i <= 6000; ++i) {
        dp[i][i] = 0;
        for(int j = 0; j < i; ++j) {
            if(i-1>=j)
                dp[i][j] = min(dp[i-1][j]+1,dp[i][j]);
            if(i-5>=j)
                dp[i][j] = min(dp[i-5][j]+1,dp[i][j]);
            if(i-2>=j)
                dp[i][j] = min(dp[i-2][j]+1,dp[i][j]);
        }
    }
    int T,n;
    scanf("%d",&T);
    while(T--) {
        scanf("%d",&n);
        int mi = inf;
        for(int i = 1; i <= n; ++i) scanf("%d",&a[i]),mi = min(mi,a[i]+1000);
        int ans = 0;
        for(int i = 1; i <= n; ++i) {
            ans += dp[a[i]+1000][mi];
        }
        for(int i = 0; i <= mi; ++i) {
            int sum = 0 ;
            for(int j = 1; j <= n; ++j) {
                sum += dp[a[j]+1000][i];
            }
            ans = min(ans,sum);
        }
        printf("%d\n",ans);
    }
    return 0;
}