HDU 4038 Stone The 36th ACM/ICPC Asia Regional Chengdu Site —— Online Contest

 

Stone

Time Limit: 3000/2000 MS (Java/Others) Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 46 Accepted Submission(s): 9

Problem Description

Given an array of integers {xi}. Each time you can apply one of the following operations to the array:
1. Choose an integer x from the array, replace it with x+1.
2. Add a new integer 1 to the array.

Define p as the product of all integers in the set. i.e. p=x1*x2*x3*...
What's the maximum possible value of p after exactly M operations?

Input

First line is a integer T (T ≤ 100), the number of test cases.
The first line of each test case contains two integers N and M, the number of integers in the initial set, and the number of operations.
The second line is N integers xi initially in the set.
1 ≤ N ≤ 100000
0 ≤ M ≤ 10^18
-10000 ≤ xi ≤ 10000

Output

For each case, you should output “Case k: ” first, where k indicates the case number and counts from one. Then the maximum product mod 1000000007.

Sample Input
4
1 1
5
3 2
1 2 3
3 2
-1 2 3
3 1
-3 -3 -3

Sample Output

Case 1: 6
Case 2: 18
Case 3: 6
Case 4: -18

Source

The 36th ACM/ICPC Asia Regional Chengdu Site —— Online Contest

Recommend

lcy

 

儘量的把數分成3。

偶數個負數不用處理，奇數個負數先把最大的一個加成0，所有0先加成正數在處理。

原來不足3的先平均的加到3。剩下的分成一個個3，如果剩下1把它加到一個3裏，剩下2單獨算一個數。

代碼：

 

#include<cstdio>   
#include<algorithm>   
#define M 1000000007   
using namespace std;  
  
typedef __int64 ll;  
ll mod(ll a,ll n,ll m)  
{  
    ll s=1;  
    while(n)  
    {  
        if(n&1)  
            s=(s%m*a%m)%m;  
        a=(a%m)*(a%m)%m;  
        n>>=1;  
    }  
    return s;  
}  
int t,tt,n,m,a,b[100005],c[100005];  
int main()  
{  
    scanf("%d",&t);  
    while(t--)  
    {  
        int i,x,n1=0,n2=0;  
        ll s=1;  
        scanf("%d%d",&n,&m);  
        for(i=0;i<n;i++)  
        {  
            scanf("%d",&a);  
            if(a<0)//負數
                c[n2++]=a;  
            else  
                b[n1++]=a;  
        }  
        if(n2&1)//奇數個負數
        {  
            sort(c,c+n2);  
            b[n1++]=c[n2-1];
        }  
        if(n1)  
        {  
            sort(b,b+n1);
            if(b[0]<0)
            {
                x=min(m,-b[0]);
                m-=x;
                b[0]+=x;
            }
            for(i=0;m&&i<n1;i++)  
                if(!b[i])  
                {  
                    b[i]++;  
                    m--;  
                }  
            i=0;  
            while(m)  
            {  
                int f=1;  
                for(i=0;i<n1&&m;i++)  
                    if(b[i]<3)  
                    {  
                        f=0;  
                        b[i]++;  
                        m--;  
                    }
                if(f)  
                {  
                    if(m==1)  
                    {  
                        b[0]++;  
                        m--;  
                    }  
                    break;  
                }  
            }  
        }  
        if(m>1)  
        {  
            if(m%3==0)  
                s=mod(3,m/3,M);  
            else if(m%3==2)  
                s=mod(3,m/3,M)*2;  
            else  
                s=mod(3,m/3-1,M)*4;  
        }  
        for(i=0;i<n1;i++)
            s=(s*b[i])%M;
        for(i=0;i<n2-(n2&1);i++)
            s=(s*c[i])%M;  
        printf("Case %d: %I64d\n",++tt,s%M);  
    }  
}