HDU/HDOJ 4038 2011成都赛区网络赛H题
Stone
Time Limit: 3000/2000 MS (Java/Others) Memory Limit: 65768/65768 K (Java/Others)Total Submission(s): 130 Accepted Submission(s): 26
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?
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
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
这个题的思路很简单，就是尽量多的凑够3
然后在这个大的思路前提下，进行各种讨论
非常考察选手的模拟能力。
这个题的情况非常多。我是把数据分为了正数和负数来处理的。
中间要考虑负数的个数是奇数还是偶数
然后分开处理
细节就不多说啦，看代码吧
我的代码：
#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<stdlib.h>
using namespace std;
typedef __int64 ll;
const ll mod=1000000007;
ll x[100005];
ll po[100005];
ll ne[100005];
ll num1,num2;
bool cmp(ll a,ll b)
{
return a>b;
}
ll power(ll p,ll n,ll mod)
{
ll sq=1;
while(n>0)
{
if(n%2==1)
sq=(sq%mod)*(p%mod)%mod;
p=(p%mod)*(p%mod)%mod;
n=n/2;
}
return sq%mod;
}
int main()
{
ll i;
ll n,m,t,T,ans;
scanf("%I64d",&T);
for(t=1;t<=T;t++)
{
num1=0,num2=0;
ans=1;
scanf("%I64d%I64d",&n,&m);
for(i=1;i<=n;i++)
{
scanf("%I64d",&x[i]);
if(x[i]>=0)
{
num1++;
po[num1]=x[i];
}
else
{
num2++;
ne[num2]=x[i];
}
}
sort(po+1,po+1+num1,cmp);
sort(ne+1,ne+1+num2,cmp);
if(num2&1)
{
if(m<abs(ne[1]))
{
ne[1]=ne[1]+m;
for(i=1;i<=num1;i++)
ans=(ans*po[i])%mod;
for(i=2;i<=num2;i++)
ans=(ans*abs(ne[i]))%mod;
ans=(ans*ne[1])%mod;
printf("Case %I64d: %I64d\n",t,ans);
}
else if(m==abs(ne[1]))
{
printf("Case %I64d: 0\n",t);
}
else
{
m=m-abs(ne[1])-1;
ne[1]=1;
num1++;
po[num1]=ne[1];
sort(po+1,po+1+num1,cmp);
for(i=1;i<=num1;i++)
{
if(po[i]==0)
{
m=m-1;
po[i]=1;
}
if(m==0)
break;
}
if(m==0)
{
for(i=1;i<=num1;i++)
ans=(ans*po[i])%mod;
for(i=2;i<=num2;i++)
ans=(ans*abs(ne[i]))%mod;
// ans=(ans*ne[1])%mod;
printf("Case %I64d: %I64d\n",t,ans);
}
sort(po+1,po+1+num1,cmp);
if(m>0)
{
bool flag=true;
while(1)
{
flag=true;
for(i=num1;i>=1;i--)
{
if(po[i]<3)
{
flag=false;
po[i]=po[i]+1;
m=m-1;
}
if(m==0)
break;
}
if(flag)
break;
if(m==0)
break;
}
if(m==0)
{
for(i=1;i<=num1;i++)
ans=(ans*po[i])%mod;
for(i=2;i<=num2;i++)
ans=(ans*abs(ne[i]))%mod;
// ans=(ans*ne[1])%mod;
printf("Case %I64d: %I64d\n",t,ans);
}
if(m>0)
{
ll tmp=m%3,uk;
if(m==1)
{
// ne[1]=ne[1]+1;
po[num1]=po[num1]+1;
for(i=1;i<=num1;i++)
ans=(ans*po[i])%mod;
for(i=2;i<=num2;i++)
ans=(ans*abs(ne[i]))%mod;
// ans=(ans*ne[1])%mod;
printf("Case %I64d: %I64d\n",t,ans);
}
else
{
if(tmp==1)
{
for(i=1;i<=num1;i++)
ans=(ans*po[i])%mod;
for(i=2;i<=num2;i++)
ans=(ans*abs(ne[i]))%mod;
// ans=(ans*ne[1])%mod;
m=m-4;
uk=power(3,m/3,mod);
uk=(uk*4)%mod;
ans=(ans*uk)%mod;
printf("Case %I64d: %I64d\n",t,ans);
}
else if(tmp==2)
{
for(i=1;i<=num1;i++)
ans=(ans*po[i])%mod;
for(i=2;i<=num2;i++)
ans=(ans*abs(ne[i]))%mod;
// ans=(ans*ne[1])%mod;
m=m-2;
uk=power(3,m/3,mod);
uk=(uk*2)%mod;
ans=(ans*uk)%mod;
printf("Case %I64d: %I64d\n",t,ans);
}
else
{
for(i=1;i<=num1;i++)
ans=(ans*po[i])%mod;
for(i=2;i<=num2;i++)
ans=(ans*abs(ne[i]))%mod;
// ans=(ans*ne[1])%mod;
uk=power(3,m/3,mod);
ans=(ans*uk)%mod;
printf("Case %I64d: %I64d\n",t,ans);
}
}
}
}
}
}
else
{
for(i=1;i<=num1;i++)
{
if(po[i]==0)
{
m=m-1;
po[i]=1;
}
if(m==0)
break;
}
if(m==0)
{
for(i=1;i<=num1;i++)
ans=(ans*po[i])%mod;
for(i=1;i<=num2;i++)
ans=(ans*abs(ne[i]))%mod;
// if(num2>=1)
// ans=(ans*ne[1])%mod;
printf("Case %I64d: %I64d\n",t,ans);
}
sort(po+1,po+1+num1,cmp);
if(m>0)
{
bool flag=true;
while(1)
{
flag=true;
for(i=num1;i>=1;i--)
{
if(po[i]<3)
{
flag=false;
po[i]=po[i]+1;
m=m-1;
}
if(m==0)
break;
}
if(flag)
break;
if(m==0)
break;
}
if(m==0)
{
for(i=1;i<=num1;i++)
ans=(ans*po[i])%mod;
for(i=1;i<=num2;i++)
ans=(ans*abs(ne[i]))%mod;
// if(num2>=1)
// ans=(ans*abs(ne[1])%mod;
printf("Case %I64d: %I64d\n",t,ans);
}
if(m>0)
{
ll tmp=m%3,uk;
if(m==1)
{
po[num1]=po[num1]+1;
for(i=1;i<=num1;i++)
ans=(ans*po[i])%mod;
for(i=1;i<=num2;i++)
ans=(ans*abs(ne[i]))%mod;
// if(num2>=1)
// ans=(ans*ne[1])%mod;
printf("Case %I64d: %I64d\n",t,ans);
}
else
{
if(tmp==1)
{
for(i=1;i<=num1;i++)
ans=(ans*po[i])%mod;
for(i=1;i<=num2;i++)
ans=(ans*abs(ne[i]))%mod;
// if(num2>=1)
// ans=(ans*ne[1])%mod;
m=m-4;
uk=power(3,m/3,mod);
uk=(uk*4)%mod;
ans=(ans*uk)%mod;
printf("Case %I64d: %I64d\n",t,ans);
}
else if(tmp==2)
{
for(i=1;i<=num1;i++)
ans=(ans*po[i])%mod;
for(i=1;i<=num2;i++)
ans=(ans*abs(ne[i]))%mod;
// if(num2>=1)
// ans=(ans*ne[1])%mod;
m=m-2;
uk=power(3,m/3,mod);
uk=(uk*2)%mod;
ans=(ans*uk)%mod;
printf("Case %I64d: %I64d\n",t,ans);
}
else
{
for(i=1;i<=num1;i++)
ans=(ans*po[i])%mod;
for(i=1;i<=num2;i++)
ans=(ans*abs(ne[i]))%mod;
// if(num2>=1)
// ans=(ans*ne[1])%mod;
uk=power(3,m/3,mod);
ans=(ans*uk)%mod;
printf("Case %I64d: %I64d\n",t,ans);
}
}
}
}
}
}
return 0;
}