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): 130Accepted 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?

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

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; }