Break the Chocolate

Time Limit: 20 Sec  Memory Limit: 256 MB

题目连接

http://acm.hdu.edu.cn/showproblem.php?pid=4112

Description

Benjamin is going to host a party for his big promotion coming up.
Every
party needs candies, chocolates and beer, and of course Benjamin has
prepared some of those. But as everyone likes to party, many more people
showed up than he expected. The good news is that candies are enough.
And for the beer, he only needs to buy some extra cups. The only problem
is the chocolate.
As Benjamin is only a ‘small court officer‘ with
poor salary even after his promotion, he can not afford to buy extra
chocolate. So he decides to break the chocolate cubes into smaller
pieces so that everyone can have some.
He have two methods to break
the chocolate. He can pick one piece of chocolate and break it into two
pieces with bare hand, or put some pieces of chocolate together on the
table and cut them with a knife at one time. You can assume that the
knife is long enough to cut as many pieces of chocolate as he want.
The
party is coming really soon and breaking the chocolate is not an easy
job. He wants to know what is the minimum number of steps to break the
chocolate into unit-size pieces (cubes of size 1 × 1 × 1). He is not
sure whether he can find a knife or not, so he wants to know the answer
for both situations.

Input

The first line contains an integer T(1<= T <=10000), indicating the number of test cases.
Each
test case contains one line with three integers N,M,K(1 <=N,M,K
<=2000), meaning the chocolate is a cube of size N ×M × K.

Output

For each test case in the input, print one line: "Case #X: A B", where X is the test case number (starting with 1) , A and B are the minimum numbers of steps to break the chocolate into N × M × K unit-size pieces with bare hands and knife respectively.

Sample Input

2 1 1 3 2 2 2

Sample Output

Case #1: 2 2 Case #2: 7 3

HINT

题意

有两种切法，一种是一次切一块，一种是一次可以切多块，然后问你在两种情况下，最少切多少下

题解：

第一种就毫无疑问，就是 a*b*c-1，第二种脑补一下，很显然是二分切

然后小心爆int，然后就好了

代码：

//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 50051
#define mod 10007
#define eps 1e-9
int Num;
char CH[20];
//const int inf=0x7fffffff;   //нчоч╢С
const int inf=0x3f3f3f3f;
/*

inline void P(int x)
{
    Num=0;if(!x){putchar(‘0‘);puts("");return;}
    while(x>0)CH[++Num]=x%10,x/=10;
    while(Num)putchar(CH[Num--]+48);
    puts("");
}
*/
inline ll read()
{
    ll x=0,f=1;char ch=getchar();
    while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}
    while(ch>=‘0‘&&ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}
    return x*f;
}
inline void P(int x)
{
    Num=0;if(!x){putchar(‘0‘);puts("");return;}
    while(x>0)CH[++Num]=x%10,x/=10;
    while(Num)putchar(CH[Num--]+48);
    puts("");
}
//**************************************************************************************
int deal(int x)
{
    int cnt=0;
    while(1)
    {
        if(1<<cnt>=x)
            return cnt;
        cnt++;
    }
}
int main()
{
    //freopen("test.txt","r",stdin);
    int t=read();
    for(int cas=1;cas<=t;cas++)
    {
        ll a,b,c;
        a=read(),b=read(),c=read();
        printf("Case #%d: %lld %lld\n",cas,a*b*c-1,deal(a)+deal(b)+deal(c));
    }
}