题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4112
解题思路：第一种：n*m*k-1, 第二种：ceil(log(n)/log(2))+ceil(log(m)/log(2))+ceil(log(k)/log(2)).
1 /************************************************************************** 2 user_id: SCNU20102200088 3 problem_id: hdoj 4112 4 problem_name: Break the Chocolate 5 **************************************************************************/ 6 7 #include <algorithm> 8 #include <iostream> 9 #include <iterator> 10 #include <iomanip> 11 #include <sstream> 12 #include <fstream> 13 #include <cstring> 14 #include <cstdlib> 15 #include <climits> 16 #include <bitset> 17 #include <string> 18 #include <vector> 19 #include <cstdio> 20 #include <cctype> 21 #include <ctime> 22 #include <cmath> 23 #include <queue> 24 #include <stack> 25 #include <list> 26 #include <set> 27 #include <map> 28 using namespace std; 29 30 //线段树 31 #define lson l,m,rt<<1 32 #define rson m+1,r,rt<<1|1 33 34 //手工扩展栈 35 #pragma comment(linker,"/STACK:102400000,102400000") 36 37 const double EPS=1e-9; 38 const double PI=acos(-1.0); 39 const double E=2.7182818284590452353602874713526; //自然对数底数 40 const double R=0.5772156649015328606065120900824; //欧拉常数:(1+1/2+...+1/n)-ln(n) 41 42 const int x4[]={-1,0,1,0}; 43 const int y4[]={0,1,0,-1}; 44 const int x8[]={-1,-1,0,1,1,1,0,-1}; 45 const int y8[]={0,1,1,1,0,-1,-1,-1}; 46 47 typedef long long LL; 48 49 typedef int T; 50 T max(T a,T b){ return a>b? a:b; } 51 T min(T a,T b){ return a<b? a:b; } 52 T gcd(T a,T b){ return b==0? a:gcd(b,a%b); } 53 T lcm(T a,T b){ return a/gcd(a,b)*b; } 54 55 /////////////////////////////////////////////////////////////////////////// 56 //Add Code: 57 /////////////////////////////////////////////////////////////////////////// 58 59 int main(){ 60 std::ios::sync_with_stdio(false); 61 //freopen("in.txt","r",stdin); 62 //freopen("out.txt","w",stdout); 63 /////////////////////////////////////////////////////////////////////// 64 //Add Code: 65 int Case,n,m,k,i; 66 scanf("%d",&Case); 67 for(i=1;i<=Case;i++){ 68 scanf("%d%d%d",&n,&m,&k); 69 LL a=(LL)n*m*k-1; 70 int b=ceil(log(n)/log(2))+ceil(log(m)/log(2))+ceil(log(k)/log(2)); 71 printf("Case #%d: %I64d %d ",i,a,b); 72 } 73 /////////////////////////////////////////////////////////////////////// 74 return 0; 75 } 76 77 /************************************************************************** 78 Testcase: 79 Input: 80 2 81 1 1 3 82 2 2 2 83 Output: 84 Case #1: 2 2 85 Case #2: 7 3 86 **************************************************************************/