hdu 3573(数学+贪心)

Buy Sticks

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 868    Accepted Submission(s): 392

Problem Description

Imyourgod
need 3 kinds of sticks which have different sizes: 20cm, 28cm and 32cm.
However the shop only sell 75-centimeter-long sticks. So he have to cut
off the long stick. How many sticks he must buy at least.

 

Input

The first line of input contains a number t, which means there are t cases of the test data.
There
will be several test cases in the problem, each in one line. Each test
cases are described by 3 non-negtive-integers separated by one space
representing the number of sticks of 20cm, 28cm and 32cm. All numbers
are less than 10^6.

 

Output

The
output contains one line for each line in the input case. This line
contains the minimal number of 75-centimeter-long sticks he must buy.
Format are shown as Sample Output.

 

Sample Input

2
3 1 1
4 2 2

 

Sample Output

Case 1: 2
Case 2: 3

 

Author

imyourgod (Special Thanks : crackerwang & Louty)

 

Source

2010 ACM-ICPC Multi-University Training Contest（14）——Host by BJTU

 

题意:需要 x 根20cm,y根 28cm,z根 32cm,问最少需要多少根 75 cm的木棍才可以都分出来?

题解:贪心的去选,先选3根的,选不了了才能选2根的,最后如果还有剩才能选一根的.

这些情况分别是

20 20 28

20 20 32

20 20 20

32 28

28 28

32 32

20

28

32

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#define LL long long
using namespace std;
int main()
{
    int tcase,t = ;
    scanf("%d",&tcase);
    while(tcase--)
    {
        int x,y,z;
        scanf("%d%d%d",&x,&y,&z);
        int ans = ,a,b,c;
        if(x>=&&y>=){ /// 20 20 28
            a = x/,b=y;
            c = min(a,b);
            x=x-*c,y=y-c;
            ans+=c;
        }
        if(x>=&&z>=){ ///20 20 32
            a = x/,b = z;
            c = min(a,b);
            x=x-*c,z=z-c;
            ans+=c;
        }
        if(x>=){ ///20 20 20
            a = x/;
            x = x%;
            ans+=a;
        }
        if(y>=&&z>=){ ///28 32
            c = min(y,z);
            ans+=c;
            y-=c;
            z-=c;
        }
        if(y>=){ ///28 28
            c = y/;
            y = y%;
            ans+=c;
        }
        if(z>=){///32 32
            c = z/;
            z = z%;
            ans+=c;
        }
        if(x||y||z) ans++;
        printf("Case %d: %d\n",t++,ans);
    }
    return ;
}