hdu 3573(数学+贪心)

Posted 2020-08-11 AC菜鸟机

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 = 1;
    scanf("%d",&tcase);
    while(tcase--)
    {
        int x,y,z;
        scanf("%d%d%d",&x,&y,&z);
        int ans = 0,a,b,c;
        if(x>=2&&y>=1){ /// 20 20 28
            a = x/2,b=y;
            c = min(a,b);
            x=x-2*c,y=y-c;
            ans+=c;
        }
        if(x>=2&&z>=1){ ///20 20 32
            a = x/2,b = z;
            c = min(a,b);
            x=x-2*c,z=z-c;
            ans+=c;
        }
        if(x>=3){ ///20 20 20
            a = x/3;
            x = x%3;
            ans+=a;
        }
        if(y>=1&&z>=1){ ///28 32
            c = min(y,z);
            ans+=c;
            y-=c;
            z-=c;
        }
        if(y>=2){ ///28 28
            c = y/2;
            y = y%2;
            ans+=c;
        }
        if(z>=2){///32 32
            c = z/2;
            z = z%2;
            ans+=c;
        }
        if(x||y||z) ans++;
        printf("Case %d: %d\n",t++,ans);
    }
    return 0;
}